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• David CivilEngineer

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Chair of Structural Mechanics Department of Civil, Geo and Environmental Engineering Technical University of Munich

Structural Dynamics Prof. Dr.-Ing. Gerhard Müller summer term 2019

Contents List of Symbols

XI

1 Introduction

1.1 1.2

Overview . . . . . . . . . . . . . . General classification . . . . . . . 1.2.1 Demands and loads . . . . 1.2.2 Frequency ranges . . . . . 1.2.3 Procedures for the prediction

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2 Description of harmonic oscillations

2.1 2.2

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Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Application to time harmonic oscillations . . . . . . . . . . . . . . . . . . . 2.2.1 Complex representation . . . . . . . . . . . . . . . . . . . . . . . . .

3 Fourier transformation

3.1 3.2

Periodic signals . . . . . . . . . . . . . . . . . . . . 3.1.1 Representation with real coefficients . . . . . 3.1.2 Representation with complex coefficients . . Fourier integral . . . . . . . . . . . . . . . . . . . .

4.2 4.3

Single 4.1.1 4.1.2 4.1.3

degree of freedom system . . . . . . . . . . . Equilibrium of forces . . . . . . . . . . . . . Virtual work . . . . . . . . . . . . . . . . . . The principle of Hamilton . . . . . . . . . . 4.1.3.1 Undamped systems . . . . . . . . . 4.1.3.2 Introduction of non-conservative forces 4.1.3.3 Lagrange multiplicators . . . . . . . Multiple degree of freedom systems . . . . . . . . . 4.2.1 Equilibrium of forces . . . . . . . . . . . . . 4.2.2 Principle of Hamilton . . . . . . . . . . . . . Damping models . . . . . . . . . . . . . . . . . . . . 4.3.1 Linear viscoelastic damping . . . . . . . . . . 4.3.2 Ideal hysteretic damping . . . . . . . . . . . 4.3.3 Coulomb damping . . . . . . . . . . . . . . .

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4 Setting up the equation of motion

4.1

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II

4.4

4.3.4 Aeroelastic damping . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 Free vibration of linear systems with single degree of freedom

5.1 5.2

The undamped case . . . . . . . . . . . . . . . . . The damped case . . . . . . . . . . . . . . . . . . 5.2.1 Velocity proportional (viscous) damping . . 5.2.1.1 Quasi periodic case (D < 1) . . . 5.2.1.2 Aperiodic limit case (D = 1) . . . 5.2.1.3 Aperiodic case (D > 1) . . . . . . 5.2.2 Relations between different damping measures

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Modelling of the excitation . . . . . . . . . . . . . . . . . . 6.1.1 Harmonic excitation . . . . . . . . . . . . . . . . . . 6.1.2 Periodic excitation . . . . . . . . . . . . . . . . . . . 6.1.3 Transient load . . . . . . . . . . . . . . . . . . . . . 6.1.4 Irregular, random loads . . . . . . . . . . . . . . . . 6.1.5 Loads influenced by the parameters of adjacent systems Frequency content . . . . . . . . . . . . . . . . . . . . . . . Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . Classification in standards . . . . . . . . . . . . . . . . . .

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6 Classification of Excitations

6.1

6.2 6.3 6.4

7.2

7.3 7.4 7.5

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 “Quasi-static”and dynamic processes . . . . . . . . 7.1.2 Time and frequency representation . . . . . . . . . . Harmonic excitation for systems with viscous damping . . . 7.2.1 Harmonic forced excitation . . . . . . . . . . . . . . 7.2.1.1 Steady state solution . . . . . . . . . . . . 7.2.1.2 Transient response . . . . . . . . . . . . . 7.2.1.3 Resonance case . . . . . . . . . . . . . . . 7.2.1.4 Transmissibility . . . . . . . . . . . . . . . 7.2.2 Harmonic root point excitation . . . . . . . . . . . . 7.2.2.1 Equation of motion . . . . . . . . . . . . . 7.2.2.2 Particular solution (forced vibration) . . . 7.2.2.3 Transmissibility . . . . . . . . . . . . . . . Harmonic forced excitation for hysteretically damped systems Arbitrary periodic excitation . . . . . . . . . . . . . . . . . 7.4.1 Decomposition into harmonics . . . . . . . . . . . . Aperiodic excitation . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Time domain . . . . . . . . . . . . . . . . . . . . . . 7.5.1.1 Unit impulse response function . . . . . .

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7 Forced vibration of linear systems with single degree of freedom

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Contents

III

7.5.2 7.5.3

7.5.1.2 Superposition, Duhamel-Integral . . . . 7.5.1.3 Response spectra . . . . . . . . . . . . Exkursus: Kelvin-Voigt model . . . . . . . . . . Frequency domain . . . . . . . . . . . . . . . . . 7.5.3.1 Response to harmonic excitation . . . . 7.5.3.2 Response to aperiodic excitation . . . .

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8 Free vibration of linear systems with multiple degrees of freedoms

8.1 8.2

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Preliminary remark: Maxwell-Betti theorem . . . . . . . . . . . . . . . . . Solution of the homogeneous system of equations . . . . . . . . . . . . . . . 8.2.1 Undamped MDOF systems . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Consideration of the initial conditions . . . . . . . . . . . . . . . . . 8.2.3 Damped MDOF systems . . . . . . . . . . . . . . . . . . . . . . . . 8.2.3.1 The quadratic eigenvalue problem . . . . . . . . . . . . . . 8.2.3.2 The standard form of the damped eigenvalue problem . . . 8.2.3.3 Rayleigh damping . . . . . . . . . . . . . . . . . . . . . . . 8.2.3.4 Caughey damping model using additional orthogonality relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9 Forced vibration of linear systems with multiple degree of freedoms

9.1 9.2 9.3 9.4

Direct solution for harmonic loads . Modal analysis . . . . . . . . . . . . Time integration methods . . . . . . Summary . . . . . . . . . . . . . . .

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10 Impedance models

10.1 Preliminary definitions . . . . . . . . . . . 10.2 Dynamic stiffness and impedance . . . . . 10.3 Dynamic stiffnesses and impedances in passive 10.3.1 Mass impedance . . . . . . . . . . . 10.3.2 Conservative forces . . . . . . . . . 10.3.2.1 Potential . . . . . . . . . . 10.3.2.2 Spring impedance . . . . . 10.3.3 Non-conservative forces . . . . . . . 10.3.3.1 Damping impedance . . . 10.3.3.2 Loss factor . . . . . . . . . 10.4 Parallel and serial connection . . . . . . . . 10.4.1 Parallel connnection . . . . . . . . . 10.4.2 Serial connnection . . . . . . . . . . 10.5 Arbitrary connections . . . . . . . . . . . . 10.6 Impedances for selected systems . . . . . .

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. . . . . . . . . . systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

IV

11 Solutions for selected continuous systems

183

11.1 String, longitudinal and torsional rod . . . . . . . . . . . . . . . . . . . . . 11.1.1 Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2.1 Solution via a separation approach . . . . . . . . . . . . . 11.1.2.2 Direct solution via a wave approach . . . . . . . . . . . . . 11.2 Euler-Bernoulli beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2 Free vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2.1 Solution via a separation approach . . . . . . . . . . . . . 11.2.2.2 Solution via a wave approach . . . . . . . . . . . . . . . . 11.2.3 Displacement approaches - spectral elements based on analytic solutions 11.2.3.1 Beams without loads and harmonic excitation at the boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.3.2 Loaded beam elements - particular solution . . . . . . . . . 11.3 Extended beam theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Thin plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Eigenfrequencies and mode shapes of selected systems . . . . . . . . . . . . 11.6 Modal analysis for loaded Euler-Bernoulli beam . . . . . . . . . . . . . . . 11.6.1 Euler-Bernoulli beam under step load . . . . . . . . . . . . . . . . . 11.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Approximate solutions of continuous systems

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13 Random vibrations

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12.1 Discretization of continuous systems . . . . . . . . . . . . 12.1.1 The principle of virtual work . . . . . . . . . . . . 12.1.2 Lumped mass matrix and static condensation . . . 12.1.3 Principle of Hamilton . . . . . . . . . . . . . . . . 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 13.2 Description of random variables . . . . . . . . . . . . 13.2.1 Single random variable . . . . . . . . . . . . . 13.2.2 Two random variables . . . . . . . . . . . . . . 13.3 Description of stochastic processes . . . . . . . . . . . 13.3.1 Probability distribution . . . . . . . . . . . . . 13.3.2 Moment functions . . . . . . . . . . . . . . . . 13.3.3 Stationarity of stochastic processes . . . . . . . 13.3.4 Ergodicity of stochastic processes . . . . . . . 13.3.5 The Gaussian process . . . . . . . . . . . . . . 13.3.6 Frequency domain analysis of stochastic processes 13.3.7 Stochastic response of linear dynamic system .

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Contents

V

14 Aeroelastic vibrations (wind)

259

14.1 General information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.1 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.2 Description of wind . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.2.1 Wind velocity . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.2.2 Temporal and spatial structure of wind velocity measurements 14.1.3 Temporal structure of the wind velocity . . . . . . . . . . . . . . . . 14.1.3.1 Power spectral density . . . . . . . . . . . . . . . . . . . . 14.1.3.2 Turbulence intensity . . . . . . . . . . . . . . . . . . . . . 14.1.4 Spatial structure of the wind velocity . . . . . . . . . . . . . . . . . 14.1.4.1 Atmospheric boundary layer - wind profile . . . . . . . . . 14.1.4.2 Turbulences . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Wind induced vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.1 Gust induced vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.2 Vortex induced vibrations . . . . . . . . . . . . . . . . . . . . . . . . 14.2.2.1 Occurence of vortices . . . . . . . . . . . . . . . . . . . . . 14.2.2.2 Frequency of vortices . . . . . . . . . . . . . . . . . . . . . 14.2.2.3 Forces related to vortices . . . . . . . . . . . . . . . . . . . 14.2.2.4 Assessment of vortex induced vibrations . . . . . . . . . . 14.2.3 Galloping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.4 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.5 Flutter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.6 Interference effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Earthquakes

279

15.1 General information and terms . 15.1.1 Magnitude . . . . . . . . 15.1.2 Intensity . . . . . . . . . 15.1.3 Calculation procedures . 15.2 Response spectrum analysis (RSA) 15.2.1 Introduction . . . . . . . 15.2.2 Design codes . . . . . . . 15.2.3 Constructive notes . . . . 15.3 Capacity design . . . . . . . . . 15.3.1 Basics . . . . . . . . . . 15.3.2 Reduction factors . . . . 15.3.3 The capacity design . . .

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16 Numerical time step procedures

16.1 General remarks . . . . . . . . 16.1.1 Overview . . . . . . . . 16.1.2 Explicit - Implicit . . . 16.1.3 Errors . . . . . . . . .

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VI

16.2 Newmark-β method . . . . . . . . . . . 16.2.1 General approach . . . . . . . . 16.2.2 Constant acceleration over the time 16.2.3 Predictor/ corrector approach . 16.2.4 Stability considerations . . . . . 16.3 The central difference method . . . . . 16.3.1 Derivation of w¨ and w˙ out of w 16.3.2 General approach . . . . . . . . 16.3.3 Start procedure . . . . . . . . . 16.4 Non-linear problems . . . . . . . . . . .

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Contents

VII

List of Figures 1.1 1.2 1.3 2.1 2.2 2.3 2.4 2.5

Two dimensional probability density function for the demand S and the resistance R and limit-state function . . . . . . . . . . . . . . . . . . . . . . . . Requirements for highly sensitive equipment given in VDI 2038 [2012]. . . Prediction methods in structural dynamics . . . . . . . . . . . . . . . . . .

3 4 6

Description of a complex number . . . . . . . . . . . . . . . . . . . . . . . . Visualization of multiplication of two complex numbers . . . . . . . . . . . Point on unit circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Point on unit circle (f0 = 1) with phase shift . . . . . . . . . . . . . . . . . Representation of complex pointers with different contributions of cosine and sine part. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The complex exponential function evolves harmonically in the complex plane over time. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Graphical representation of complex Fourier series coefficients.   Fourier transform of the rectangular function g(t) = g0 Π Tt0 .

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3.3

Fourier transform of the rectangular function g 0 (t) = h0 Π

3.4 3.5 3.6

and T00 = T20 . Fourier transform Fourier transform Fourier transform

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4.1 4.3

External and reaction forces on single mass point. . . . . . . . . . . . . . . Illustrations of different viscoelastic models. . . . . . . . . . . . . . . . . .

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5.1 5.2 5.3 5.4

Free body cut for the unforced (free) SDOF system . . . . . . . Homogeneous solution of SDOF-system . . . . . . . . . . . . . . Damped oscillation over one period T with w0 = 1 and ϕ = 0. . Homogeneous solution of the SDOF system for the aperiodic case D

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6.1 6.2 6.3 6.4 6.5 6.6

Transient, triangular load . . . . . . . . . . . . . . Representation of an aperiodic signal using an infinite Periodic signal and harmonic decomposition . . . Transient, triangular load . . . . . . . . . . . . . . Representation of an aperiodic signal using an infinite Illustration of a stochastic process . . . . . . . . .

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t T00



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with h0 = 2g0

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. . . . return . . . . . . . . return . . . .

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VIII

6.7 6.8

Illustration of the autocorrelation function . . . . . . . . . . . . . . . . . . Synchronization of pedestrians with the structural vibration, courtesy of M. Schneider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 Frequency content characterization in a emission–transmission–imission system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 Common methods in strucutral dynamics and vibro-acoustics and their applicability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7.2 7.4 7.5 7.6 7.8 7.13 7.14 7.15 7.16 7.18 7.21 7.22 7.24 7.25 7.26

Displacement solution wp (t) in the complex plane. . . . . . . . . . . . . . . Transfer function in the complex plane. . . . . . . . . . . . . . . . . . . . . Transfer function of the SDOF system over frequency . . . . . . . . . . . . Particular solution in the complex plane. . . . . . . . . . . . . . . . . . . . Transient response in resonance for D = 0.1. . . . . . . . . . . . . . . . . . Transfer function for linear hysteretic damping in the complex plane. . . . Amplification function harmonically foced SDOF system with linear hysteretic damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example for an arbitrary periodic load time history. . . . . . . . . . . . . Square wave with period T . . . . . . . . . . . . . . . . . . . . . . . . . . . Unit impulse response function for varying percentage of critical damping D Response spectrum of SDOF system for rectangular excitation . . . . . . . Triangular shock on column with varying length . . . . . . . . . . . . . . . Response spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Displacement response response of Kelvin-Voigt model. . . . . . . . . . . . Response of SDOF system to arbitrary excitation. . . . . . . . . . . . . . .

8.1 8.2

Solution for the 2DOF system. Solution for the 2DOF system.

9.1

Displacement w1+ for the 2DOF system under harmonic excitation, lution, undamped system . . . . . . . . . . . . . . . . . . . . . . Displacement w1+ for the 2DOF system under harmonic excitation, lution, damped system . . . . . . . . . . . . . . . . . . . . . . . Solution for the 2DOF system under two step loads . . . . . . .

9.2 9.3

80 80 81 82 91 94 95 96 100 106 107 108 110 115 119 120 121 123 127

. . . . . . . . . . . . . . . . . . . . . . . . 141 . . . . . . . . . . . . . . . . . . . . . . . . 147 direct so. . . . . . 154 direct so. . . . . . 154 . . . . . . 160

10.3 Physical quantities f (t) and v(t). . . . . . . . . . . . . . . . 10.8 Voigt-Kelvin-Modell. . . . . . . . . . . . . . . . . . . . . . . 10.9 SDOF impedance model for different values of damping ratio 10.10Elastic element impedance model for different values of damping 10.11Elastic element impedance model for different values of damping

. . . . . . . . . . . . . . . . . . . . . . . . ratio . . . ratio . . .

164 174 178 179 181

11.1 11.2 11.3 11.4

. . . .

184 186 191 191

Forces on differential elements. . . . . . . . . Boundary conditions . . . . . . . . . . . . . . Part of the wave solution. . . . . . . . . . . . Wave traveling in the positive x-direction. . .

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. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

List of Figures

IX

12.1 Nodal forces and displacement degrees of freedom . . . . . . . . . . . . . . 229 12.2 Functions for unit displacements/rotations - cubic approach . . . . . . . . . 231 12.3 Nodal forces for the beam element . . . . . . . . . . . . . . . . . . . . . . . 234 13.5 Autocorrelation function φXX for the above process in terms of time shift. 13.6 Five realizations of the stochastic process with X0 following the standard normal distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7 from [Petersen 2000, p.1220, Fig. 34] . . . . . . . . . . . . . . . . . . . . . 13.8 Comparison of different processes from [Petersen 2000, p. 1228, Fig. 43]) . . 13.9 Autocorreation function and spectral density for the white noise process [Petersen 2000, p. 1228, Fig. 43]) . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Measured wind velocities at three different heights [Holmes 2007, p. 56] . . 14.2 Frequenzspektrum der horizontalen Windgeschwindigkeit (1957), Messungen in Brookhaven, New York [Burton et al 2001, p. 12], [Petersen 2000, p. 596] . 14.3 aus Ruscheweyh, Dynamische Windwirkung an Bauwerken Vol. 2 (1982), S. 16, Abb. 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Static and dynamic part of the wind velocity . . . . . . . . . . . . . . . . . 14.5 Comparison of different models: z = 10 m, v¯10 = 20 m s , z0 = 0.05 (dimensionless), κ = 0.006 (open grassland, grassland); above: log-log illustration of the dimensionless turbulence spectra; below: Comparison for low frequencies [Runtemund 2013] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 Example for Davenport spectrum . . . . . . . . . . . . . . . . . . . . . . . 14.7 Vortices, from [Petersen 2000] . . . . . . . . . . . . . . . . . . . . . . . . . 14.8 Different states of vortex shedding . . . . . . . . . . . . . . . . . . . . . . . 14.9 vgl: Bachmann, Hugo, Vibration Problems in Structures (1997), in: Vibrations in across-wind direction induced by vortex-shedding, S. 203, Abb. H.10 . . 14.10Examples for the aerodynamic coefficient cs (α) . . . . . . . . . . . . . . . . 14.11Profiles at which Galloping can occur . . . . . . . . . . . . . . . . . . . . . 14.12Flutter, [Ruscheweyh 1982, Vol. 2 (1982), S. 118, Abb. 4.67] . . . . . . . . 15.1 Terminology of an earthquake, from [Flesch 1993, p. 203, Fig. 7.1] . . . . . 15.2 Frequency of occurrence of magnitudes for the whole earth (a) and the Rhine valley (b) from [Flesch 1993, p. 203, Fig. 7.2] . . . . . . . . . . . . . . . . . 15.3 Exemplary accelerogram given in terms of the acceleration relative to the gravitational acceleration on earth g. . . . . . . . . . . . . . . . . . . . . . . . 15.4 SDOF system with different natural frequencies and damping ratios for the response spectrum analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5 Response spectra for absolute and pseudo-absolute acceleration, relative and pseudo-relative velocity and relative displacement . . . . . . . . . . . . . . 15.6 Design response spectrum according to the DIN 4149. . . . . . . . . . . .

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249 249 251 255 256 261 262 264 265

266 268 269 270 273 274 275 277 280 281 282 285 286 288

List of Figures

X

List of Symbols Greek Symbols α, β α γ δ(x) δ   η ϑ λ Λ µ ν ξ Π ρ σ σ τ ϕ φ φ Φ Ω ω

angle; also coefficients (of the Rayleigh damping) vector of coefficients shear strain; also coefficient (of the Newmark-β method) Dirac-delta function decay constant vector of strains mean-square error; also strain loss factor; also kinematic viscosity propagating angle eigenvalue (scalar); also wave length diagonal matrix of squared eigenfrequencies, i.e., [Λ]ii = λi = ωi2 mass per unit length; also ductility Poissons ratio dimensionless coordinate, i.e., ξ = xl ; also percentage of critical damping potential density internal force vector; also vector of stresses stress; also standard deviation time; also shear stress Phase eigenvector (cross-sectional) rotation matrix of eigenvectors circular frequency (of the load) radial frequency i.e., ω = 2πf

Roman Symbols A, B

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amplitudes

XI

A A a ak B0 bk C c D d E F {·} F −1 {·} f (x), g(x) f f G g H H(t) h(t) ˜ h(t) h I i i K k L(t) l M M m N N n Pr(E) Q q R Re

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cross section flexibility matrix acceleration coefficient of the Fourier series bending stiffness coefficient of the Fourier series damping matrix damping constant; also wave velocity damping ratio, i.e. percentage of critical damping D = 2√ckm diameter Young’s modulus; also energy Fourier transformation inverse Fourier transformation function/ signal of variable x force; also frequency f = T1 force vector shear modulus gravitational acceleration horizontal force Heaviside function impulse response function transfer function, Fourier transform of the impulse response function thickness rotatory inertia; also intensity imaginary unit, i.e., i2 = −1 rotatory inertia per unit length stiffness matrix stiffness Lagrange function length moment; also magnitude (of an earthquake) mass matrix mass normal force shape function vector number of DOFs probability of event E lateral force velocity pressure (dynamic pressure) Resistance Reynolds number

List of Symbols

XII

r S T T (t) t U (t) Ub u V v Var(X) W (x) W w w Z X x Y (x) Y y Z z z z∗

absolute value of a complex number; also reaction force; also radius demand; also spectral density; also Strouhal number time period kinetic energy time potential energy; also unit-step function transfer function displacement lateral force; also amplification function velocity variance of X space function work displacement vector displacement impedance random variable real part of a complex number; also Cartesian spatial coordinate space function random variable imaginary part of a complex number; also Cartesian spatial coordinate impedance state space vector complex number; also Cartesian spatial coordinate complex conjugate of complex number z

Mathematical Operations ∂ ∂x d dx

()T w˙ w¨ w0 w00 ,w00 , . . .

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partial differential operator with respect to x differential operator with respect to x Fourier transformation in frequency domain inverse Fourier transformation in time domain matrix transpose the first derivative of w with respect to time the second derivative of w with respect to time the first derivative of w with respect to the spatial coordinate the second, third,. . . derivative of w with respect to the spatial coordinate

List of Symbols

XIII

Acronyms BEM CDF DOF FEM FT IFT MDOF ODE PDE PDF SDOF SEA VDI

boundary element method cumulative distribution function degree of freedom finite element method Fourier transformation inverse Fourier transformation multi eegree of freedom ordinary differential equation partial differential equation probability density function single degree of freedom statistical energy analysis Verein Deutscher Ingenieure

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List of Symbols

XIV

1 Introduction 1.1 Overview With increasing slenderness of structures, growing comfort demands, higher demands for low vibrational influence of machinery and a dense arrangement of buildings, the requirements of structural dynamic investigations increase. The foundations of structural dynamics were found by Newton, Euler, d’Alembert, Lagrange and Hamilton in the 17th and 18th century. The insight, that bridges demand special treatment, has surely been in the mind of the builders before the collapse of the Braughton bridge in 1831. However, the influence of moving loads on bridges has been treated since the middle of the 19th century. Noteworthy are the names of Willis, Zimmermann and Timoshenko, Fryba, Bachmann, Grundmann. The names Geiger und Rausch are connected with machinery foundations, that are supported in order to reduce the elastic dynamic forces. Regarding the installation conditions of sensitive equipment the relevant publications by Ungar, Heiland has to be mentioned. The topic soil-structure interaction for linear loading processes was extensively treated, e.g. by Wolf and Gazetas. For the topic of earthquakes there are a number of research groups in the USA and Japan. This field of work is closely connected to the names of Biot, Housner, Newmark and Rosenblueth. In Germany also since 1951, after the earthquake of Euskirchen with the name of Meskouris, and in Switzerland with the name of Bachmann. Wind induced vibrations especially rose to general awareness since the collapse of the Tacoma Narrows bridge in 1940. Here the name Ruscheweyh is to be mentioned. The treatment of irregular and stochastic excitations is connected to the names Freudenthal, Shinuozuka, Lin, Grundmann, Schueller, Soize. Consideration of structure borne sound, i.e., higher frequency vibrations, were in particular treated by Cremer, Heckl and Lyon. The mentioning of these names is only exemplarily — as the overview over the technical rules and literature — and shall facilitate the introduction of the various topics for the reader.

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1

1.2 General classification 1.2.1 Demands and loads Dynamic loads acting on buildings can be relevant with respect to structural safety, fatigue strength, or structural serviceability, depending on the impact strength and the demand. The demands and loads therefore depend on the specific task, distinguished as follows: • Particularly strong dynamic loads, e.g. impact from – earthquakes, – explosions, – airplane crash, or – vehicle impact have to be treated with respect to the damage and the collapse of the structures (structural safety). • Dynamic loads that occur repeatedly regularly and do not cause damages are impacts from – wind, – pedestrians, – moving persons, or – moving loads. These have to be examined with respect to fatigue (fatigue strength) and serviceability of the structure. Demands on the serviceability can also exist for various impacts with low strength, e.g. for vibrations due to railway or street traffic, vibrations due to building operations, vibrations due to moving persons, or due to machinery induced dynamic forces. They do not have to be assessed with respect to fatigue or structural safety, but with respect to comfort requirements or the useability of sensitive equipment. Whereas non-linear approaches are necessary for the assessment of structural safety under one-time extreme load impacts (that are partly rehashed by linear replacement procedures for practical applications), for problems concerning the fatigue strength and the serviceability, in general procedures linear material laws are applied. The semi-probabilistic safety concept in the context of structural safety sets minimal probabilities of failure Pr(R < S), with the resistance of the structure R and the demand S. For the serviceability much lower vibration limits exist, for which the exceedance is usually less critical, as it is not linked to danger to life, but to a non-compliance with comfort requirements

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1 Introduction

2

of the people in the building or with requirements for the operation of sensitive equipment. Thereby, larger exceedance probabilities compared to the structural safety concept can be accepted (comp. Fig. 1.1b).

(a) Structural safety

(b) Serviceability

Figure 1.1: Two dimensional probability density function for the demand S and the resistance R and limit-state function (identity R = S ) from Müller [2009].

Since the approaches and results from dynamic prognoses can be tested via measurements of the construction’s response especially for the assessment of fatigue and serviceability, measurement procedures play a significant role for the demand S as well as the resistance R value. Also it is often useful to increase the prediction accuracy by means additional measurements in the planing and building phase. Thus, the uncertainties linked to the prediction for the proof of structural safety under static loads are significantly less verifiable through measurements than the uncertainties in structural dynamics. Especially for problems concerning the serviceability structural dynamic investigations are not necessary for all building projects. In VDI 2038 [2012] criteria are summarized in order to decide exemplarily to what extent dynamic investigations are necessary. Fig. 1.2 exemplarily depicts current requirements given in VDI 2038 [2012].

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1 Introduction

3

1000

VC-A

100 VC-B VelocityH[µHm/s]

VC-C VC-D

10 VC-E Nano-D Nano-E Nano E-F

1

0.1 10

1

1000

100 FrequencyH[Hz]

VC-A

VC-B

VC-C

VC-D

VC-E

Nano-D

Nano-E

Nano E-F

Figure 1.2: Requirements for highly sensitive equipment given in VDI 2038 [2012].

1.2.2 Frequency ranges In structural dynamics the following frequency ranges are distinguished: • vibrations: 0 Hz (static deformation) to 80 Hz • structure borne sound: from around 16 Hz to multiple kHz The frequency content of a time varying signal can be assessed with the help of signal analysis (comp. Cha. 3)

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1 Introduction

4

z y

sound waves z

Reradiated sound

x, y, z

Vibrations

Waves in solids Emission Vibrations 1 2

Transmission

Immission

Structure borne sound

5 10 20 5080100 Frequency in Hz

1000

1.2.3 Procedures for the prediction For the numerical prediction of noise and vibration there are a variety of methods available, which are depicted in Fig. 1.3. In engineering practice the different methods are typically applied depending on the system, the frequency and the type of problem (Fig. 1.3). Especially uncertainties in the modelization and the necessary detailing of the response, influence considerably the choice of the method. The methods are not competing with each other, but are complementary, so that there is a "Marriage à la mode“ (Zienkiewicz et al [1977]) - "entre plusiers personnes“ - not only applying jointly the Boundary and Finite Element method, but also analytical methods, the discretization by the help of lumped mass systems, Structural Fuzzy (Soize [1993]; Weaver [1997]) and the Statistical Energy Analysis SEA. Thus, there is a demand for hybrid methods, bringing together the various approaches for either different types of sub-systems or different frequency ranges (Langley and Bremner [1999]; Langley [1990]).

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1 Introduction

5

hybrid approaches

analytical algorithms unlimited systems limited simple boundary conditions

Mu osc ltib illa ody tor

M BE

A SE

high

M FE

frequency complex

low

Figure 1.3: Prediction methods in structural dynamics

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1 Introduction

6

2 Description of harmonic oscillations This chapter gives an introduction to the description of harmonic oscillations. Arbitrary harmonic functions with a phase shift can be described through sine and cosine functions. Computations with those functions can be greatly simplified through the use of a complex description. To this end, we first introduce complex numbers together with important rules of computation and subsequently treat the description of harmonic oscillations.

2.1 Complex numbers Consider a complex number z ∈ C z = x + iy

(2.1)

where x = Re (z) and y = Im (z) are the real and imaginary part of z. We define the complex conjugate of z as z ∗ = x − iy . With this, we introduce the following identities, that define the real and imaginary part of a complex number using complex conjugation z + z∗ 2 z − z∗ Im (z) = . 2i Re (z) =

(2.2) (2.3)

We can also state z in polar form z = r (cos ϕ + i sin ϕ)

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(2.4)

7

Im z − z∗ Im

−z ∗

z

y ϕ −ϕ

x

Re

i Im z

z z + z ∗ Re

ϕ −ϕ

Re z −y

z∗

z∗

Figure 2.1: Geometric description of a complex number. On the left hand side, the complex number z with real part x and imaginary part y is depicted. It’s complex conjugate z ∗ can be geometrically interpreted as the reflection of z about the real axis. The right hand side gives a geometric interpretation of Eqs. (2.2) and (2.3). The sum z + z ∗ always results in a real outcome, whereas z − z ∗ always gives a purely imaginary result.

and from comparison with Eq. (2.1) we find x = r cos ϕ y = r sin ϕ . Now, consider Euler’s formula eiϕ = cos ϕ + i sin ϕ

(2.5)

which introduces the complex exponential function. With this, we can simply describe the complex number z as z = reiϕ

(2.6)

where r is the absolute value of the complex number z = x + iy = reiϕ , |z| =

q

x2 + y 2 =

√ zz ∗ = r

(2.7)

and ϕ is the argument or phase of z and is given by the relation tan(ϕ) =

Im z . Re z

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(2.8)

2 Description of harmonic oscillations

8

As the inverse relation is not uniquely defined throughout the range iof ϕ, using the arctanh function, the phase of z is just uniquely defined within the interval ϕ ∈ − π2 , π2 . To circumvent this, we introduce the function atan2, ϕ = arg z = atan2 (Im z, Re z) = atan2 (y,x) ,

(2.9)

with

atan2 (y,x) =

   y  arctan  x     arctan y + π    x   arctan y − π

for x > 0

π     2  π   − 2   

for x = 0 and y > 0 for x = 0 and y < 0 for x = 0 and y = 0 .

x

undefined

for x < 0 and y > 0 for x < 0 and y ≤ 0

(2.10)

From Eq. (2.5) we can further deduce 







cos ϕ = Re eiϕ

sin ϕ = Im eiϕ . Through the symmetry properties cos (−ϕ) = cos (ϕ) and sin (−ϕ) = − sin (ϕ) and using (2.4), we find for the complex conjugate ∗

z ∗ = reiϕ = r(cos ϕ + i sin ϕ)∗ = r (cos ϕ − i sin ϕ) = r (cos (−ϕ) + i sin (−ϕ)) = re−iϕ which as a side result gives ∗

(ez )∗ = e(z) . A major advantage of the exponential notation in Eq. (2.6) is the simplification of multiplication of complex numbers z1 z2 = r1 r2 eiϕ1 eiϕ2 = r1 r2 ei(ϕ1 +ϕ2 ) .

(2.11)

From this we conclude, that multiplication with eiϕ2 corresponds to a rotation of the vector in the complex plane by ϕ2 . Therefore, the multiplication of a complex number z with the π imaginary unit i = ei 2 results in iz = ireiϕ = rei 2 eiϕ = ei(ϕ+ 2 ) π

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π

(2.12)

2 Description of harmonic oscillations

9

which can be interpreted as a rotation of

π 2

or 90◦ in the complex plane.

Im

Im π i = ei 2 z = eiϕ

z1 = eiϕ1 π

iz = ei(ϕ+ 2 ) z2 = eiϕ2

ϕ2

ϕ1

π 2

Re

ϕ1 +

π 2

Re

ϕ1

ϕ1 + ϕ2 z1 z2

Figure 2.2: Visualization of multiplication of two complex numbers. The left hand side shows the multiplication of two complex numbers with unit amplitude |z1 | = |z2 | = 1. The resulting complex number z1 · z2 is the complex number that is rotated by ϕ1 + ϕ2 in the complex plane. The special case of z2 = i is depicted on the right hand side. In the complex plane, the purely imaginary number z2 = i is rotated by π2 with respect to the real axis. Thus, multiplication with i always results in rotation of π2 .

For the multiplication of two complex numbers, we find the following identities for the absolute value and argument of the resulting complex number



|z1 z2 | = r1 r2 eiϕ1 eiϕ2 = r1 r2 = |z1 ||z2 |

(2.13)

arg z1 z2 = arg r1 r2 eiϕ1 eiϕ2 = arg r1 r2 ei(ϕ1 +ϕ2 ) = ϕ1 + ϕ2 .

(2.14)

2.2 Application to time harmonic oscillations We now want to use the above results to find descriptions for arbitrary time-harmonic signals. From the harmonic addition theorem we know that sums of sine and cosine functions with arbitrary amplitudes and phase shifts but the same frequency can be reduced to a single cosine or sine function. To reduce complexity, we consider a signal f (t) that is composed of a cosine and sine function with amplitudes f01 and f02 , respectively. f (t) = f01 cos (ωt) − f02 sin (ωt)

(2.15)

where t denotes time, and ω is the radial frequency ω = 2πf . The subtraction of the sine term is explained later on.

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2 Description of harmonic oscillations

10

y

x,y yP = sin(ωt)

yP

P π 2

P0 x ωt

π

3 π 2

2π ωt

xP xP = cos(ωt)

Figure 2.3: Point on unit circle. The location of a point on the unit circle can be given in Cartesian coordinates (xP ,yP ) or polar coordinates through the angle ωt. If we express (xP ,yP ) in terms of ωt, we note that the coordinates xP and yP vary harmonically.

Consider now a unit-circle in the x-y-plane. As the circle has the radius 1, every point on the circle can be solely described by an angle ωt. Furthermore, a point P can also be characterized in terms of its Cartesian coordinates (xP ,yP ). This is depicted in Fig. 2.3, from where we also deduce xP = cos (ωt)

(2.16)

yP = sin (ωt) .

(2.17)

The x- and y-coordinates are also depicted in Fig. 2.3 in dependency on the angle ωt. We can also interpret the coordinates as the projection of the vector P on x- and y-axis, respectively. Here, we find the two parts, that make up our signal from Eq. (2.15), up to the amplitudes f01 and f02 . Using the harmonic addition theorems cos (α ± β) = cos (α) cos (β) ∓ sin (α) sin (β)

(2.18)

sin (α ± β) = sin (α) cos (β) ± cos (α) sin (β)

(2.19)

we further introduce the identities f01 cos (ωt) − f02 sin (ωt) = f0 [cos (ωt) cos (ϕ) − sin (ωt) sin (ϕ)] = f0 cos (ωt + ϕ)

(2.20)

f02 cos (ωt) + f01 sin (ωt) = f0 [cos (ωt) sin (ϕ) + sin (ωt) cos (ϕ)] = f0 sin (ωt + ϕ)

(2.21)

with f01 = f0 cos (ϕ)

(2.22)

f02 = f0 sin (ϕ)

(2.23)

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2 Description of harmonic oscillations

11

or inversely f0 =

q

2 2 f01 + f02

(2.24)

ϕ = atan2 (f02 ,f01 ) .

(2.25)

With this, we can equivalently state a harmonic signal using a single phase-shifted sine or cosine term. Hereinafter, we choose the representation using f0 cos (ωt + ϕ) as it complies with Eq. (2.15). Further, we define ϕ∗ as the complementary angle to ϕ, such that ϕ + ϕ∗ = π2 and state Eq. (2.21) in terms of ϕ∗ as f0 sin (ωt + ϕ∗ ) = f0 [cos (ωt) sin (ϕ∗ ) + sin (ωt) cos (ϕ∗ )]

(2.26)

Using the identities π sin ϕ = sin − ϕ = cos ϕ 2   π ∗ cos ϕ = cos − ϕ = sin ϕ 2 ∗





(2.27) (2.28)

to further transform Eq. (2.26), it reads f0 sin (ωt + ϕ∗ ) = f0 [cos (ωt) sin (ϕ∗ ) + sin (ωt) cos (ϕ∗ )] =

(2.29)

= f0 [cos (ωt) cos (ϕ) + sin (ωt) sin (ϕ)] =

(2.30)

= f01 cos (ωt) + f02 sin (ωt)

(2.31)

= f0 cos (ωt − ϕ) .

(2.32)

The same can be stated for the following. f0 cos (ωt + ϕ∗ ) = −f0 sin (ωt − ϕ) .

(2.33)

Both, a sine and cosine with negative phase shift ϕ are depicted in Fig. 2.4. In the next step, we represent the harmonic signals using complex arithmetic. This shall be discussed in the following section.

2.2.1 Complex representation We now consider the complex number z = x+iy. From Sec. 2.1 we know that we can represent z using Euler’s formula. In the following we want to show that we can represent the signal from Eq. (2.15) by a complex exponential and its complex conjugate. For that we separately

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2 Description of harmonic oscillations

12

y

x,y

yP = sin(ωt − ϕ)

xP 0 ϕ yP

π 2

x

xP

3 π 2

π

2π ωt

P

ωt P0

yP0 ϕ

xP = cos(ωt − ϕ)

Figure 2.4: Point on unit circle (f0 = 1) with phase shift

apply some operations to the sine and cosine term. First, the term 12 f01 (i sin (ωt) − i sin (ωt)) is added to f01 cos (ωt), as it does not alter the equation. 

f01 cos (ωt) = f01

1 1 (cos (ωt) + i sin (ωt)) + (cos (ωt) − i sin (ωt)) 2 2



(2.34)

i 1 h = f01 e iωt + e −iωt . 2

(2.35)

We note, that we can easily apply Euler’s formula to convert the expression to the complex exponential form. The same is attempted for the sine part. To this end, we add 12 f02 (i cos(ωt)− i cos(ωt)) to f02 sin (ωt). f02 sin (ωt) = f02



1 1 (sin (ωt) + i cos (ωt)) + (sin (ωt) − i cos (ωt)) . 2 2 

(2.36)

Now, the conversion to the complex exponential form requires some more alterations to the above equation. For this purpose, we factor the imaginary unit i out of the right hand side. "

= if02

1 2

sin (ωt) 1 + cos (ωt) + i 2 !

sin (ωt) − cos (ωt) i

!#

.

(2.37)

We further use that division by the complex number i can be rewritten by multiplication with i −i due to the fact that 1i = ii2 = −1 = −i. = if02

Structural Dynamics summer term 2019



1 1 (cos (ωt) − i sin (ωt)) − (cos (ωt) + i sin (ωt)) . 2 2 

(2.38)

2 Description of harmonic oscillations

13

Now, we can again apply Euler’s formula to obtain i 1 h = −i f02 e iωt − e −iωt . 2

(2.39)

To obtain a full representation of Eq. (2.15), we combine results from Eqs. (2.35) and (2.39). f (t) = f01 cos (ωt) − f02 sin (ωt)

(2.40)

i i 1 h 1 h = f01 e iωt + e −iωt + i f02 e iωt − e −iωt 2 2 1 1 = [f01 + if02 ] e iωt + [f01 − if02 ] e −iωt {z } {z } |2 |2 fb+

(2.41) (2.42)

fb−

= fb+ e iωt + fb− e −iωt =

f0 iϕ iωt f0 −iϕ −iωt e e + e e . 2 2

(2.43)

We note, that an arbitrary harmonic signal with circular frequency ω is composed of two complex exponential parts, that are complex conjugates of each other 

fb+ e iωt

∗



= fb+∗ e iωt

∗

= fb− e −iωt

(2.44)

as fb+∗ = fb− 

e iωt

∗

= e −iωt .

(2.45) (2.46)

Depending on whether we have a pure sine or cosine signal we can state for the coefficients fb+ and fb− : pure cosine signal → fb+ and fb− are real pure sine signal → fb+ and fb− are imaginary

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2 Description of harmonic oscillations

14

Im

Im

fb+ = f01

fb

+

Re

(a) Pure cosine, real

Im

= if02

fb+ = f01 + if02 Re

(b) Pure sine, imaginary fb+

fb+

Re

(c) Mixed, arbitrary complex fb+

Figure 2.5: Representation of complex pointers with different contributions of cosine and sine part. The left hand side complex number fb+ only has contributions of the cosine part f01 and is therefore purely real. Vice versa the middle illustration shows the complex number that only consists of a sine contribution f02 and is therefore purely imaginary. In a general case, the complex pointer shows an arbitrary orientation, depending on f01 and f02 .

Im 1 z1 Re

z2 1

π −1

π 2

3π 2

2π

3π 5π 2

7π 2

ωt 4π z3

−1 Figure 2.6: The complex exponential function evolves harmonically in the complex plane over time. It is z1 = e iωt (—); z2 = Re(z1 ) = cos(ωt), (—); z3 = i Im(z1 ) = i sin(ωt), (—).

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2 Description of harmonic oscillations

15

3 Fourier transformation 3.1 Periodic signals At first deterministic signals are considered. We consider random signals in Cha.13. We distinguish between transient signals that are only present within a limited time windows, and periodic signals that have a repetition frequency (period). After the time T a periodic signal f (t) repeats, such that it holds: f (t) = f (t ± nT ) .

(3.1)

This family of signals can be represented by a series of sine and cosine functions. This representation is called Fourier series. Subsequently we discuss how to determine such representations.

3.1.1 Representation with real coefficients An arbitrary periodic function can be represented by a Fourier series under the condition that it is integrable on one period [0,T ], i.e. the integral ZT

f (t) dt

(3.2)

0

exists. Then the Fourier series reads f (t) =

∞ a0 X + (ak cos (kω0 t) + bk sin (kω0 t)) 2 k=1

(3.3)

with ωk = kω0 and ω0 = 2π . For the above equality the series has to converge for k → ∞. T The coefficients ak , bk are determined such that the mean-square error ε between the given

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16

function f (t) and the series representation is minimal. The error reads ε=

ZT 0

!2

∞ a0 X f (t) − − (ak cos (ωk t) + bk sin (ωk t)) 2 k=1

dt .

(3.4)

In order to find an optimal solution we differentiate the error ε with respect to the unknown coefficients and set the resulting expressions to zero: ∂ε ! =0 ∂aj

(3.5)

∂ε ! = 0. ∂bj

(3.6)

Applying the chain rule, and neglecting the case j = 0 at first, we find T

Z ∂ε = −2 ∂aj 0

T

Z ∂ε = −2 ∂bj 0

∞ a0 X f (t) − − (ak cos (ωk t) + bk sin (ωk t)) cos (ωj t) dt = 0 2 k=1

(3.7)

∞ a0 X f (t) − − (ak cos (ωk t) + bk sin (ωk t)) sin (ωj t) dt = 0 . 2 k=1

(3.8)

!

!

For j,k > 0, we find that ZT 0

ZT 0

ZT

 0

sin(ωk t) sin(ωj t) dt =  T

2

 0

cos(ωk t) cos(ωj t) dt =  T

2

j= 6 k j = k 6= 0,

(3.9)

j 6= k j = k 6= 0,

(3.10)

cos(ωk t) sin(ωj t) dt = 0,

(3.11)

0

ZT

cos(ωj t) dt =

0

ZT

sin(ωj t) dt = 0.

(3.12)

0

Thus, Eqs. (3.7) and (3.8) reduce to  T  Z T −2  f (t) cos (ωk t) dt − ak  0

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2

=0

(3.13)

3 Fourier transformation

17

 T  Z T −2  f (t) sin (ωk t) dt − bk 

2

0

= 0.

(3.14)

The coefficients ak , bk can then be determined: T 2Z ak = f (t) cos (ωk t) dt T

(3.15)

T 2Z bk = f (t) sin (ωk t) dt . T

(3.16)

0

0

For the case k = 0, we have T

Z ∂ε =2 ∂a0 0

∞ a0 X 1 f (t) − − (ak cos (ωk t) + bk sin (ωk t)) dt = 0 . 2 2 k=1

!

(3.17)

The integrals over the sine and cosine functions vanish, so that it follows ZT

f (t) dt −

0

a0 T = 0. 2

(3.18)

The coefficient a0 then reads T 2Z f (t) dt . a0 = T

(3.19)

0

The Fourier coefficients are completely determined by Eqs. (3.19), (3.15) and (3.16). Since Eq. (3.15) is equivalent to Eq. (3.19) for k = 0 since cos(0) = 1, the expression for a0 is often not explicitly stated.

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3 Fourier transformation

18

Example 3.1

Consider the following wave f (t) with amplitude f0 and period T .

f f0

t

T

The considered signal is an odd function, i.e., f (t) = −f (−t). Through this, we can immediately state that all ak = 0 and thus all cosine terms in the series vanish. This leaves us with the evaluation of the coefficients bk . Using Eq. (3.8), we find:

bk =

2 T

 T  T 2 Z Z    f0 sin(ωk t) dt + (−f0 ) sin(ωk t) dt   0

T 2

T

2 4 1 = − f0 cos(ωk t) T ωk 0   4f0

for k = 1,3,5, . . . for k = 0,2,4, . . .

=  kπ 0

Then, the Fourier series representation of f (t) reads f (t) =

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4f0 π

1 sin (ωk t) . k=1,3,... k ∞ X

3 Fourier transformation

19

3.1.2 Representation with complex coefficients In general it is easier to represent a periodic function by a Fourier series in complex notation

f (t) =

∞ X

cbk e iωk t .

(3.20)

k=−∞

Therein the coefficients cbk are determined by cbk

1ZT = f (t)e −iωk t dt . T 0

(3.21)

By application of Euler’s formula e iϕ = cos(ϕ) + i sin(ϕ), we find cbk =

1ZT 1ZT 1 f (t)e −iωk t dt = f (t) (cos(iωk t) − i sin(iωk t)) dt = (ak − ibk ) T 0 T 0 2

(3.22)

where the last step can be found by comparing with Eqs. (3.15) and (3.16). Furthermore for the coefficient with negative k we find: cb−k =

1 (ak + ibk ) . 2

(3.23)

Thus, cbk and cb−k are complex conjugate: cb−k = cb∗k .

(3.24)

For k = 0 it holds cb0 =

a0 . 2

(3.25)

We can check the identity of the real and complex Fourier representations by inserting Eqs. (3.22) and (3.23) into Eq. (3.20): f (t) =

∞ X k=−∞

Structural Dynamics summer term 2019

cbk e iωk t

=

∞ X

cbk e iωk t + cb∗k e −iωk t .

(3.26)

k=1

3 Fourier transformation

20

We split the summation from −∞ to ∞ by applying the knowledge that the coefficients for negative k are the complex conjugate values of the coefficients with positive k. 1 f (t) = cb0 e 0t +

1 (abk − ibk ) (cos(ωk t) + i sin(ωk t)) + k=1 2 ∞ X

+

1 (abk + ibk ) (cos(ωk t) − i sin(ωk t)) . (3.27) k=1 2 ∞ X

This can finally be shown to be ∞ a0 X (ak cos(ωk t) + bk sin(ωk t)) f (t) = 2 k=1

(3.28)

which is equal to Eq. (3.22). Im(cbk ) cb4

cb−3

Re(cbk ) cb−2 cb−1 Re cbk

cb−4

cb3

Im(cbk )

r

r

cb1 cb2 odd

even

Figure 3.1: Graphical representation of complex Fourier series coefficients. Here, exemplarily three terms are depicted in the complex plane, and the corresponding discrete spectrum over r for real and imaginary part of the b ck are given. Due to the complex conjugate property of the complex coefficients, we can easily deduce that the real parts are always be even with respect to r while the imaginary parts are always odd with respect to r.

The following expression summarize the relations between the coefficients of the real and complex representation of the Fourier series. a0 2 a0 = 2cb0 cb0 =

1

1 (ak − ibk ) 2 ak = cbk + cb−k cbk =

1 (ak + ibk ) 2 bk = i (cbk − cb−k ) .

cb−k =

(3.29) (3.30)

Here we see that due to the form of the Fourier series, where the sine term is added rather than subtracted, as was introduced in the representation of harmonic signals in Cha. 2, the coefficient of the positive time exponential function has a negative inaginary part.

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3 Fourier transformation

21

3.2 Fourier integral The Fourier transformation can be interpreted as a limiting case of the Fourier series for signals with an infinite period T → ∞. By this non periodic functions the representation of non-periodic functions can be approached. We insert the definition of the coefficients cbk from Eq. (3.21) into the Fourier series Eq. (3.20) and obtain ∞ X

f (t) =

e

ikω0 t

k=−∞

Z T 0

!

f (t)e

and introduce ∆ω = ω0 =

2π . T

−ikω0 t

dt

1 T

(3.31)

With

2π = dω T →∞ T

lim ∆ω = lim

(3.32)

k∆ω → ω

(3.33)

T →∞

one finds for the limiting case T → ∞: ∞ ∞ 1 Z Z f (t)e −iωt dt e iωt dω f (t) = 2π



−∞



−∞

and further f˜(ω) =

Z∞

f (t)e −iωt dt

(3.34)

−∞ ∞ 1 Z ˜ f (t) = f (ω)e iωt dω . 2π

(3.35)

−∞

Eq. (3.34) is called Fourier transform (FT), Eq. (3.35) is called inverse Fourier transform (IFT). An alternative representation using f = g˜(f ) =

Z∞

ω 2π

is given by

g(t)e −i2πf t dt

(3.36)

g˜(f )e i2πf t df .

(3.37)

−∞

g(t) =

Z∞ −∞

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3 Fourier transformation

22

The functions f (t) and f˜(ω) and g(t) and g˜(f ) are called a Fourier transform pair. ˜ The Fourier  transform can also be written in operator notation as F (f (t)) = f (ω) and F −1 f˜(ω) = f (t). Often, the Fourier transform of f (t) is denoted by an uppercase F {f } = F (ω). We use the introduced notation to avoid confusion with the convention that random variables are denoted by an uppercase letter (comp. Chapter 13). Example 3.2: Fourier transform of the rectangular function

The following rectangular function is given t g(t) = g0 Π T0 



=

   g0

−T0 < t < T0 t = ±T0 else

g0 /2    0

where Π(t) denotes the unit rectangular function. By applying Eq. (3.34), we find the Fourier transform of g˜(t). ZT0

g˜(f ) =

g0 e −i2πf t dt =

−T0



= g0 

ZT0







cos(2πf t) dt − ig0  



−T0



sin(2πf t) dt 

−T0

|

= 2g0 T0

ZT0

{z

=0

}

sin(2πf T0 ) . 2πf T0

T0 ) The function sin(2πf is also known as the sinus cardinalis function sinc(2πf T0 ). The 2πf T0 value at f = 0, g˜(0) = 2g0 T0 corresponds to the area under the signal in the time domain. The Fourier transform pair g(t) and g˜(f ) is depicted in Fig. 3.2

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3 Fourier transformation

23

g(t)

g˜(f )

I = 2g0 T0

2g0 T0 g0 t −T0

f

T0

1 T0

1 2T0

Figure 3.2: Fourier transform of the rectangular function g(t) = g0 Π



t T0



3 2T0

. The Fourier

transform g˜(f ) has the maximum amplitude 2g0 T0 at f = 0. Zero crossings n . are given by f = 2T 0

If we now alter the function g(t) such that the signal is half as long with T00 = with double amplitude h0 = 2g0 , such that t g 0 (t) = h0 Π T00

!

2t = 2g0 Π T0 



=

   2g0

g0    0

− T20 < t < t = ± T20 else

T0 2

, but

T0 2

we find the Fourier transform g˜0 (f ), which is depicted in Fig. Eq. (3.3). I 0 = 2h0 T00 = 2g0 T0

g 0 (t)

g˜0 (f ) 2h0 T00 = 2g0 T0

h0 = 2g0 t T00 =

T0 2

f 1 2T00

Figure 3.3: Fourier transform of the rectangular function g 0 (t) = h0 Π and T00 = T20 . 2h0 T00 = 2g0 T0

= 

t T00

2 2T0



=

1 T0

with h0 = 2g0

The Fourier transform g˜0 (f ) has the maximum amplitude I 0 = = I at f = 0. Zero crossings are given by f = Tn0 .

Through shortening the signal in the time domain, the Fourier transform becomes wider in the frequency domain. The first zero crossing of g˜0 (f ) is twice as high as for

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3 Fourier transformation

24

the original signal g˜(f ). The maximum amplitude does not change, as the area under the signal does not change either. In the limit case T0 → 0 the Fourier transform results in I = 2g0 T0

Z∞

δ (t) dt = 2g0 T0

−∞

with the Dirac-delta function defined, such that δ(t) = 0 for t 6= 0 Z∞

δ(t) dt = 1 .

−∞

The Fourier transform of a Dirac impulse becomes a constant function in the frequency domain. This is depicted in Fig. 3.4. g 0 (t)

g˜0 (f ) 2g0 T0 δ(t)

2g0 T0

t

f

Figure 3.4: Fourier transform of the impulse function g 0 (t) = g0 T0 δ(t). The Fourier transform g˜0 (f ) is the constant function with g˜0 (f ) = 2g0 T0 .

Example 3.3: Fourier transform of the sine function

We consider the sine function g(t) = g0 sin(2πf0 t)

(3.38)

and its Fourier transform g˜(f ) = i

Structural Dynamics summer term 2019

g0 [−δ(f − f0 ) + δ(f + f0 )] 2

(3.39)

3 Fourier transformation

25

which is taken without further proof. The Fourier transform pair g(t), g˜(f ) is depicted in Fig. 3.6. g˜(f ) + f0 )

i g20 δ(f

g(t) sin(2πf0 t)

g0 f0

t

f

−f0 1 f0

−i g20 δ(f − f0 )

Figure 3.5: Fourier transform of the sine function g(t) = g0 sin(2πf0 t).

We can show that g˜(f ) (Eq. (3.39)) is the Fourier transform of g(t) (Eq. (3.38)), by applying the inverse Fourier transformation (Eq. (3.37)) to Eq. (3.39). Z∞

g(t) =

g˜(f )e i2πf t df =

−∞ ∞ Z∞ ig0  Z i2πf t = −δ(f − f0 )e df + δ(f + f0 )e i2πf t df  = 2





−∞

−∞

ig0 −e i2πf0 t + e −i2πf0 t = 2 ig0 = [− cos(2πf0 t) − i sin(2πf0 t) + cos(2πf0 t) − i sin(2πf0 t)] = 2 ig0 = [−2i sin(2πf0 t)] = g0 sin(2πf0 t) 2 =

h

i

which is the signal we defined in Eq. (3.38).

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3 Fourier transformation

26

Example 3.4: Fourier transform of the cosine function

We consider the cosine function g(t) = g0 cos(2πf0 t)

(3.40)

and its Fourier transform g˜(f ) = i

g0 [δ(f − f0 ) + δ(f + f0 )] 2

(3.41)

which is taken without further proof. The Fourier transform pair g(t), g˜(f ) is depicted in Fig. 3.6.

g0 δ(f 2

g˜(f ) + f0 )

g(t)

g0 δ(f 2

− f0 )

cos(2πf0 t)

g0

t

f −f0

f0 1 f0

Figure 3.6: Fourier transform of the cosine function g(t) = g0 cos(2πf0 t).

We can show that g˜(f ) (Eq. (3.39)) is the Fourier transform of g(t) (Eq. (3.38)), by applying the inverse Fourier transformation (Eq. (3.37)) to Eq. (3.41). Z∞

g(t) =

g˜(f )e i2πf t df =

−∞ ∞ Z∞ ig0  Z i2πf t δ(f − f0 )e df + δ(f + f0 )e i2πf t df  = = 2



−∞



−∞

i ig0 h i2πf0 t e + e −i2πf0 t = 2 ig0 = [cos(2πf0 t) + i sin(2πf0 t) + cos(2πf0 t) − i sin(2πf0 t)] = 2

=

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3 Fourier transformation

27

=

ig0 [2 cos(2πf0 t)] = g0 cos(2πf0 t) 2

which is the signal we defined in Eq. (3.40).

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3 Fourier transformation

28

4 Setting up the equation of motion 4.1 Single degree of freedom system The single degree of freedom (SDOF) system plays a significant role in many problems in structural dynamics. This importance stems from the fact, that many engineering systems or system components can be approximated by SDOF systems. The mechanical behaviour is described by the equation of motion, which is an ordinary differential equation. We summarize important methods to obtain this differential equation in the following.

4.1.1 Equilibrium of forces The equation of motion of the SDOF system follows from the equilibrium of forces. They are composed out of the external forces, the gravity force, and the reaction forces, that result from the motion of the system. Additional to the static reaction forces (restoring force), we consider the d’Alembert inertia forces and, possibly, damping forces. The equations of motion are determined upon the deflected system applying the principle of d’Alembert. To this end, a free body cut is applied around the mass point of the SDOF system, as depicted in Fig. 4.1. fe

fe - external force fg

w,w, ˙ w¨

fg - gravity force

fK

fK - elastic restoring force

fC

fC - damper force (formulated as velocity-proportional in the following)

fI

fI - d’Alembert inertia force

Figure 4.1: External and reaction forces on single mass point.

Structural Dynamics summer term 2019

29

Thus, the equilibrium results in −fK − fC − fI + fe + fg = 0.

(4.1)

Inserting the expressions for the single forces, we obtain mw¨ (t) + cw˙ (t) + kw (t) = fe (t) + fg .

(4.2)

For the free vibration problem, i.e., fe (t) = 0 and neglecting the gravity force, the equation of motion simplifies to mw¨ (t) + cw˙ (t) + kw (t) = 0.

(4.3)

We obtain a differential equation, where together with the displacement w(t), also the first two derivatives with respect to time, namely the velocity w(t) ˙ and the acceleration w(t), ¨ appear.

Example 4.1: Examples for SDOF systems

«««< .mine «««< .mine In the following figure, some examples for SDOF models describing real systems are depicted. w m wc

EI

k

m µ=0 l EI

µ=0 m w

c)

b)

a)

w

l

GIT

m µ=0

EI

Iθ

kφ

φ

d)

Structural Dynamics summer term 2019

e)

4 Setting up the equation of motion

30

Cases a) to d):

Case e): Iθ φ¨ + cφ φ˙ + kφ φ = m(t)

mw¨ + cw˙ + kw = f (t) with w(t) the displacement. The characteristic parameters are

with φ(t) the rotation. The characteristic parameters are

m . . . mass

Iθ . . . Rotatory intertia

c . . . damping constant

cφ . . . torsional damping constant

k . . . stiffness

kφ . . . torsional stiffness

For example, for c), the stiffness term For example, for e), the stiffness term 3EI k is k = l3 . kφ is kφ = GIl t . ||||||| .r59 In the following figure, some examples for SDOF models describing real systems are depicted. w m wc

EI

k

m µ=0 l EI

µ=0 m w

c)

b)

a)

w

l

GIT

m µ=0

EI

Iθ

kφ

φ

d)

Structural Dynamics summer term 2019

e)

4 Setting up the equation of motion

31

Cases a) to d):

Case e): Iθ φ¨ + cφ φ˙ + kφ φ = m(t)

mw¨ + cw˙ + kw = f (t) with w(t) the displacement. The characteristic parameters are

with φ(t) the rotation. The characteristic parameters are

m . . . mass

Iθ . . . Rotatory intertia

c . . . damping constant

cφ . . . torsional damping constant

k . . . stiffness

kφ . . . torsional stiffness

For example, for c), the stiffness term For example, for e), the stiffness term 3EI k is k = l3 . kφ is kφ = GIl t . ======= In the following figure, some examples for SDOF models describing real systems are depicted. w m wc

EI

k

m µ=0 l EI

µ=0 m w

c)

b)

a)

w

l

GIT

m µ=0

EI

Iθ

kφ

φ

d)

Structural Dynamics summer term 2019

e)

4 Setting up the equation of motion

32

Cases a) to d):

Case e): Iθ φ¨ + cφ φ˙ + kφ φ = m(t)

mw¨ + cw˙ + kw = f (t) with w(t) the displacement. The characteristic parameters are

with φ(t) the rotation. The characteristic parameters are

m . . . mass

Iθ . . . Rotatory intertia

c . . . damping constant

cφ . . . torsional damping constant

k . . . stiffness

kφ . . . torsional stiffness

For example, for c), the stiffness term For example, for e), the stiffness term 3EI k is k = l3 . kφ is kφ = GIl t . »»»> .r61 ||||||| .r51 In the following figure, some examples for SDOF systems are depicted. w m wc

EI

k

m µ=0 l EI

µ=0 m w

c)

b)

a)

w

l

GIT

m µ=0

EI

Iθ

kφ

φ

d)

Structural Dynamics summer term 2019

e)

4 Setting up the equation of motion

33

Cases a) to d):

Case e): Iθ φ¨ + cφ φ˙ + kφ φ = M (t)

mw¨ + cw˙ + kw = F (t) with w(t) the displacement. The characteristic parameters are

with φ(t) the rotation. The characteristic parameters are

m . . . mass

Iθ . . . Rotatory intertia

c . . . damping constant

cφ . . . torsional damping constant

k . . . stiffness

kφ . . . torsional stiffness

For example, for c), the stiffness term For example, for e), the stiffness term 3EI k is k = l3 . kφ is kφ = GIl t . ======= In the following figure, some examples for SDOF models describing real systems are depicted. w m wc

EI

k

m µ=0 l EI

µ=0 m w

c)

b)

a)

w

l

GIT

m µ=0

EI

Iθ

kφ

φ

d)

Structural Dynamics summer term 2019

e)

4 Setting up the equation of motion

34

Cases a) to d):

Case e): Iθ φ¨ + cφ φ˙ + kφ φ = m(t)

mw¨ + cw˙ + kw = f (t) with w(t) the displacement. The characteristic parameters are

with φ(t) the rotation. The characteristic parameters are

m . . . mass

Iθ . . . Rotatory intertia

c . . . damping constant

cφ . . . torsional damping constant

k . . . stiffness

kφ . . . torsional stiffness

For example, for c), the stiffness term . k is k = 3EI l3 »»»> .r61

For example, for e), the stiffness term kφ is kφ = GIl t .

4.1.2 Virtual work Analogously the equations of motions can be obtained using the principle of d’Alembert, where a suitable virtual displacement is imposed on the system. Then the principle of virtual work is written. The general procedure is shown by means of the following example. Example 4.2: Setting up the ODE using the principle of virtual work

The following system is excited by a dynamic force F (t). Gravity forces are not considered. f (t)

m

starr c

l/2

k

l/2

l/2

l/2

l/2

The system is kinematic and has one degree of freedom. The system in the deformed state is depicted below.

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4 Setting up the equation of motion

35

wm wc

ϕ2 ϕ1

wG

wk

The kinematic relations are: l = ϕ2 l 2 wm = ϕ1 l wG = ϕ1

→

ϕ1 = ϕ2 2

l wm = 2 2 l wm wF = wG = ϕ1 = 2 2 ϕ1 l 1 wm l = ϕ1 l = . wk = ϕ2 = 2 2 2 4 4

wc = ϕ 1

Hence the following forces act on the system: f (t)

wm wc m w¨m

c w˙ c

wf

wk k wk

Next, a virtual displacement is superimposed upon the deformed system.

δwm δwc f (t)

wm wc m w¨m

c w˙ c

wf

δwk δwG

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wk k wk

4 Setting up the equation of motion

36

For the system to be in equilibrium, the sum of virtual work of all forces has to vanish, i.e. δW = 0. Thus, −mw¨m δwm − cw˙ c δwc − kwk δwk + f (t)δwF = 0 mw¨m δwm + c

wm δwm δwm w˙ m δwm +k = f (t) . 2 2 4 4 2

With δwm 6= 0 it follows 1 1 1 mw¨m + cw˙ m + kwm = f (t) . 4 16 2 This equation corresponds to the differential equation of the system with respect to the vertical displacement of the point mass m.

4.1.3 The principle of Hamilton In the following we shall introduce the principle of Hamilton. Also refer to the lecture on Continuum Mechanics. 4.1.3.1 Undamped systems

Alternatively to the principle of virtual work, the equations of motion can be derived through the principle of Hamilton. As derived in the course Continuum Mechanics, Hamilton’s principle states that for conservative systems (where forces have a potential) among all kinematically possible movements the movement will arise that minimizes the time integral over the Lagrange-function L(t) = T (t) − U (t).

(4.4)

with the kinetic energy Ekin (t) = T (t) and the potential energy Epot (t) = U (t). For fixed conditions at t0 und t1 , I=

Zt1

L (w,w) ˙ dt = extremum .

(4.5)

t0

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4 Setting up the equation of motion

37

w(t) δw

t0

t1

t

Example 4.3

An undamped SDOF system is deflected from its static position to initial conditions, and is then released. There are infinitely many possibilities, which can satisfy the boundary conditions. For each of these possibilities, the velocity w, ˙ can be determined at each time t. Thereby also the Lagrange function can be determined and the above integral can be evaluated. The function, for which the integral gives an absolute extremum (minimum), i.e., the integral that is smaller than any other possible integrals, is the function that will really occur. As a second boundary condition we consider the observation that the SDOF system shows the static displacement after the time t = T /4, i.e., w( T4 ) = 0. The system parameters are chosen as follows. m = 1,

k = 1,

ω = 1,

T = 2π .

Initial conditions: w0 = w(t = 0) = −1

vstart = v(t = 0) = 0 .

We investigate three different approaches for the displacement function w(t): • 1. approach: quadratic parabola • 2. approach: cosine function We now evaluate the Lagrange function for the three approaches: • 1. approach. We choose the following quadratic parabola, complying with our observation at t = T4 , w0 w(t) = w0 − 4 2 t2 , π 

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4 Setting up the equation of motion

38

and derive the velocity, kinetic and potential energy, and the Lagrange funtion. w0 dw(t) =− 8 2 t dt π   1 1 w0 2 T (t) = mv(t)2 = m 8 2 t2 2 2 π     1 w0 2 2 1 2 U (t) = kw(t) = k w0 − 4 2 t 2 2 π L(t) = T (t) − U (t) . 

v(t) =



The evaluation of Eq. (4.5) thus reads T

Z4

L(t) dt = 5,534 · 10−3 .

0

1 T(t)

0.5

U(t) L(t)

0 -0.5

2 w(t) v(t)

0 0.5 1 1.5

1 0 -1

2

0 0.5

1 1.5

2

• 2. approach. We choose the following cosine function, again complying with our observation at t = T4 , s



k  w(t) = w0 cos  t m and derive the velocity, kinetic and potential energy, and the Lagrange function. s

s



dw(t) k k  v(t) = = −w0 sin  t dt m m s



k k  1 1 T (t) = mv(t)2 = mw02 sin2  t 2 2 m m s



1 1 k  U (t) = kw(t)2 = kw02 cos2  t 2 2 m

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4 Setting up the equation of motion

39

L(t) = T (t) − U (t) . The evaluation of Eq. (4.5) thus reads T

Z4

L(t) dt = 0 .

0

1

0.5 T(t) U(t) L(t)

w(t) v(t)

0 -0.5

0 0.5

-1

2

1 1.5

0

0 0.5

t

1 1.5 t

2

The computation calculus for determining the function w(t) is the variational calculus (comp. continuum mechanics). From the variational calculus Eulers differential equation results as δI =

Zt1 t0

d ∂T (w) ˙ ∂U (w) − δw dt = 0 . − ∂w dt ∂ w˙ !

As δw can be chosen arbitrary, e.g., always such that it has the same sign as the expression in the braces, the integral can only be equal to zero, if the brace itself becomes zero. Thus, the differential equation for the SDOF system follows to: d ∂T (w) ˙ ∂U (w) + = ∂w dt ∂ w˙

∂

h

1 kw2 2

∂w

i

h

i

1 2 d ∂ 2 mw˙ + = kw + mw¨ = 0 . dt ∂ w˙

4.1.3.2 Introduction of non-conservative forces

Without non-conservative forces (e.g., damping forces or exterior forces that cannot be described by a potential), the Hamilton’s principle states I=

Zt1

(T − U ) dt = extremum

t0

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4 Setting up the equation of motion

40

Zt1  t0

1 1 mw˙ 2 − kw2 dt . 2 2 

The variation δw (t) can be chosen arbitrary but needs to obey the boundary conditions: Zt1

(mw˙ · δ w˙ − kw · δw) dt .

t0

Integrating parts ( u0 v = − uv 0 + [uv]) results in: R

Zt1 t0

R

t1 (−mw¨ · δw − kw · δw) dt + [mwδw] ˙ t0

|

{z

Principle of virtual work

}

where δw is chosen to vanish at t = t0 and t = t1 (δw can be chosen arbitrarily). Apparently, the virtual work, done by inertia and spring forces, can be described via the variation of kinetic and potential energy T and U . The difference T −U is called the Lagrangian L of the system. Thus, the well-known formulation for the equilibrium can be derived: −mw¨ − kw = 0 . This expression can be expanded by non conservative forces in the context of the principle of virtual work and the equilibrium: −mw¨ − kw − cw˙ + f = 0 . Principle of virtual work: (−mw¨ − kw − cw˙ + f )δw(t) = 0 . For arbitrary δw(t), one obtains: Zt2

(−mw¨ − kw − cw˙ + f )δw dt + [mw˙ · δw]tt21 = 0 |

t1

Zt2

{z

=0

}

[(mw˙ · δ w˙ − kw · δw) − cw˙ · δw + f · δw] dt = 0

t1

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4 Setting up the equation of motion

41

Zt2

(δ(T − U ) − cwδw ˙ + f δw) dt = 0

t1

where δw is the variation applied to the system. Thus, Hamilton’s principle can be extended by the work Wnc of non-conservative forces: w(t) δw

t0

Wnc =

t1

t

Zw1

(−cw˙ + f ) dw .

w0

The variation of Wnc results in: δWnc = (−cw˙ + f )δw which permits to extend the Lagrangian as: I=

Zt1

(T − U + Wnc ) dt = extremum .

t0

This implies: δ

Zt1

L dt + δ

t0

Structural Dynamics summer term 2019

Zt1

Wnc dt = 0 .

t0

4 Setting up the equation of motion

42

Example 4.4

f (t)

m c

Zt1

δ

L dt + δ

Zt1

k

w(t)

Wnc dt = 0

t0

t0

with L=T −U and T

kinetic energy

U

potential energy

Wnc

non-conservative (external forces, damping) work .

• Conservative forces 1 U = kw2 − mgw 2 1 T = mw˙ 2 2 1 1 L = T − U = mw˙2 − kw2 + mgw 2 2 δWnc = f (t)δw − cwδw ˙ . Applying Hamilton’s principle delivers: Zt1

δ

t0

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L dt = δ

Zt1 t0

1 ˙2 1 2 mw − kw + mgw dt 2 2

4 Setting up the equation of motion

43

=

Zt1

mwδ ˙ w˙ − kwδw + mgδw dt .

t0

• Expanded with non-conservative forces Zt1

Zt1

mwδ ˙ w˙ dt −

⇒

kwδw dt +

t1 mwδw ˙ t0

−

Zt1

mwδw ¨ dt −

t1

Zt1



t0

|

t0

|

{z

(a)

}

Zt1

kwδw dt +

Zt1

(f (t) − cw˙ + mg) δw dt = 0

t0

t0

t0

⇒ mwδw ˙ +

(f (t) − cw˙ + mg) δw dt = 0

t0

t0

t0

Zt1



− mw¨ − kw + f (t) − cw˙ + mg δw dt = 0 . {z

(b)

}

The variation δw can be chosen such that (a) disappears, i.e. δw(t0 ) = δw(t1 ) = 0, according to the derivation of Hamilton’s principle. Furthermore (b) must be zero independent from δw in order to satisfy the equation. Thus (b), states the equation of motion. By applying the variation calculus, the same is obtained by the Euler-Lagrange-equation: ˙ w) δWnc ˙ w) ∂L(w, d ∂L(w, + + =0 dt ∂ w˙ ∂w δw d − mw˙ − kw + mg + f (t) − cw˙ = 0 dt − mw¨ − kw + mg + f (t) − cw˙ = 0 .

−

4.1.3.3 Lagrange multiplicators

The principle of Hamilton is especially advantageous, as further unknowns can be introduced in the equations very easily. Often, the number of unknowns does not correspond to the number of independent degrees of freedom. In these cases, further constraints are formulated, that can then be considered through the method of Lagrange multipliers.

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4 Setting up the equation of motion

44

Example 4.5: Chain structure

l1

l2

l3

l4

l5

It is advantageous, to describe the kinetic and potential energy through the nodal degrees of freedom (4 · 2 = 8 displacements ui , vi , or q1 . . . q8 ). Through the bearing not all degrees of freedom can be chosen independently, as the lengths of the single chain links l1 , l2 , . . . , l5 stay constant. Thus, it holds L = T − U = L(q1 , . . . , q8 , q˙1 , . . . , q˙8 ) . The geometrical constraints l(q1 , q2 ) = l1 l(q1 , q2 , q3 , q4 ) = l2 ... l(q7 , q8 ) = l5 are written in the form f (q1 , q2 ) = l(q1 , q2 ) − l1 = 0 f (q1 , q2 , q3 , q4 ) = l(q1 , q2 , q3 , q4 ) − l2 = 0 ... f (q7 , q8 ) = l(q7 , q8 ) − l5 = 0. The principle of variation with the additional constraints is formulated as follows: ˜ = L(q1 , . . . , q8 , q˙1 , . . . , q˙8 ) + λ1 f (q1 , q2 ) + λ2 f (q1 , q2 , q3 , q4 ) + . . . + λ5 f (q7 , q8 ) L ˜ ˜ ∂L d ∂L + = 0, with i = 1 . . . 8 dt ∂ q˙i ∂qi ˜ ˜ ˜ d ∂L ∂L ∂L − + = 0 → = 0, with j = 1 . . . 5 dt ∂ λ˙ j ∂λj ∂λj −

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4 Setting up the equation of motion

45

Example 4.6: Rigid beam, one degree of freedom, two unknowns w1 and w2

l1 w1

1 T = mw˙ 22 2

l2

m

starr

k

w2

1 U = kw12 2

Geometric constraint: w2 w1 =− l2 l1 δ

Zt1  t0

→ f = w2 l1 + w1 l2 = 0

1 1 mw˙ 22 − kw12 + λ(w2 l1 + w1 l2 ) dt = 0 2 2 

˜ ˜ ∂L d ∂L + = 0 → −mw¨2 + λl1 = 0 dt ∂ w˙2 ∂w2 ˜ ˜ d ∂L ∂L Variation δw1 : − + = 0 → kw1 + λl2 = 0 dt ∂ w˙1 ∂w1 ˜ ∂L ˜ d ∂L + = 0 → w2 l1 + w1 l2 = 0 Variation δλ : − dt ∂ λ˙ ∂λ Variation δw2 :

−

(4.6) (4.7) (4.8)

Multiplication of Eq. (4.6) with Eq. (4.7) results in −ml2 w¨2 + kl1 w1 = 0

X



M =0 .

With Eq. (4.8) w2 = −

w1 l2 l1

follows ml22 w¨1 + kl12 w1 = 0.

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4 Setting up the equation of motion

46

Example 4.7: Elastic beam, Ritz-approach

The application of the principle of Hamilton for discretized, continuous systems is explained by means of the following example. m

x EI,µ

EI,µ k l

xi Kinetic energy: T =

Zl 0

1 2 1 µw˙ (x) dx + mw˙ 2 (xi ) . 2 2

Potential energy (without consideration of the gravitational potential energy): U=

Zl 0

1 1 EI(w00 (x))2 dx + kwi2 . 2 2

Boundary condition: wi = w(xi )

→ f = wi − w(xi ) = 0 .

Lagrangian: ˜= L

Zl  0

1 2 1 1 1 µw˙ (x) − EI(w00 (x))2 dx + mw˙ i2 − kwi2 + λ (wi − w(xi )) . 2 2 2 2 

Ritz-approach: w(A, x, t) = A sin

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πx sin (ωt) l

4 Setting up the equation of motion

47

wi = B sin (ωt) λ = λ0 sin (ωt) . Furthermore it holds: πx cos (ωt) w(x) ˙ = ωA sin l π2 πx w (x) = − 2 A sin sin (ωt) l l 00

→

Zl 0

→

1 w˙ 2 dx = lω 2 A2 cos2 (ωt) 2 Zl

2

(w00 (x)) dx =

0

1 π4 2 2 A sin (ωt) 2 l3

w˙ i = ωB cos (ωt) . With the help of the Ritz approach, the integral over the Lagrange function (depending on A, B, and λ0 ) can be expressed through ˜= L

ZT

˜ L(A, B, λ0 ) dt =

0

1 π4 1 1 πxi 1 = µlω 2 A2 − EI 3 A2 + mω 2 B 2 − kB 2 + λ0 B − A sin 4 4 l 2 2 l "



#

T . 2

Minimization of the integral (principle of Hamilton) is then done by ˜ ∂L 1 1 π4 πxi 2 = µlω A − EI 3 A − λ0 sin =0 ∂A 2 2 l l ˜ ∂L = mω 2 B − kB + λ0 = 0 ∂B ˜ πxi ∂L = B − A sin = 0. ∂λ0 l

(4.9) (4.10) (4.11)

Inserting Eq. (4.10) into Eq. (4.9) results in 1 πxi 1 π4 πxi µlω 2 A − EI 3 A + mω 2 B sin − kB sin = 0. 2 2 l l l B can be expressed using Eq. (4.11) (B = A sin πxl i ): 1 1 π4 πxi πxi µlω 2 − EI 3 + mω 2 sin2 − k sin2 = 0. 2 2 l l l

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4 Setting up the equation of motion

48

Solving for ω 2 returns the Rayleigh quotient 2

ω ≤

2 ωest

=

4 1 EI πl3 + k sin2 πxl i 2 1 µl + m sin2 πxl i 2

=

Epot, max Ekin, max/ω 2

.

4.2 Multiple degree of freedom systems Hereafter different possibilities for setting up the equations of motion of multiple degree of freedom (MDOF) systems are discussed. Here, the equilibrium of forces, the principle of Hamilton, and the principle of virtual work are applied.

4.2.1 Equilibrium of forces The equations of motion for MDOF systems can be obtained through cutting the single system components free and writing the corresponding equilibrium conditions. This is presented hereafter for a oscillator with a mass damper. Example 4.8: 2DOF system - equilibrium

m1 w¨1

m1 w1 k2 (w2 − w1 )

k2 k1 2

k1 2

m2

k2 (w1 − w2 ) k1 w 2 1

m2 w¨2

k1 w 2 1

w2 The equilibrium of vertical forces reads: m1 w¨1 + k1 w1 + k2 (w1 − w2 ) = 0 m2 w¨2 + k2 (w2 − w1 ) = 0 .

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4 Setting up the equation of motion

49

In matrix notation the equilibrium reads m1 0 0 m2

"

#"

w¨1 k + k2 −k2 + 1 w¨2 −k2 k2 #

"

#"

#

w1 =0 w2

or ¨ + Kw = 0 . Mw

Example 4.9: Frame structure with rigid bars and massless columns

This example gives an introduction to a simplified modeling of continuous frame structures as MDOF systems. For the depicted frame structure we assume that the vertical columns are massless (µ = 0). The total floor masses are concentrated in the bars and the bars are assumed to be rigid. All external forces act on the bars.

i+1

xi

mi+1 u¨i+1

Element

mi u¨i

i wi i−1

x1

mi−1 u¨i−1 µ=0

w1

The vertical columns are described by a Euler-Bernoulli beam approach: EIw00000 = 0 as µ = 0 .

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4 Setting up the equation of motion

50

The displacement of the respective beam is then given by (with the local coordinate x): w(x) = C0 + C1 x + C2 x2 + C3 x3 w0 (x) = C1 + 2C2 x + 3C3 x2 w00 (x) = 2C2 + 6C3 x w000 (x) = 6C3 . The four unknown of the displacement C0 , . . . , C3 can be expressed by the displacements and rotations at the boundaries of the vertical beam element, i.e. through the nodal unknowns wl , φl , wu and φu (l for “lower” side, u for “upper” side). This results in w(x) = wl + φl x + 3

wu − wl 2 φu + 2φl 2 φu + φl 3 wu − wl 3 x − x + x −2 x 2 2 l l l l3

and with ξ = xl : 















w(x) = 1 − 3 ξ 2 + 2 ξ 3 wl + l ξ − 2 ξ 2 + ξ 3 φl + 3 ξ 2 − 2 ξ 3 wu + l −ξ 2 + ξ 3 φu . The nodal forces can be obtained from the derivatives of the displacement field as M = −EIw00 and Q = −EIw000 . Under consideration of the definition of positive forces corresponding to the following figure: node Vr Mu beam element x Ml Vl

w

node

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4 Setting up the equation of motion

51

the nodal forces result to: 6 6 4 2 φ − φu , w + w + l l u l2 l l2 l   6 6 2 4 Mr = − [−EIw00 (l)] = EI 2 wl + φl − 2 wu + φu , l l l l   12 6 12 6 000 Vl = − [−EIw (0)] = EI 3 wl + 2 φl − 3 wu + 2 φu , l l l l   12 6 12 6 000 Vr = [−EIw (l)] = EI − 3 wl − 2 φl + 3 wu − 2 φu , l l l l Ml = [−EIw00 (0)] = EI





and for the special case of φl = φu = 0 to: Vl = 12

EI (wl − wu ) l3

Vr = 12

EI (−wl + wu ) . l3

With the full description of the beam elements, the interaction within the complete structure can be described by the equilibrium conditions at the nodes. From the assumption of rigid horizontal system components (floors), we can conclude, that the rotations at the nodes vanish and that the horizontal displacements on the left and right side are the same. For that reason the whole rigid horizontal system component can be cut free and described by the equilibrium condition. In order to consider the interaction of all elements, the displacement field and respective nodal end forces have to considered in global coordinates. (i+1)

ui = wu(i) = wl

.

In the equilibrium Fx = 0 for each floor i, the inertia forces of the bar and external forces are considered: P

Vl

(i+1)

mi u¨i

Vl

(i+1)

fi Vr(i)

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Vr(i)

4 Setting up the equation of motion

52

Transition condition: wi = ui . Then the equilbirium condition reads: X

H=0:

EI EI EI 2 −12 3 ui−1 + 2 · 12 3 ui − 12 3 ui+1 + m¨ ui = fi l l l Ki,i−1 ui−1 + Kii ui + Ki,i+1 ui+1 + m¨ ui = fi 



Ku + M¨ u=f with Ki,i−1 = −24

EI l3

Kii = 48

EI l3

Ki,i+1 = −24

EI . l3

Analogously this can be done for all floors. We need to consider that for the highest floor no contribution V (i+1) appears.

4.2.2 Principle of Hamilton Analogously to the SDOF systems discussed in Sec. 4.1, the principle of Hamilton can also be applied to MDOF systems. This is presented in the following example. Example 4.10: 2DOF - Hamilton’s principle

1

0 l1 φ1

2 kφ l2

k

φ2

m Iθ

The kinematic relations read w1 = φ1 l1 w2 = φ1 l1 + φ2 l2 .

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4 Setting up the equation of motion

53

The kinetic and potential energy can thus be expressed as 2 1 1 1 1  T = mw˙ 22 + Iθ φ˙ 22 = m φ˙ 1 l1 + φ˙ 2 l2 + Iθ φ˙ 22 2 2 2 2 1 1 1 1 U = kw12 + kφ (φ2 − φ1 )2 = k(φ1 l1 )2 + kφ (φ2 − φ1 )2 2 2 2 2 2 1 1 ˙ 1 1 L = T − U = m φ1 l1 + φ˙2 l2 + Iθ φ˙22 − kl12 φ21 − kφ (φ2 − φ1 )2 . 2 2 2 2

The Euler-Lagrange equations read d ∂L(φ1 ,φ2 ,φ˙1 ,φ˙2 ) ∂L(φ1 ,φ2 ,φ˙1 ,φ˙2 ) =0 + dt ∂φ1 ∂ φ˙1 d ∂L(φ1 ,φ2 ,φ˙1 ,φ˙2 ) ∂L(φ1 ,φ2 ,φ˙1 ,φ˙2 ) − + = 0. dt ∂φ2 ∂ φ˙2 −

Under consideration of ∂L(φ1 ,φ2 ,φ˙ 1 ,φ˙ 2 ) = −kl12 φ1 + kφ (φ2 − φ1 ) ∂φ1 i h ∂L(φ1 ,φ2 ,φ˙ 1 ,φ˙ 2 ) ˙ 1 l1 + φ˙ 2 l2 l1 φ = m ∂ φ˙1 ∂L(φ1 ,φ2 ,φ˙ 1 ,φ˙ 2 ) = −kφ (φ2 − φ1 ) ∂φ2 i h ∂L(φ1 ,φ2 ,φ˙ 1 ,φ˙ 2 ) ˙ 1 l1 + φ˙ 2 l2 · l2 + Iφ φ˙ 2 φ = m ∂ φ˙ 2 the Euler-Lagrange equations result in −ml1 (φ¨1 l1 + φ¨2 l2 ) − kl12 φ1 + kφ (φ2 − φ1 ) = 0 −ml2 (φ¨1 l1 + φ¨2 l2 ) − Iθ φ¨2 − kφ (φ2 − φ1 ) = 0. Again, the equations are written in matrix notation. "

ml12 ml1 l2 ml1 l2 ml22 + Iθ

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#"

φ¨1 kl12 + kφ −kφ + −kφ kφ φ¨2 #

"

#"

#

φ1 =0 φ2

4 Setting up the equation of motion

54

or ¨ + Kw = 0 . Mw

In the preceding examples, damping forces were neglected. As will be discussed later on, the treatment of damping is often not straightforward (comp. Sec. 4.3). In general, some assumptions are made that lead to a damping matrix C (comp. Chap. 8). Then the equation of motion reads ¨ + Cw ˙ + Kw = 0 Mw

(4.12)

4.3 Damping models 4.3.1 Linear viscoelastic damping In order to know how typically damping models are applied in engineering applications, we first start by briefly looking at the transition from the theory of elasticity to the theory of viscoelasticity. For this, we consider a spring that perfectly follows Hooke’s law, i.e., force and displacement are dependent in a linear relationship: f (t) = ku(t)

(4.13)

with the proportionality constant k. Under a harmonically varying displacement u(t) = u0 cos(ωt), the force is given by: f (t) = ku0 cos(ωt) = f0 cos(ωt)

(4.14)

and thus, force and displacement oscillate in phase. The force-displacement curve is depicted in Fig. 4.2a. However, in real applications we observe that there is a phase shift between force and displacement, showing that the model of a spring is not sufficiently describing the real behavior. The force-displacement curve results in figures like 4.2b, also known as hysteresis curveRor loop. With the knowledge, that the area within the hysteresis loop is a measure of the work F (t)v(t) dt done by the oscillating force, we can conclude that for the ideal elastic spring no energy is dissipated during a harmonic cycle. As the dissipation of energy (or rather the conversion of mechanical into thermal energy or heat) is inherent to all mechanical phenomena, the elastic model is not sufficient for an accurate description of the physical processes, when it comes to the dynamic responses. A more realistic force-displacement curve also considering non-linearities under cyclic oscillation is given in Fig. 4.2b.

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4 Setting up the equation of motion

55

f

f slightly nonlinear

highly nonlinear u

u

(b) More realistic hysteresis loops, considereing non-linearities, adapted from Nashif et al [1985]

(a) Ideal linear elastic spring, Hooke’s law

Figure 4.2: Force-displacement curves, hysteresis loops

In general, the processes that lead to damping require more complex constitutive models, because usually the governing effects are highly nonlinear. Two categories of damping are commonly defined: external and internal damping. External damping stems from boundary effects and is often associated with friction, such as between structural components. Internal damping on the other hand is induced by the structural material. For internal damping, the damping mechanisms can be linked to internal reconstructions, due to, e.g. magnetic or thermal effects Nashif et al [1985]. An extension of the above mentioned theory of elasticity to cope with such phenomena is the theory of viscoelasticity, which we will introduce briefly. For a linear viscoelastic material that is subject to time-dependent variations of stress and strain, the most general relation between stress and strain is given by the following partial differential equation Snowdon [1968] d d2 dn a0 + a1 + a2 2 + . . . + an n + . . . f = dt dt dt !

d d2 dn = b0 + b1 + b2 2 + . . . + bn n + . . . u. (4.15) dt dt dt !

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4 Setting up the equation of motion

56

This relation is also known as the generalized standard model Nashif et al [1985], which relates forces and displacements or, in an alternative setup, stresses and strains. It holds for arbitrary orders and describes every system consisting of springs and viscous dampers. If we consider all terms up to order n = 1, we obtain the adapted standard linear model: a0 f + a1

df du = b0 u + b1 . dt dt

(4.16)

We now consider a few simple models, that are all derived from Eq. (4.15), where we also give the corresponding form with the spring constant k and the viscous damping coefficient c. • Hooke model. This model corresponds to the purely elastic Hooke’s law with and all ai = bi = 0 for i ≥ 1. a0 f = b 0 u

→

f = ku .

b0 a0

=k

(4.17)

In this case all time derivatives vanish. • Newton model. This model corresponds to the viscous dashpot with a0 , b1 6= 0 and all other ai , bi equal to zero. a0 f = b 1

du dt

→

f = cu˙ .

(4.18)

• Maxwell model. This model corresponds to the a spring and dashpot in serial connection with a0 , a1 , b1 6= 0 and all other ai , bi equal to zero. The forces in the spring and the damper are equal, thus a0 f + a1

du df = b1 dt dt

→

c f + f˙ = cu˙ . k

(4.19)

• Kelvin-Voigt model. This model corresponds to the a spring and dashpot in parallel connection with a0 , b0 , b1 6= 0 and all other ai , bi equal to zero. The displacements in the spring and the damper are equal, thus a0 f = b 0 u + b 1

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du dt

→

f = ku + cu˙ .

(4.20)

4 Setting up the equation of motion

57

c c

k

k

c

k (a) Hooke model

(b) Newton model

(c) Maxwell model

(d) KelvinVoigt model

Figure 4.3: Illustrations of different viscoelastic models derived from the generalized standard model.

We now consider the generalized standard model for the case of harmonic oscillations. To this end, we insert f (t) = fbe iωt and u(t) = ube iωt into Eq. (4.15): d d2 dn a0 + a1 + a2 2 + . . . + an n + . . . fbe iωt = dt dt dt !

d2 dn d = b0 + b1 + b2 2 + . . . + bn n + . . . ube iωt . (4.21) dt dt dt !

We can cancel e iωt on both sides. Furthermore, we can write the n-th derivative of e iωt as dn iωt e = (iω)n e iωt dtn

(4.22)

and thus find for Eq. (4.21): 



a0 + a1 (iω) + a2 (iω)2 + . . . + an (iω)n + . . . fb = 



= b0 + b1 (iω) + b2 (iω)2 + . . . + bn (iω)n + . . . ub. (4.23) Then, the ratio of force and displacement in the frequency domain fb b0 + b1 (iω) + b2 (iω)2 + . . . + bn (iω)n + . . . = = kb = k (1 + iη) . 2 n ub a0 + a1 (iω) + a2 (iω) + . . . + an (iω) + . . .

(4.24)

Here, we made use of the fact, that the ratio of any two arbitrary complex numbers can again be written as a complex number. In general, now k(ω) and η(ω) are frequency dependent values. The relationship between force and displacement is now, analogously to Hooke’s law, interpreted as a complex spring, that represents the complete viscoelastic behavior. This important result shows, that for harmonic oscillations, any connection of spring and damper elements can be represented by a frequency dependent complex spring.

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Example 4.11: Complex spring of Kelvin-Voigt model

Consider again the Kelvin-Voigt model, with the differential equation cu˙ + ku = f . For time harmonic oscillations f (t) = fbe iωt and u(t) = ube iωt , it results c(iω)ube iωt + k ube iωt = fbe iωt . Then, the complex spring kb is found as kb =

cω fb =k 1+i ub k 



.

Thus, the imaginary part η(ω) = cω is a linear function of the frequency, whereas the E real part k is constant throughout the frequency range. It is preempted here, that this is often not found to be a realistic approximation to the behavior observed in engineering materials.

Consider the Newton model depicted in Fig. 4.3b. The governing

The Newton model

equation is f = cu˙ .

(4.25)

Assuming time harmonic oscillation u(t) = u0 sin(ωt), it is u(t) ˙ = u0 ω cos(ωt) and we obtain f = cωu0 cos(ωt) .

(4.26)

The work dissipated in one cycle of oscillation then is Wd =

ZT

f (t)v(t) dt =

0

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ZT 0

cω 2 u20 cos2 (ωt) = cω 2 u20

T = πcωu20 . 2

(4.27)

4 Setting up the equation of motion

59

Furthermore, we derive the hysteresis loop for q the Newton model. To this end, we start with Eq. (4.26) and insert the identity cos(ωt) = 1 − sin2 (ωt): q

f (t) = cωu0 1 − sin2 (ωt) .

(4.28)

Taking the square on both sides: 



f 2 (t) = c2 ω 2 u20 1 − sin2 (ωt) . As sin(ωt) = f (t) cωu0

!2

u(t) , u0

(4.29)

we can write

u(t) =1− u0

!2

(4.30)

which can be further written as f (t) cωu0

!2

u(t) + u0

!2

= 1.

(4.31)

This equation describes an ellipse and represents the hysteresis loop, i.e., the relationship between instantaneous force versus instantaneous displacement, of the single dashpot element. Furthermore, the area within an ellipse, described by (x/a)2 + (y/b)2 = 1 is given by A = πab, thus for the hysteresis loop, we find A = πcωu0 u0 = πcωu20 = Wd .

(4.32)

We note, that the area within the hysteresis loop corresponds to the work dissipated within one cycle of motion (Fig. 4.4a). The Kelvin-Voigt model

The governing equation is

Furthermore, consider the Kelvin-Voigt model depicted in Fig. 4.3d.

f = ku + cu˙ .

(4.33)

Assuming time harmonic oscillation u(t) = u0 cos(ωt + ϕ), it is u(t) ˙ = u0 ω sin(ωt + ϕ) and we obtain f = ku0 cos(ωt + ϕ) + cωω sin(ωt + ϕ).

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(4.34)

4 Setting up the equation of motion

60

f f cωu0

ku0 cωu0 u

u u0

u0

(a) Dashpot, Newton model

(b) nonlinear hysteresis loops, adapted from Nashif et al [1985]

Figure 4.4: Force-displacement curves, hysteresis loops

To find an expression for the work dissipated in one cycle of motion, we separate the force terms and first investigate the work done by the spring force fk (t) = ku0 cos(ωt + ϕ) Wd,k =

ZT

fk (t)v(t) dt =

−kωu20

0

ZT

cos(ωt + ϕu ) sin(ωt + ϕu ) dt = 0 .

(4.35)

0

No energy is dissipated in the spring, which is as expected, since the spring behaves ideally linear elastic. Furthermore the work done by the damping force fd (t) = cωu0 sin(ωt + ϕ) is Wd,d =

ZT

fd (t)v(t) dt =

ZT

0

cω 2 u20 sin2 (ωt + ϕ) = cω 2 u20

0

T = πcωu20 . 2

(4.36)

We obtain the same solution as for the Newton model in Eq. (4.27). Analogously, we derive the hysteresis loop for the Kelvin-Voigt model. To this end, we start with Eq. (4.34) and insert q 2 the identity cos(ωt + ϕ) = 1 − sin (ωt): q

f (t) = ku0 cos(ωt + ϕ) + cωu0 1 − cos(ωt + ϕ) .

(4.37)

Rearranging the equation, and taking the square on both sides leads to f (t) − ku0 cos(ωt + ϕ) cωu0

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!2

= 1 − cos(ωt + ϕ).

(4.38)

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Because cos(ωt + ϕ) = f (t) − ku(t) cωu0

!2

u(t) , u0

we can write

u(t) =1− u0

!2

(4.39)

which can be further written as f (t) − ku(t) cωu0

!2

u(t) + u0

!2

= 1.

(4.40)

This equation again describes an ellipse and represents the hysteresis loop, i.e., the relationship between the instantaneous force versus the instantaneous displacement of the single KelvinVoigt element. The ellipse is now tilted by an angle, which is given in Fig. 4.4b. Furthermore, the area within this ellipse still corresponds to the work dissipated in the damper element during one cycle of motion.

4.3.2 Ideal hysteretic damping For many engineering materials, experiments show, that the amount of dissipated energy under harmonic cycling does not depend on the frequency of oscillation, but rather is a function of the maximum displacement. This motivated the creation of the concept of linear hysteretic damping, often just termed hysteretic damping or structural damping. The term hysteretic damping can be misleading, in the sense that also other forms of damping produce hysteresis loops as was shown for viscous damping in the preceding section. The linear hysteretic damping force is defined in the frequency domain by f˜D (ω) = iηk sgn(ω)˜ u(ω) .

(4.41)

We observe that the damping force is now proportional to the displacement u˜(ω) but in phase with the velocity due to the factor i. It has been found that the linear hysteretic damping model is noncausal (Crandall [1961]), that is, the system responds before it is excited by a force, which leads to non-physical behavior. Nevertheless, for many applications, e.g. steadstate vibration, this error can be accepted. The damping force in the time domain is found by fD (t) = F −1 f˜D (ω) = ηkF −1 {i sgn(ω)˜ u(ω)} = ηkH {u(t)} = ηk ub(t) n

o

(4.42)

where xb(t) = H {u(t)} is the Hilbert transform of u(t), which we will not discuss further in the scope of this lecture. Nevertheless, it can be found that the energy dissipated by the ideal

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4 Setting up the equation of motion

62

linear hysteretic damping element is given by (Kausel [2017]): Wd,h =

ZT

(4.43)

fD (t)v(t) dt = πηku20 .

0

We compare this to the energy dissipated in the viscous damper element (Newton model, Eq. (4.27)), where we insert c = ηmωn and mωn = ωkn , and obtain Wd,v = πcωu20 = πηmωn ωu20 = πη

k ω ωu20 = πηk u20 . ωn ωn

(4.44)

q

k Here, we introduce the natural frequency ωn = m of the single degree of freedom system, which will be introduced in Chap. 5. The amount of energy dissipated in one cycle of harmonic motion differs by the factor ωωn for the hysteretic and the viscous damping element. In the case ωωn = 1, i.e., for resonance, the dissipated energies are equal. This relation is depicted in Fig. 4.5

Wd

viscous hysteretic

πηku20

ω −ωn

ωn −πηku20

Figure 4.5: Dissipated energy for viscous and hysteretic damper element

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4.3.3 Coulomb damping For the Coloumb damping model we use the model of a rigid block on a rigid support, where the sliding friction force in the interface is equal to the maximum static friction force. w,w˙ f (t) |fD | = µ N N µ

N |fD | = µ N

Figure 4.6: Damping due to friction

For this model, the absolute value of the reaction force fD (t) due to sliding friction does not depend on the absolute value of the vibration velocity v(t). The direction of the reaction force fD (t) follows directly from the direction of the velocity v(t) and the direction of the external force f (t) in the steady state is the same as the direction of the velocity. This indicates that a positive work is done on the system. This model is called St. Venant-model. It is sketched in Fig. 4.7, where also the respective hysteresis is depicted.

f w f (t) (a) System

(b) Hysteresis Figure 4.7: St. Venant model

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4.3.4 Aeroelastic damping Using a quadratic relation between force and velocity, the reaction forces in an adjacent fluid (gas or fluid) can be approximated. According to the Bernoulli-equation, the pressure depends quadratically on the flow velocity. D = −αv 2 (t) sgn (v(t))

(4.45)

4.4 Summary • Energy and equilibrium methods enable the determination of the governing differential equations. With the help of the Hamilton principle, the relevant equations can be elegantly stated. As energy quantities are scalar values the signs in the governing equations result from the derivation, which is less error prone and thus advantageous. Furthermore, additional constraints can be taken into account using Lagrange multipliers. • If forces have a potential, they can be derived from it. Therein, potential means, that the potential energy of the system can be uniquely determined from the current deformation state of the system. If all forces have a potential, under the deformation and subsequent reverse deformation of the system no energy is lost. • For damping and frictional forces as well as all external forces no potential exists. The Hamilton principle can then be extended for these non-conservative forces.

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5 Free vibration of linear systems with single degree of freedom In the following we will look at mechanical systems with one degree of freedom, also termed single degree of freedom (SDOF) systems. These systems are characterized by three coefficients describing the inertia force fI , the elastic restoring force fK and the damping force fC : m . . . mass k . . . stiffness c . . . damping coefficient (here, viscous damping) We assume, that the equation of motion is found by any of the methods in section 4.1, e.g. the equilibrium of forces, repeated here for convenience. All forces acting on a moving mass point in the free state (no external force) at equilibrium are considered, namely spring, inertia, and damping forces. The equation of motion is

mw¨ m k

w c

m kw

cw˙

Figure 5.1: Free body cut for the unforced (free) SDOF system

mw(t) ¨ + cw(t) ˙ + kw(t) = 0 .

(5.1)

In the following the free vibration of the SDOF-system shall be studied.

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5.1 The undamped case For a free vibration without damping, the equation of motion simplifies to mw(t) ¨ + kw(t) = 0 .

(5.2)

As no forcing term is present, no particular solution is needed to satisfy the differential equation and the homogeneous solution describes the complete solution w(t) = wh (t). This homogeneous differential equation can be solved using the following equivalent approaches. w(t) = wb+ eiωt + wb− e−iωt = w01 cos(ωt) − w02 sin(ωt) = w0 cos(ωt + ϕw ) .

(5.3)

We decide to apply the latter approach, using a single cosine function with phase shift. We now derive the second derivative with respect to time w(t). ¨ w(t) ˙ = −w0 ω sin(ωt + ϕw )

(5.4)

w(t) ¨ = −w0 ω 2 cos(ωt + ϕw ) .

(5.5)

After inserting (5.4), (5.2) can be expressed as 



−mω 2 + k w0 cos(ωt + ϕw ) = 0 .

(5.6)

As in general w0 cos(ωt + ϕw ) 6= 0, the above equation is satisfied, if −mω 2 + k = 0

(5.7)

which results in s

ω = ωn =

k = 2πfn . m

(5.8)

ωn constitutes the natural (circular) frequency or eigenfrequency of the system. The solution is depicted in Fig. 5.2 over time and as rotation in the plane, as was introduced in Chap. 2. 1 2π = fn ωn w02 tan ϕw = w01

T =

w0 =

q

2 2 w01 + w02

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(natural) period of the motion phase angle amplitude of vibration .

5 Free vibration of linear systems with single degree of freedom

67

w

w

w0

ϕ

π 2

∗

3 π 2

π

2π ωt

∗

ωt

ωt

w = w0 cos(ωt + ϕ) ϕ ωT Figure 5.2: Homogeneous solution of SDOF-system

w0 and ϕ are determined to satisfy the initial conditions w(0) = w0 and w(0) ˙ = v0 , respectively. The solution can then be given (without derivation): w (t) = w(t = 0) cos (ωn t) +

v(t = 0) sin (ωn t) . ω

(5.9)

5.2 The damped case 5.2.1 Velocity proportional (viscous) damping To solve the equation of motion considering velocity proportional damping mw(t) ¨ + cw(t) ˙ + kw(t) = 0

(5.10)

the approach for a linear homogeneous differential equation with constant coefficients b λt w(t) = we

(5.11)

is used. b λt ⇒ w(t) ˙ = v(t) = λwe

(5.12)

b λt ⇒ w(t) ¨ = a(t) = λ2 we

(5.13)

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If (5.12) and (5.13) are introduced into (5.10) as b λt = 0 (mλ2 + cλ + k)we

(5.14)

the characteristic polynom λ2 +

k c λ+ =0 m m

(5.15)

with the non-trivial solution λ1,2

c =− ± 2m

s 

c 2m

2

c ± =− 2m

s 

c 2m

2

c =− ± ωn 2m

s

−

k m

− ωn2

c 2mωn

2

−1

(5.16)

is obtained. We introduce c 2m ccrit = 2mωn c D= (= ξ) ccrit δ=

decay constant, critical damping, percentage of critical damping.

Thus, Eq. (5.16) can be written as λ1,2 = −δ ± ωn

s

c

2

ccrit

−1

√ = −δ ± iωn 1 − D2

(5.17)

Depending on the value D, this equation has different types of solutions. D < 1,

complex root:

√ λ1,2 = −δ ± iωn 1 − D2 = −δ ± iωD

(5.18)

D = 1,

root = 0 :

λ1,2 = −δ

(5.19)

D > 1,

real root:

λ1,2 = −δ ± ωn D2 − 1 = −δ ± ω ∗

√

(5.20)

The solution for the three cases is given subsequently.

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5.2.1.1 Quasi periodic case (D < 1)

After inserting Eq. (5.18) into the initial approach in Eq. (5.11) the solution is obtained as a sum of conjugate complex solutions with conjugate complex coefficients. We denote the respective amplitudes by w+ and w− . w(t) = wb+ e (−δ+iωD )t + wb− e (−δ−iωD )t 

= e −δt wb+ eiωD t + wb− e−iωD t



(5.21)

= e −δt (wb+ (cos(ωD t) + i sin(ωD t)) + wb− (cos(ωD t) − i sin(ωD t))) = e −δt ((wb+ + wb− ) cos(ωD t) + i (wb+ − wb− ) sin(ωD t)) |

{z

}

w01 =2 Re(w b+ )

|

{z

}

iw02 =2i Im(w b+ )

= e −δt (w01 cos(ωD t) − w02 sin(ωD t))

(5.22)

or w(t) = w0 e −δt cos(ωD t + ϕ) with √ ωD = 1 − D2

s

k m

1 TD = √ f 1 − D2 w2 tan ϕw = w1 w0 =

q

2 2 w01 + w02

damped natural (circular) frequency (natural) period of the motion phase angle initial amplitude at the beginning of the vibration .

w0 and ϕw , or w01 and w02 , or wb+ and wb+ are determined such that the solution satisfies the initial conditions w(t = 0) = w0 and w(t ˙ = 0) = v0 . The resulting formulation in form of Eq. (5.22) is given without further derivation as "

w (t) = e

−δt

v0 + δw0 w0 cos (ωD t) + sin (ωD t) . ωD #

(5.23)

The resulting vibration is depicted exemplarily in Fig. 2.4 for the case w0 = 1 and v0 = 0.

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Im(2w+ )

w 1 − e−δT

ωt

Re(2w+ )

π 2

3 π 2

π

2π ωD t

e−δt 1 − e−δT Figure 5.3: Damped oscillation over one period T with w0 = 1 and ϕ = 0.

In this App the time response of a damped single degree of freedom system is visualized. The mass m, stiffness k, damping constant c and initial velocity v0 are parameters which can be changed and the initial displacement w0 = 0. By starting the oscillation, the system oscillates about the static equilibrium position. http://apps.bm.bgu.tum.de:5006/Damped_oscillator

From the definition of the damped natural frequency, we find (from Chopra [1995]) ωD =

√

ωD ω

1 − D2 ω

ωD = (1 − D2 )ω 2 

ωD ωn

2

1



 ωD 2 ω

+ D2 = 1

+ D2 = 1 .

D 1 For small values√of damping D < 0.2, the ratio of damped and undamped natural frequency is close to one ( 1 − 0.22 ≈ 0.98). 5.2.1.2 Aperiodic limit case (D = 1)

The case D = ωδn is the so-called critical damping case, for which c = ccrit = 2mωn . This is the limit above which the system does not oscillate but decays in an aperiodic movement. As the two eigenvalues of the problem in Eq. (5.19), λ1,2 = −δ, coincide, the solution approach

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5 Free vibration of linear systems with single degree of freedom

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has to be altered. In general, the solution for the SDOF with eigenvalues of multiplicity 2 is given as w(t) = w1 e −δt + w2 te −δt = e−δt (w1 + w2 t) .

(5.24)

The constant coefficients w1 and w2 are determined to fulfill the initial conditions w(0) = w0 and w(0) ˙ = v0 . If the default parameters are used, i.e. the mass m = 8 kg and the stiffness N k = 50 m , then ωn = 2,5 rad and ccrit = 40 Ns . Thus the aperiodic limit case s m . can be visualized when the damping constant is set to c = ccrit = 40 Ns m http://www.bm.bgu.tum.de/lehre/interactive-apps/ damped-oscillator/

5.2.1.3 Aperiodic case (D > 1)

For the aperiodic case with D > 1, the solution reads w(t) = w1 e (−δ+ω

∗ )t

+ w2 e (−δ−ω

∗ )t

∗

= w1 e −δt e ω t + w2 e −δt e −ω ∗



= e−δt w1 e ω t + w2 e −ω

∗t

∗t



(5.25)

with ω∗ =

√

s

D2 − 1

k . m

The constant coefficients w1 and w2 are determined to fulfill the initial conditions w(0) = w0 and w(0) ˙ = v0 . The solution is exemplarily depicted in Fig. 5.4 for the aperiodic and aperiodic limit case.

5.2.2 Relations between different damping measures The values describing the damping of a system can be expressed via a series of measures. Tab. 5.1 provides an overview of commonly used values and their relations.

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v0 > 0

w (t)

v0 = 0 v0 = −w0 (δ + ω ∗ ) w0 t

v0 < −w0 (δ + ω ) ∗

Figure 5.4: Homogeneous solution of the SDOF system for the aperiodic case D > 1. The solution is depicted for different initial conditions. For all these cases, no oscillatory is present.

Table 5.1: Overview over the commonly used damping values and their relation.

mechanical resistance

decay constant

c

δ

percentage of critical damping

c

=

c

2mδ

D √ 2D km

δ

=

δ

ωn D

D

=

c 2m

√c 2 km

δ ωn

D

Λ

=

√ 2πc 4km−c2

η

=

√c km

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√ 2πδ 2

ωn −δ 2 δ πf0

logarithmic decrement Λ q

2Λ

km 4π 2 +Λ2

loss factor η √ η km η ω 2 n

√ ωn Λ 4π 2 +Λ2 √

η 2

Λ 4π 2 +Λ2

√2πD 1−D2

Λ

2D

√ 2Λ 4π 2 +Λ2

√

πη 1−(η/2)2

η

5 Free vibration of linear systems with single degree of freedom

73

6 Classification of Excitations It is of benefit to group the various excitations in order to be able to find favorable solution strategies for each of those problems. Hereafter, types of excitation that are treated in structural dynamics problems are introduced.

6.1 Modelling of the excitation In structural dynamics, excitations from time varying forces or displacements are causal for occurring motions. Nevertheless, the excitations—depending on the the considered system that is “cut free” from the physical environment—are not always modeled as forces. Relevant excitations in structural dynamics from earthquakes, traffic induced vibrations, vibrations from industrial facilities in the neighborhood, building work or detonations, are usually applied via root-point excitation, i.e., displacements (comp. Sec. 7.2.2). Thereby, the causal (force) excitation, e.g., induced by dynamic forces (emission) between a moving train wheel and the track including the transmission into the soil and to the foundation, is described indirectly via the motion of the soil, e.g. at the foundation, by a displacement excitation. The concept of impedances, with which also large substructures can be described and possibly measured, thereby helps for the consideration of the coupling (e.g., Tab. 10.1). Nevertheless, often the excitation is directly modeled by dynamic forces, e.g. resulting from the operation of fixed or moving machinery with direct influence on the structure, from building work, traffic, wind, shock waves, from explosions, or from moving persons. Vibrations, that result from time varying or, for moving loads, spatially varying parameters (e.g., a periodically varying stiffness under a train wheel on and between the railway sleepers), are termed as parametric excitations, as they depend on the systems’ parameters (especially the stiffness) of the excited structure. They are either described by differential equations with time dependent stiffness and/or damping values (solution in the time domain), indirectly by substitute forces, obtained from simple models, or by time varying systems, for calculations in the time domain. Furthermore, the temporal structure of the excitation influences the modeling. Depending on it, it is distinguished between transient loads (e.g. explosion loads, falling weights, impact loads, jumping persons), periodic loads (e.g. machines with rotating parts, effect of unroundness of railway wheels, loads from moving persons, stick-slip vibrations, which appear,

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e.g., for friction processes, pile shaking, dynamic building ground compression, rotating rollers, etc.) and stochastic loads (e.g. wind loads, earthquake loads, traffic induced vibrations, loads from mills and crushers). A non-periodic, transient load, can be replaced by a periodic load with an infinite period T , moving from the Fourier series to the Fourier integral (comp. Chap. 3). However, due to the inevitable discretization in the time and frequency representation, an infinite period is not suitable. For transient signals described in the frequency domain, the computational temporal repetition length T can be chosen such that the response between the repeating transient events has decayed. Then, discrete spectra may be applied. f

t Figure 6.1: Transient, triangular load

f

t T →∞

t=0

T →∞

Figure 6.2: Representation of an aperiodic signal using an infinite return period

For linear time invariant system—or system which allow linearization around a static initial state—the advantage of a representation in the frequency domain can be made use of. The periodic excitation can the be represented by a sum of harmonic loads. The description of the vibration response to stochastic loads is done for a few problems, e.g., for earthquake engineering, by means of response spectra. (comp. Sec. 7.5.1.1). They process the stochastic excitation for engineering applications, such that approximate deterministic calculations can be carried out. Independently, treatment of the stochastic processes can be done with the help of the probability theory (comp. Chap. 13) In VDI 2038 [2012] the most important dynamic excitations for buildings, components and other plant components are classified. Subsequently, we give a short overview over the different categories of temporal structures.

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6.1.1 Harmonic excitation Within the model of a harmonic excitation it is assumed that the load has been acting on the system since infinite time. This requirement can be relaxed by requiring the time since the start of the excitation to be much larger than the eigenperiod of the system T , so that Tt  1. Then it can be assumed, that the system is in the steady state vibration. This means, that the transient response, which is present after the beginning of the excitation, has sufficiently decayed. Examples for harmonic excitations are: • Machines with rotating components • Paper factories (large rolls) • Rotating machinery • Wind (to certain extent) A harmonic load f (t) can be given as f (t) = f0 cos(Ωt + ϕ), with amplitude of load in N

f0 Ω = 2πf = f=

Ω 2π

T =

1 f

2π T

circular frequency (of the load) in rad s frequency (of the load) in Hz = s1

=

2π Ω

period of the motion in s phase shift angle in rad

ϕ

6.1.2 Periodic excitation In the model representation of a periodic excitation we also assume that the load has been acting on the system since infinite time. This type of excitation is, however, not characterized through a single excitation frequency, but a superposition of a number of harmonic excitations. Thus, harmonic excitations can be seen as a special case of a periodic excitation. The Calculation of the individual harmonics carried out by a Fourier transformation as explained in Chap. 3. Examples for periodic excitations are: • Rotating machinery • Propellers • Surface irregularities of wheels of trains or out-of-roundness of truck tires • Pedestrians

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• Stick–slip vibrations • Sheet pile shaking, dynamic building ground compression Example 6.1: Decomposition of periodic load into harmonics

Figure 6.3 shows a load record and its decomposition in individual harmonic signals. f (t) f= T

T = f1 t

f1 sin (Ω0 t + ϕ1 )

t

f2 sin (2Ω0 t + ϕ2 )

t

f3 sin (3Ω0 t + ϕ3 )

T + T 2

T 2

+ T 3

f2

f3

T T 3 3

+ ... Figure 6.3: Periodic signal and harmonic decomposition

Example 6.2: Short-time Fourier transformation

In the below figure, a short-time Fourier transformation of the acceleration at the axlebearing of an ICE is depicted. The single lines in the colored representation show the excitation frequencies. In this illustration a line pattern with distinct excitation frequencies becomes apparent (e.g. for around 60, 80, 100, 120 Hz). This excitation stems from the unroundness of the wheels. Periodic loads then act on the ground as

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Time [s]

well as the vehicle.

Frequency [Hz]

6.1.3 Transient load Transient loads are only present in the system during a limited time window. Examples for transient excitations are: • Explosions • Shock or crash • Impact • Hammers • Jumping Fig. 6.4 shows a triangular load signal over time as an example for a transient excitation. Aperiodic transient loads can be substituted by periodic loads with an infinite period. This is linked to switching from the Fourier-series to the Fourier-integral (comp. Chap. 3).

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f

t Figure 6.4: Transient, triangular load

f

t T →∞

t=0

T →∞

Figure 6.5: Representation of an aperiodic signal using an infinite return period

6.1.4 Irregular, random loads A great number of loading processes show neither periodic parts nor can they be described through a transient time function exactly. The randomness inherent in the excitation requires a different description. Examples for irregular, random excitations are: • Wind (only to certain extent) • Train-induced vibration in the soil (to certain extent) • Vibrations in vehicles through irregularities on the ground • Earthquakes

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79

Figure 6.6: Illustration of a stochastic process

Figure 6.7: Illustration of the autocorrelation function

This type of excitation can be characterized by a probabilistic description (comp. Chap. 13). Typically the load is described through the autocorrelation function, its power density through the Power Spectral Density. Here, it is of great importance to to identify and describe the significant characteristics of the excitation as basis for suitable methods for the acquisition (measurement) as well as for the description of the excitation function.

6.1.5 Loads influenced by the parameters of adjacent systems Some excitation processes result from the interaction of two systems, e.g. the load of vehicles, that drive over a (flexible) bridge or over a (flexible) railway track. Some load components can be neglected, if e.g. the flexibility of the bridge or the railway would be infinitely small, i.e. the stiffness infinitely large. Vibration processes resulting from this type of coupling are also called parametric vibrations, because they depend on the parameters of the adjoining coupled systems. Examples for loads associated with interaction are: • Vehicles on a bridge or railway track • Pedestrians that synchronize with the structural vibration

Figure 6.8: Synchronization of pedestrians with the structural vibration, courtesy of M. Schneider

• Fluid-Structure-Interaction

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6.2 Frequency content The signals that typically occur in engineering problems can have various frequency characteristics. Fig. 6.9 depicts a typical scenario. A train induces oscillation in the ground and the surrounding air. Through wave motion the disturbances radiate from the track as airborne and solid waves. Vibrations are induced into adjacent structures, like the building depicted. Linked with these vibrations sound is radiated, termed reradiated sound. Sound waves that directly cause vibrations inside the house cause primary sound. To summarize, multiple paths occur. Depending on the frequency range of the occurring signals, the oscillations are termed differently, depending on the specific background (e.g. structure-borne sound, vibrations).

z y

sound waves z

Reradiated sound

x, y, z

Vibrations

Waves in solids Emission Vibrations 1 2

Transmission

Immission

Structure borne sound

5 10 20 5080100 Frequency in Hz

1000

Figure 6.9: Frequency content characterization in a emission–transmission–imission system.

6.3 Methods The choice of methods for structural dynamic or vibro-acoustic predictions depends on the specific problem setting. Fig. 6.10 gives an classification of commonly used methods.

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hybrid approaches

analytical algorithms unlimited systems limited simple boundary conditions

Mu osc ltib illa ody tor

M BE M FE

A SE

high frequency

complex

low

Figure 6.10: Common methods in strucutral dynamics and vibro-acoustics and their applicability.

6.4 Classification in standards The dynamic load through machinery with rotating masses (engines, fans, turbines, vibrators, centrifuges, generators, rotary pumps, etc.) can be well represented well in the frequency domain with respect to their excitation frequency that depends on the rotational speed. To some extent, from the moving masses the excitation forces can directly be concluded (e.g., DIN [2004a]). If this is not possible, the excitation forces can be obtained via measurements at the machine foundations, together with impedance (comp. Chap. 10) measurements at the support points. Hints are given in DIN 4024-1 [1988]; DIN 4024-2 [1991]. The indirect evaluation of the forces by measurements is described in VDI 2038 [2012]. Machinery with oscillating masses (piston engines and pumps, saws, but also bells) are partly well describable with respect to the induced, periodic forces (e.g., DIN 4178 [2005]). The measurement based determination of forces is, e.g., described in DIN [2004b]; VDI 2038 [2012]. The excitation is usually described in the frequency domain, whereby also natural vibrations (homogeneous solutions) at impact processes, e.g., for fast (emergency) switch off, have to be considered. Machinery with non-periodically moving masses (forging hammers, presses, jacking systems, packing machines, placement machines, grinding machines, milling machines) lead to transient excitations, that can be either represented in the time or frequency domain. Especially for hammer and falling weights, the excitation can be deduced from simple mechanical models considering the masses and geometries. For machinery with complex inner movement processes usually measurements are necessary.

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In the context of traffic induced loads the investigations, carried out for railway induced excitations, are gathered in simplified substitute models (see e.g. Thompson [2008]; Wettschureck et al [2012]). Here, usually measurements are carried out, but are interpreted in knowledge of the mechanical phenomena—from the parametric excitation resulting from the periodic changes of the track stiffness due to discrete supports of the tracks by sleepers, over the displacement induced periodic excitation from the unroundness of the wheels, the stochastic excitation from the roughness of the track surface and the track geometry, the excitation from vibrations of the vehicle and the track, up to the excitation from the moved self weight (z.B. Wettschureck et al [2012]; Müller [2011]). Remarks for road traffic induced vibrations are given, e.g in Melke [1995]; VDI 2038 [2012]. Pedestrian induced excitations can be modeled by periodic loads. An overview can be found in Butz and Distl [2008]. In Sahnaci [2013]; VDI 2038 [2012] a comprehensive overview over the state of the art is given. Dynamic excitations for building work and detonations include impact-like and periodic processes, for which an overview with mechanical models can be found in VDI 2038 [2012]; Müller-Boruttau and Breitsamter [2000]. Dynamic wind excitations range from stochastic broadband loads from turbulences (gust induced vibrations) over periodic loads from vortex excitation to self-excited vibrations. For and overview it is referred to, e.g., Peil and Clobes [2008]; Ruscheweyh [1982]. The single load types are normatively edited in DIN EN 1991-1-4:2010-129 [2010]; ISO 4354 [2009]. For the modeling of the highly dynamic, transient excitations in case of shock loads and earthquake loads, it is referred to the corresponding contributions in this volume. In the following table, the most important dynamic excitations for buildings, components and other plant components are summarized, that are of of relevance in practical applications (from VDI 2038:2010-11):

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1

2

3

4

5

6

Excitation Operational • Machinery with rotating masses • Machinery with oscillating masses • from fluids in components • transport in buildings, floor conveyors • bell ringing Traffic-induced • railway traffic • road traffic • ship traffic • air traffic Pedestrian-induced • Walking, running • jumping, dancing • seesawing building work • shaking, compressing • vibratory driving • impact driving • chiselling work • blasts • demolition, load crash • work traffic Wind-induced • gusts • vortex shedding • galloping (self-excited vibration) • flutter, interference Incidents, special loads • earthquakes • blast waves • impact from airplanes / flying object • impact from ships / vehicles • explosions of highly energetic tanks • machinery accident

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Remark

Code

periodic periodic periodic - pressure oscillations Transient - shocks transient periodic

DIN 4024/ VDI 3838 VDI 3842

transient transient periodic - under water sound periodic – low frequency sound

DIN 4150-2 DIN 4150-2

periodic - pedestrian bridges periodic - ceilings, corridors periodic – stands periodic/ impacts periodic impacts / periodic impacts / periodic impacts impacts transient / impacts stochastic periodic amplifying vibration amplifying vibration transient

DIN 4178

ISO 10137

DIN 4150

DIN 1055-4 ISO 4354

DIN 4149

6 Classification of Excitations

84

railway induced vibrations description by

force excitation

vehicle vibration

interaction excitation

vehicle-track interaction wheel motion from sleeper vibrations in buildings on the basis of measurements in the free field

base point excitation harmonic time function periodic transient

wheel roughness moving load model (axle distance, wagon length) track impacts

stochastic

track unroundness

point source

vibrations in the near field vibrations in the far field

spatial funct. line sources

pedestrians

machinery

wind

constr. work

area source stochastic distribution

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spread of the qualitiy of wheels/bogies

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85

7 Forced vibration of linear systems with single degree of freedom 7.1 Introduction 7.1.1 “Quasi-static” and dynamic processes The system response to dynamic loads depends on the temporal structure of the load process. Very slow processes can be treated as “quasi-static” like “infinitely” slowly applied loads. In this case, the forces caused by the temporal changes of the processes (comp. Chap. 10, inertia and damping forces, but not forces from sliding friction), are small compared to the static reaction forces of the structure and can be neglected. For “quickly” preceding processes, the forces caused by the temporal changes in those processes have to be considered. The distinction between “quasi-static” and dynamic processes depends on the ratio between the displacement dependent reaction forces and the inertia, damping, and friction forces caused by the temporal changes. As shown in Chap. 10, the lowest eigenfrequencies of a structure provides insight about the ratio of inertia and damping forces related to the static reaction forces under time varying loads. Whereas for very soft and heavy structures, i.e., for structures with comparably low displacement dependent reaction forces and high inertia (i.e., very small lowest eigenfrequency) also relatively slow excitations, e.g., slow walking of a person, already have to modeled as dynamic loading, for stiffer and lighter structures (e.g. a steel girder with short span width and comparably high lowest eigenfrequency) it is sufficient, to consider the same load as “quasi-static”. Thus, for an engineering evaluation of the influence of the temporal structure on the vibration response, the temporal structure of the load has to be assessed considering the dynamic properties of the system. This is usually done via a representation in the frequency domain (spectral representation). If a load at rest is applied very fast and “abruptly”, larger inertia forces occur compared to a slower, “softer” loading. Here the time interval, within which the load is applied, plays a significant role. This can be either considered by a spectral approach (comp. Chap. 3) or by a representation of the response vibration with respect to the ratio of the characteristic time intervals of the loading and the eigenperiods of the system, obtained from the eigenfrequencies (comp. Sec. 7.5.1.1; response spectra: Figs. 7.21 and 7.24).

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7.1.2 Time and frequency representation Every representation of the excitation as a function of time, i.e., in the time domain, can, without loss of information, be transformed in an equivalent representation in the frequency domain (spectral representation). The representation in the frequency domain simplifies the solutions of the underlying partial differential equations considerably. Mathematically, this corresponds with solutions of the equations of motion obtained by sine and cosine approaches. In many practical cases it is sufficient, to limit the evaluation of the structural response to the frequency domain (to spectra). Every part of the spectrum can thereby be described as a harmonic function (by a sine or cosine function over time with the radial frequency Ω = 2πf ). In the frequency domain, the temporal structure of the excitation for dynamic problems with respect to the expected phenomena can usually be characterized more clearly in a mechanical and physical sense. Thus, for example, resonance effects can be immediately analyzed and the sensitivity of the result to changes in the load and system’s parameters can be narrowed down. However, for the determination of the time function, e.g., for assessing the maximum displacement over time, a transformation into the time domain is necessary, i.e., a superposition of the single spectral parts. For a consideration in the time domain, the state variables (displacements, velocities, strains, stresses) are directly represented as functions over time. Maximum state variables are thereby directly obtained. Material laws and kinematic relations describe the relations between stresses and strains and displacement and strains, respectively, whereby the state variables are considered as time dependent. If the material and/or the kinematic is described by non-linear relations (e.g., non-linear stress-strain relationship or theory of 2-nd order), the linear superposition of the responses to different load parts is not valid and the principle of superposition cannot be applied. In the case of non-linear systems, therefore, a treatment in the time domain is necessary. If the influence of the non-linearities can be neglected with respect to the vibration response, e.g., due to very small vibration values, superposing the static initial state, a linearization is possible. Then the principle of superposition can still be applied and the solution of the dynamic process can be determined in the frequency domain, after the static initial state has been described by non-linear analysis. The solution is then finally given by the superposition of the single spectral parts. As well as representations in the time as well as the frequency domain have great significance and therefore considered parallelly.

7.2 Harmonic excitation for systems with viscous damping In this Section we will deduce the solutions for harmonically forced SDOF-systems with viscous damping. Two cases are of particular interest here, namely the SDOF-system with external,

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harmonic loading, and the base point excited SDOF-system.

7.2.1 Harmonic forced excitation In the following we investigate the SDOF-system under a harmonically varying load as a basis for the representation in the frequency domain. f (t) m k

m w¨ w

c

f01 cos Ωt − f02 sin Ωt m

kw

c w˙

Figure 7.1: Single degree of freedom system with harmonic load

For the depicted harmonically excited SDOF-system the equation of motion is given by (comp. Sec. 4.1) mw¨ + cw˙ + kw = f (t) = f01 cos (Ωt) − f02 sin (Ωt) = f0 cos(Ωt + ϕf ).

(7.1)

This linear, inhomogeneous differential equation is solved by a combination of the homogeneous solution wh (t), which is introduced in chapter 5, and a particular (inhomogeneous) solution wp (t) as w (t) = wh (t) + wp (t)

(7.2)

As introduced in chapter 2, the description of harmonic oscillations can be done in various ways. We choose to replace the right hand side of the differential equation (7.1) by a complex representation, as f (t) = f01 cos (Ωt) − f02 sin (Ωt) 1 1 = (f01 + if02 ) e iΩt + (f01 − if02 ) e −iΩt {z } {z } |2 |2 fb+

= fb+ e iΩt + fb− e −iΩt

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fb−

(7.3)

7 Forced vibration of linear systems with single degree of freedom

88

A detailed derivation of the above was given in section 2.2.1. The resulting excitation is now a sum of two complex terms fb+ eiΩt and fb− e−iΩt . These are complex conjugate as are their amplitudes, and have the same absolute value of the frequency with opposite signs. The absolute value of the amplitude of the force f (t), f0 , therefore results in f0 =

q

2 f01

+

r

2 f02



= 2 Re fb+

2



+ Im fb+

2

(7.4)

7.2.1.1 Steady state solution

For the given linear SDOF-system, a harmonic excitation leads to harmonic particular solution, given as wp (t) = wbp+ e iΩt + wbp− e −iΩt

(7.5)

= wp01 cos (Ωt) − wp02 sin (Ωt) = wp0 cos(Ωt + ϕw )

(7.6)

where 1 (wp01 + iwp02 ) = 2 1 = (wp01 − iwp02 ) = 2

wp+ = wp−

1 wp0 e iϕw 2 1 wp0 e −iϕw 2

(7.7) (7.8)

We continue the derivation with the complex representation of wp (t) in Eq. (7.5), as it simplifies the calculations. The first two derivatives of (7.5) with respect to t are given by dwp (t) = iΩwbp+ e iΩt − iΩwbp− e −iΩt dt 2 d wp (t) w¨p (t) = = −Ω2 wbp+ e iΩt − Ω2 wbp− e −iΩt . 2 dt w˙ p (t) =

(7.9) (7.10)

If we now insert approach (7.5) together with (7.9) and (7.10) into the equation of motion (7.1) and separate the equation for the parts with e iΩt and e −iΩt , we obtain (−mΩ2 + icΩ + k)wbp+ e iΩt = fb+ e iΩt

(7.11)

(−mΩ2 − icΩ + k)wbp− e −iΩt = fb− e −iΩt

(7.12)

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After canceling the exponential on both sides of Eqs. (7.11) and (7.12), we can recast the formulation to wbp+ = Ub (+Ω)fb+ = Ub+ fb+

(7.13)

wbp− = Ub (−Ω)fb− = Ub− fb−

(7.14)

with U (Ω) the spectral transfer function (dynamic flexibility) Ub (Ω) =

1 . k − mΩ2 + icΩ

(7.15)

The inverse of the transfer function i−1

kˆ+ = Ub (Ω) h

= (k − mΩ2 + icΩ)

(7.16)

and i−1

kˆ− = Ub (−Ω) h

= (k − mΩ2 − icΩ)

(7.17)

∗ ∗ as well as Ub (−Ω) = Ub (Ω) are pairs of are called dynamic stiffnesses. Note that kˆ− = kˆ+ complex conjugate values. Furthermore, it holds ∗ wbp− = wbp+ .

(7.18)

The solution in the complex plane is depicted in Fig. 7.2. We further continue with 1 fˆ+ k − mΩ2 + ic Ω k − mΩ2 − ic Ω (f01 + i f02 ) = (k − mΩ2 )2 + (c Ω)2 2

wbp+ =

=

[(k − mΩ2 )f01 + c Ωf02 ] + i [(k − mΩ2 ) f02 − c Ωf01 ] . 2 [(k − mΩ2 )2 + (c Ω)2 ]

With this, we can then isolate real and imaginary part Re (wbp+ ) =

1 (k − mΩ2 )f01 + c Ωf02 2 [(k − mΩ2 )2 + (c Ω)2 ]

(7.19)

Im (wbp+ ) =

1 (k − mΩ2 ) f02 − c Ωf01 2 [(k − mΩ2 )2 + (c Ω)2 ]

(7.20)

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Im

2wp+ 2wp+ e iΩt wp+ wp+ e iΩt Ωt

2 Re(wp+ e iΩt )

Re

ϕw −ϕw

wp+ + wp− = 2 Re(wp+ )

−Ωt wp− e −iΩt

wp−

2wp− e −iΩt 2wp−

Figure 7.2: Displacement solution wp (t) in the complex plane. The solution wp (t) is the sum of two complex conjugate values wp+ e iΩt and wp− e −iΩt . At the time t = 0, the exponential function reduces to one, and the solution is solely the sum wp (t = 0) = wp+ + wp− . As time progresses the two complex pointers get rotated in the complex plane by the angle Ωt and −Ωt, respectively. The solution is always real, as the imaginary parts of wp+ e iΩt and wp− e −iΩt cancel out for all t.

Correspondingly, for wbp− we obtain 1 fb− k − mΩ2 − ic Ω k − mΩ2 + ic Ω (f01 − i f02 ) = (k − mΩ2 )2 + (c Ω)2 2

wbp− =

=

(k − mΩ2 )f01 − c Ωf02 − i((k − mΩ2 )f02 − c Ωf01 ) , 2 [(k − mΩ2 )2 + (c Ω)2 ]

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As the solution wp (t) - just like the force f (t) - consists of two terms that are complex conjugate, it finally results in (physically correct) real values. Using Eq. (2.13) and Eq. (2.14), we can decompose the absolute value and argument of |wbp+ | and |wbp− | into force and transfer function contributions





(7.21)







(7.22)

|wbp+ | = Ub+ fb+ |wbp− | = Ub− fb−















Because Ub+ = Ub− and fb+ = fb− , also |wbp+ | = |wbp− | holds. The corresponding amplitude wp0 in the representation of Eq. (7.6) reads

wp0 =

q

q

2 2 = 2 Re(wbp+ )2 + Im(wbp+ )2 = 2|wbp+ | + wp02 wp01

(7.23)

Furthermore, the response phase is found by ϕw = arg wbp+ = arg Ub+ + arg fb+ = ϕU+ + ϕf

(7.24)

The response wp (t) thus shows a phaseshift ∆ϕ = ϕU+ = ϕf − ϕw with respect to the force f (t). In case of a positive phase shift the response follows behind the force. Using Eqs. (7.19) and (7.19), absolute value and argument of the transfer function read b U+



1



= Ub− = q

(7.25)

(k − mΩ2 )2 + (cΩ)2





tan (∆ϕ) = tan ϕU+ =

Im Ub+









Re Ub+

=

−cΩ k − mΩ2

(7.26)

Thus, the amplitude of the response (Eq. (7.23), considering Eq. (7.21)) can be written as: wp0 =

2|wbp+ |

=

2 Ub+ fb+

=

  b 1 2 U+ f0

2

f0 =q (k − mΩ2 )2 + (cΩ)2

(7.27)

The transfer function is illustrated in the complex plane as shown in Fig. 7.4. In practice, the magnitude of the response is usually given by the frequency-dependent and real amplification function V (Ω, ωn ) = V (η) = k|U (Ω)|. V (η) = q

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k (k − mΩ2 )2 + (cΩ)2

= s

1 1−



Ω ωn

2  2

=q +

( 2D Ω)2 ωn

1 (1 − η 2 )2 + (2Dη)2

(7.28)

7 Forced vibration of linear systems with single degree of freedom

92

with the frequency ratio η = ωΩn . The amplification function gives the ratio between the maximum amplitude and the static displacement wstat = fk0 . The phase-shift ∆ϕ describes by how much the response of the system lags behind the force. For ∆ϕ = 0 the response of the system is in phase with the load. For ∆ϕ < 0 the oscillation reaches its maximum (or minimum, zero crossing) at a later time instance relative to the load. For ∆ϕ = −π = −180◦ the response and the excitation oscillate in counterphase. |V | 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0

D=0 D = 0.2 D = 0.4 Vmax =

∆ϕ −π

√1 2 D 1−D2

D=0 D = 0.2 D = 0.4

− π2 0.5

1.0 √ 1.5 2.0 ηmax = 1 − 2 D2

2.5

0.0

3.0 η

0.5

1.0

1.5

2.0

2.5

3.0 η

Figure 7.3: Amplification function for force excitation

For Ω → 0 the dynamic stiffness converges to the static spring stiffness k. For very large values of Ω the mass term −mΩ2 dominates the dynamic stiffness and the real part of the dynamic stiffness becomes negative. At resonance (η = 1, i.e. Ω = ωn ), the spring stiffness k and the mass term −mΩ2 cancel out. The response is then determined by the damping constant c. In this case the dynamic stiffness becomes purely imaginary. Example 7.1: SDOF system with cosine load

In the following, we load the SDOF system with a harmonic load that is described by a cosine function. 0 >sin (Ωt) = f cos (Ωt)  f (t) = f01 cos (Ωt) −  f02 0 which implies, that the phase ϕL = 0 and f0 = f01 , as the amplitude of the sine part f02 is zero. The resulting response amplitude is given by Eq. (7.23) in terms of the real

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Im

Im Im(U− )

Re

U− D = 0.5 Re

ϕU− ϕU+

Re(U+ ) = Re(U− )

Im(U+ )

D = 0.2

U+

D = 0.1

(a) Complex pointers of the transfer function.

(b) Nyquist plot of the transfer function for positive frequencies and different values of damping.

Figure 7.4: Transfer function in the complex plane.

and imaginary part of wbp+ . Under consideration of f02 = 0, it reads Re (wbp+ ) =

1 (k − mΩ2 ) f0 2 [(k − mΩ2 )2 + (c Ω)2 ]

(7.29)

Im (wbp+ ) =

1 −c Ω f0 . 2 [(k − mΩ2 )2 + (c Ω)2 ]

(7.30)

Inserting Eqs. (7.29) and (7.30) into Eq. (7.23), we obtain wp0 = =

q

q

2 2 wp01 + wp02 = 2 Re(wbp+ )2 + Im(wbp+ )2

v u u (k 2f0 u t 

− mΩ2 )2 + (cΩ)2

4 (k − mΩ2 )2 + (cΩ)2

2

=

1 f0 1 q =q f0 = = k (1 − η 2 )2 + (2Dη)2 (k − mΩ2 )2 + (cΩ)2 |

{z

= |U (Ω)| f0 = V (η)

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}

=|U (Ω)|

|

{z

=V (η)

}

f0 k

7 Forced vibration of linear systems with single degree of freedom

94

Ω

ωn ωt |U |

Re(U ) 1 k

Im(U ) Figure 7.5: Transfer function of the SDOF system over frequency (—) and real (—) and imaginary (—) part.

where η = ωΩn is the frequency ratio and V (η) is the amplification function. We note that the amplification function is linked to the spectral transfer function via k |U (Ω)| = V (η). Thus, the response amplitude can be calculated from the force amplitude by a multiplication with the quotient of the amplification function V (η) and the stiffness k. The response lags behind the excitation, as the imaginary part of wbp+ is negative. As for the cos-excitation the phase of the force ϕf is zero, the phase shift is equal to the phase of the response, which follows to tan (ϕw ) =

−cΩ 2Dη Im(wbp+ ) = =− 2 Re(wbp+ ) k − mΩ 1 − η2

with c = 2Dωn m (comp. Tab. 5.1). The equations of motion (7.11) and (7.12) constitute a dynamic equilibrium and can be illustrated using pointer diagrams.

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Im

Im

t −Ω2 mwp+

F+t

t iΩcwp+

t kwp+

Re

Ωt

F+ e iΩt wp+ e iΩt

Im(wp+ e iΩt ) ∆ϕ = ϕwp+ Ωt

Ωt − ∆ϕ

−Ωt

−Ωt + ∆ϕ

F+ = F−

∆ϕ = ϕwp− Im(wp− e −iΩt )

Re

Im Re

wp− e −iΩt

Ωt

F− e −iΩt F−t

t −Ω2 mwp−

t kwp−

t −iΩcwp−

Figure 7.6: Particular solution in the complex plane. The illustration shows the two pairs of complex conjugate force and displacement values f+ and f− as well as wp+ and wp− at time t. As the sine term is zero, the force term lies on the real axis at time t = 0. For the sake of clarity the pointers wp+ and wp− at time t = 0 are not depicted. Both force and response rotate in the complex plane with the rotation angle Ωt. The positive term rotates counterclockwise (mathematically positive) whereas the negative term rotates clockwise (mathematically negative). The response lags behind the force by an angle of ∆ϕ. Again the sum of both parts results in real output values. The equilibrium of external force f and internal forces fk , fc and fm in the spring, damper and mass, respectively, is depicted for the positive and negative pointer at time t (indicated by superscript (·)t , such t that, i.e. wp− = wp− e −iΩt ).

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7.2.1.2 Transient response

We are now interested in obtaining the full response of a system, on which a harmonic load starts acting at time t = 0. We assume, that for time t = 0, the system is at rest or has otherwise defined initial conditions w (t = 0) and w˙ (t = 0) = v (t = 0). Then, the full solution is given by the superposition of homogeneous solution wh (t) (depending on damping, Eqs. (5.22), (5.24), (5.25)) and the particular solution wp (t). w (t) = wh (t) + wp (t)

(7.31)

w˙ (t) = w˙ h (t) + w˙ p (t)

(7.32)

The constants w01 and w02 of the homogeneous solution are determined from the initial conditions w (t = 0) and v (t = 0) = w˙ (t = 0). After the homogeneous solution has decreased, in the steady state only the particular solution wp (t) needs to the be taken into account. Due to damping the homogeneous part of the solution wh (t) decays for increasing time. In the limit t → ∞, we observe wh (t) → 0 for the homogeneous part. Vibrations with large amplitudes can occur in the starting phase of the vibration. This holds especially for systems with low eigenfrequencies and is especially important for high frequency excitation on systems at rest. The transient response of a system initially at rest over time is exemplarily shown in Fig. 7.7.

w(t)

w(t) = wh (t) + wp (t) wp (t)

t

wh (t)

Figure 7.7: Transient response of forced SDOF system. We observe that the homogeneous part of the solution has almost completely decayed out after 5 periods. Nevertheless, the full solution shows a higher maximum response amplitude than would have been predicted by the particular solution only.

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The full solution is given without further derivation: w(t) = e −δt (w01 cos (ωD t) − w02 sin (ωD t)) +

f0 V (η) cos (Ωt + ϕw ) k

(7.33)

with: w01 = w(t = 0) −

f0 V (η) cos (ϕw ) k

f0 f0 1 v(t = 0) + w(t = 0) − V (η) cos (ϕw ) + V (η)Ω sin (ϕw ) . =− ωD k k "

w02

(7.34) #

!

(7.35)

In this App the response due to a harmonic load is visualized. The homogenous and particular solution as well as the total solution are plotted. One can visualize that the homogenous solution is important for the initial conditions and over time its influence tends to disappear according to the damping constant. http://www.bm.bgu.tum.de/lehre/interactive-apps/ single-degree-of-freedom-system/

7.2.1.3 Resonance case

For η = 1, i.e., when excitation frequency and eigenfrequency coincide (Ω = ωn ), weakly damped systems undergo very large amplitudes. In an undamped system the dynamic stiffness kb+ (comp. Eq. (7.16)) would tend to zero, as k and −mΩ2 cancel each other out, and the resulting response amplitude would become infinitely large. Often, it is of interest to know about the system response, when the excitation is only in the resonance frequency for a short time, e.g. when the rotational speed of a machine is increased and the region of resonance is passed. We begin with the derivation of the solution of the damped SDOF system in the case of resonant excitation, i.e., Ω = ωn and η = 1. Here, we assume that the force has zero phase, i.e., ϕf = 0, and it follows a cosine function f (t) = cos(ωn t). With this, the particular solution reads wp (t) =

f0 V (1) cos(ωn t + ∆ϕ). k

(7.36) 

For η = 1 the phase shift ∆ϕ is π2 . Using the identity cos α +

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π 2



= − sin (α), and inserting

7 Forced vibration of linear systems with single degree of freedom

98

V (1) =

1 , 2D

Eq. (7.36) reads

wp (t) = −

f0 sin(ωn t). 2Dk

(7.37)

The homogeneous solution to the problem reads wh (t) = e −δt (wh1 cos(ωD t) − wh2 sin(ωD t)) .

(7.38)

The total solution w(t) = wh (t) + wp (t) = e −δt (wh1 cos(ωD t) − wh2 sin(ωD t)) −

f0 sin(ωn t) 2Dk

(7.39)

has to satisfy the boundary conditions. We assume that at t = 0 the system is at rest, i.e., w0 = v0 = 0. It follows !

w(t = 0) = wh1 = 0

(7.40)

With wh1 = 0, the velocity reads w(t) ˙ = −wh2 δe −δt sin(ωD t) − wh2 e −δt ωD cos(ωD t) − ωn

f0 cos(ωn t) 2Dk

(7.41)

Under the condition w(t ˙ = 0) = 0, we obtain wh2 =

ωn f0 . ωD 2Dk

We insert ωD =

√

(7.42) 1 − D2 ωn and δ = Dωn , and write the full solution as

f0 1 f0 sin(ωD t) − sin(ωn t) 2 2Dk 1 − D 2Dk   √ f −Dωn t 2 sin(ω t) √0 e sin(ω t) − 1 − D = D n 2Dk 1 − D2

w(t) = e −Dωn t √

(7.43) (7.44) (7.45)

For small damping ratios D 0 for ω < 0

(7.75)

.

Despite the fact that there is no obvious physical interpretation, the loss factor η can still be given in terms of the damping ratio D as η = 2D.

(7.76)

The resulting Nyquist plot (representation in the complex plane) is given in the following Figure. Im

Im Im(U− )

Re

U− D = 0.5 ϕU− ϕU+

Im(U+ )

Re Re(U+ ) = Re(U− )

U+

(a) Complex pointers of the transfer function.

D = 0.2

D = 0.1 (b) Nyquist plot of the transfer function for positive frequencies and different values of damping.

Figure 7.13: Transfer function for linear hysteretic damping in the complex plane.

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|V | 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0

D=0 D = 0.2 D = 0.4 Vmax =

∆ϕ −π

1 2D

D=0 D = 0.2 D = 0.4

− π2 0.5

1.0 1.5 ηmax = 1

2.0

2.5

3.0 η

0.0

0.5

1.0

1.5

2.0

2.5

3.0 η

Figure 7.14: Amplification function harmonically forced SDOF system with linear hysteretic damping.

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7.4 Arbitrary periodic excitation 7.4.1 Decomposition into harmonics According to the principle of superposition, the steady state response (particular solution) of linear systems to a dynamic loading—that consists of different sine and cosine parts—can be obtained by the superposition of the different partial solutions. Assuming an excitation with

F (t) =

nF X

F0,i cos(Ωi t + ϕF,i ) = F0,1 cos(Ω1 t + ϕF,1 ) + F0,2 cos(Ω2 t + ϕF,2 ) + . . . ,

(7.77)

i

the particular solution of the system reads wp (t) =

nF X

w0,i cos(Ωi t + ϕw,i ) = w0,1 cos (Ω1 t + ϕw,1 ) + w0,2 cos (Ω2 t + ϕw,2 ) + . . .

(7.78)

i

with w0,i = q

F0,i

(7.79)

2

(k − mΩ2i ) + (cΩi )2

ϕw,i = ϕF,i + ϕU,i tan ϕU,i = −

(7.80)

cΩi . k − mΩ2i

(7.81)

Fig. 7.15 shows an example of an arbitrary periodic force. F

t

T Figure 7.15: Example for an arbitrary periodic load time history. The signal is repeated with period T , which corresponds to a fundamental radial frequency Ω0 = 2π T

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Reminder 7.1: Fourier series

A short repetition of Fourier series (comp. Section 3.1.1) shall be given in the following. An arbitrary periodic loading can be represented by a Fourier series f (t) =

∞ a0 X + (ak cos (kω0 t) + bk sin (kω0 t)) 2 k=1

with ωk = kω0 and ω0 = 2π . The coefficients of the individual parts are determined T in such a way, that the mean-square error between the function F (t) and the Fourier series is minimal. ε=

ZT 0

!2

∞ a0 X f (t) − − (ak cos (ωk t) + bk sin (ωk t)) 2 k=1

dt

The Fourier-coefficients follow to T 2Z ak = f (t) cos (ωk t) dt T

for k = 0,1,2, . . .

(7.82)

for k = 1,2, . . .

(7.83)

0

T 2Z bk = f (t) sin (ωk t) dt T 0

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Example 7.2: Response of SDOF system to square-wave

In this example we will derive the Fourier series of the square-wave function depicted in Fig. 7.16. f f0

t

T Figure 7.16: Square wave with period T .

The Fourier series representation of this signal is derived in Example 3.1. The considered signal is an odd function, i.e., f (t) = −f (−t), which results in all aj = 0 and thus all cosine terms in the series vanish. The Fourier coefficients bj are given by:

bj =

2 T

 T  T 2 Z Z    f0 sin(ωj t) dt + (−f0 ) sin(ωj t) dt   0

T 2

T

2 4 1 = − f0 cos(ωj t) T ωj 0

=

  4f0 jπ

0

for j = 1,3,5, . . . for j = 0,2,4, . . .

Then, the Fourier series representation of f (t) reads f (t) =

∞ 4f0 X 1 sin (ωj t) . π j=1,3,... j

Using the amplification function (comp. Eq. (7.28)), the amplitude of the response of

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the SDOF system can be given as w0 = V (η)wstat = V (η)

f0 . k

Further considering the phase shift between load and response, we obtain for the displacement response: w(t) =

4F0 X 1 1 q sin(jω0 t + ϕw,j ) . π j=1,3,... j (k − m(jω0 )2 )2 + (cjω0 )2

with tan ϕw,j = −

ckω0 . k − m(kω0 )2

For the superposition of the individual harmonic parts the phase angle of each harmonic has be taken into account. For a damped system with c 6= 0 they will vary between 0 and π. The following illustrations show the decomposition of load and response into the individual harmonics. Below, the resulting series representation is depicted for maximum j of 7. We assume that no damping is present, thus the phase shift can easily be incorporated as being either 0 or −π. Furthermore, the system is such that the eigenfrequency ωn corresponds to four times the fundamental frequency ω0 .

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f (t)

(1 + 3 + 5) (1)

f0

(1 + 3) (1 + 3 + 5 + 7)

t T 2

w(t)

(1 + 3)

(1 + 3 + 5 + 7)

(1)

(1 + 3 + 5)

t T 2

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Fourier spectrum of the external force

f (ωj ) 4F0 π

with eigenfrequency ωn = 4ω0

(1)

( 13 )

( 51 )

3

1

( 17 )

j=

7

5

V (Ω)

ωj ω0

Amplification function V (Ω = ωj ) = √(k−mωk2 )2 +(cω j

2,908 1

f (Ω) V k 4F0 π

1,216

1,079

0,389 3

1

(1,079)

F ourier spectrum of the response (only amplitude spectrum)

(0,969)

(0,234)

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3

j

7 5 4 (resonance)

(ωj )

1

2 j)

5

(0,056) 7

j

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7.5 Aperiodic excitation 7.5.1 Time domain 7.5.1.1 Unit impulse response function

Additionally to the response of the SDOF-system to harmonic loads, we can also directly determine the response to an impulse. The impulse of a force that acts upon the system during a time frame ∆t reads − ∆t 2

I0 =

Z

f (t) dt

(7.84)

− ∆T 2

f (t) f (t) = I0 δ(t)

∆t

t

Figure 7.17: Force impulse

If the limit ∆t tends to be zero, while the integral retains a finite value, the force must tend to infinity, which can be written in a symbolic manner using the δ-Dirac-distribution. f (t) = I0 δ(t) .

(7.85)

The impulse then follows to − ∆t 2

− ∆t 2

I = lim

Z

∆t→0 − ∆t 2

f (t) dt = lim I0

Z

∆t→0

δ(t) dt = I0 .

(7.86)

− ∆t 2

At the start of vibration (undeformed configuration), all other forces in a SDOF system (the spring force kw, the damping force cw) ˙ are negligible in comparison to f (t). The system is at

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rest, i.e., w(t = 0) = 0. The conservation of momentum reads − ∆t 2

+ mv0+ =  m 0−

Z

f (t) dt = I0 ,

− ∆t 2

with the initial velocity before the impact v0− and the velocity after the impact v0+ . If additionally v(t− = 0) = v0− = 0, then mv0+ = I0 → v0 = v0+ =

I0 . m

For the case of the undercritically damped SDOF system with the initial velocity v0 = obtain the following homogeneous response w(t) = I0 h(t)

I0 , m

we

(7.87)

where h(t) is the unit impulse response function, defined as √  1 h(t) = q e−Dωn t sin 1 − D 2 ωn t . km(1 − D2 )

(7.88)

Fig. 7.18 shows the unit impulse response function for varying percentage of critical damping D. √ kmh(t) 1 D=0 D = 0.7

D = 0.05 D = 0.2 ωn t

−1 Figure 7.18: Unit impulse response function for varying percentage of critical damping D; D = 0 (—), D = 0.05 (–·–), D = 0.2 (– –), D = 0.7 (· · ·). For larger D the response decays more quickly.

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7.5.1.2 Superposition, Duhamel-Integral

For linear systems, the response to an arbitrary loading can be calculated by the superposition of the responses to a sequence of impulses. In order to determine the response at the time t the responses caused by all impulses f (τ ) dτ are summed up. At time t the response to an impulse at time τ is equal to dI h(t − τ ) where dI = f (τ ) dτ . The total response can finally be obtained as the sum of all responses that have acted until time t. w(t) =

τ =t Z

f (τ )h(t − τ ) dτ .

τ =0

The principle is depited in Fig. 7.19. f

w

f (τ )dτ

h(t − τ )

f (t) t τ dτ

t

t−τ

τ t

Figure 7.19: Superposition of responses to single impulses. The left hand side figure shows the impulse dI acting on the system at time τ , dI = f (τ ) dτ . This impulse causes the response that is given by Eq. (7.87) and depicted on the right hand side. The unit impulse response function h(t) is shifted to the time τ and scaled with F (τ ). Due to this we subtract τ in the argument of h(t − τ ).

Due to the fact, that h(t) = 0 for t ≤ 0 1 , we can rewrite Eq. (??) as follows . w(t) =

Z∞

f (τ )h(t − τ ) dτ

(7.89)

τ =0

Furthermore, we consider f (τ ) = 0 for τ ≤ 0, and thus w(t) =

Z∞

f (τ )h(t − τ ) dτ

(7.90)

−∞

1

This follows from causality. An impulse cannot cause a response before its impact.

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The given mathematical operation in the equation is called convolution. For the viscously damped SDOF system, the Duhamel integral reads: w(t) =

1

Zt

q

km(1 − D2 ) −∞

f (τ )e−Dωn (t−τ ) sin

√



1 − D2 ωn (t − τ ) dτ .

(7.91)

For the undamped SDOF system Eq. (7.91) simplifies to t

1 Z w(t) = √ f (τ ) sin (ωn (t − τ )) dτ . km −∞

(7.92)

Note that by entering the impulse response function of the SDOF system, we can adjust the limits of integration. This is because h(t) = 0 for t < 0, i.e. no response can occur before an excitation (causality). In the definition of the impulse response function this is implicitly assumed without further specification. 2 Example 7.3: Undamped SDOF system under step load

We calculate the response of a linear SDOF system subjected to the step load depicted in Fig. 7.20. f (t) f0

t0

t

Figure 7.20: Step load for a SDOF system

The Duhamel integral reads t < t0 :

w(t) = f0

Zt 0

2

t

f0 Z h(t − τ ) dτ = √ sin (ωn (t − τ )) dτ km 0

A more rigorous definition could be done by the consideration of the cases t < 0 and t ≥ 0 or by multiplication with the Heaviside function.

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t

f0 f0 1 cos (ωn (t − τ )) = (1 − cos (ωn t)) =√ k km ωn 0

t > t0 :

w(t) = f0

Zt0

h(t − τ ) dτ =

0

f0 [cos ωn (t − t0 ) − cos (ωn t)] . k

(7.93) (7.94)

In general, the solution can also be found by already found solutions to the differential equation of the SDOF system. For this the response is determined superimposing results from two time-shifted opposite load jumps. In principle the solution can also be found by separating the ranges, and solving the system separately under consideration of the initial and transfer conditions. In the analysis often only the maximum displacment wmax is of interest. Once the duration of the excitation exceeds half the length of the period, which implies t0 > T2n , wmax reaches the value wmax = 2fk0 within the interval 0 < t ≤ t0 . In all other cases, the maximum value occurs at t1 > t0 and can be found by finding the maximum in Eq. (7.94). Thus, we state t > t0 : w =

f0 [cos (ωn (t − t0 )) − cos (ωn t)] k

!

w˙ = 0 wmax =

→ t1

→ wmax = w(t1 )

2f0 t0 sin ωn k 2 



.

(7.95) (7.96) (7.97)

Note that as the system is undamped, the corresponding maximum value may be found repeatedly.

7.5.1.3 Response spectra

The results to peak value searches are presented in so called response-, impact-, or shock response-spectra. The response spectrum shows the maximum response value depending on the eigenfrequency or eigenperiod of the system, without noting the corresponding time of occurrence. Typically, a characteristic time of excitation, relative to the eigenperiod Tn = f1n , is shown dimensionless on the abscissa. The ordinate then usually shows the maximum dynamic amplitude relative to the static displacement. With this the maximum response quantity can be solely deduced in dependence on eigenfrequency and damping. Figure 7.21 exemplarily shows the response spectrum of the undamped SDOF system under rectangular excitation of Fig. 7.20 in example 7.3.

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kwmax F0

2

t > t0

t < t0

0,5

t0 Tn

1

Figure 7.21: Response spectrum of SDOF system for rectangular excitation with duration t0 and amplitude f0 using Eq. (7.94) and Eq. (7.97). The graph depicts the maximum response value as a function of the ratio of load duration and eigenperiod of the system. For a load duration t0 shorter than half of the eigenperiod of the system, the maximum value will be reached after the load has stopped acting and vice versa, for t0 longer than half of the eigenperiod of the system, the maximum value will be reached before the load has stopped acting.

Response spectra are also used for the characterization of stochastic excitations, e.g. in earthquake engineering (DIN [2006]). In the case of root point excitation the force term f (t) needs to be replaced withqthe load term of −m w¨e (t) of Eq. (7.62) in the Duhamel integral in Eq. (??). k Considering ωn = m , the integral for determining the realtive displacement simplifies to wr (t) =

τ =t Z τ =0

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−w¨e (τ )

1 q

ωn (1 − D2 )

e−Dωn (t−τ ) sin

√



1 − D2 ωn (t − τ ) dτ .

(7.98)

7 Forced vibration of linear systems with single degree of freedom

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Example 7.4: Triangular shock

The response spectrum can also be given as a function of Tn/t0 . As an example consider a massless, flexible column with varying length and attached mass at its top end. A load acts on the mass point. For increasing length of the system, the stiffness decreases, leading to decreasing eigenfrequencies and thus, increasing eigenperiods. f (t) f0 t t0

f (t) increasing Tn Figure 7.22: Triangular shock on column with varying length

Depending on the ratio of eigenperiod to load duration, the maximum response value is given in the following diagram. V =

max w f0 k

V3

max V = 1,25 V1

V2

Tn = 0

0,1

0,56

1

5

10

Tn t0

2π ωn

[log]

Figure 7.23: Response spectrum of SDOF system for triangular excitation with duration t0 and maximum amplitude f0 . The graph depicts the maximum response value as a function of the ratio of the eigenperiod of the system Tn and the load duration t0 .

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Response spectra for the SDOF system under different impact time histories are depicted in Fig. 7.24 (cf. Petersen [1996]). V 2.0

V 2.0

V 2.0

1.5

1.5

1.5

1.0

1.0

1.0

0.5 0

0.5

t t0

0

1.0 2.0 3.0 4.0 t0 Tn

V 2.0

0

0.5

t t0

0

1.0 2.0 3.0 4.0 t0 Tn

V 2.0

0

1.5

1.5

1.0

1.0

1.0

0

0.5

t t0

0

1.0 2.0 3.0 4.0

0

1.0 2.0 3.0 4.0

t0 Tn

1.0 2.0 3.0 4.0 t0 Tn

0.5

t t0

0

0

V 2.0

1.5

0.5

t t0

0

t t0

0

1.0 2.0 3.0 4.0

t0 Tn

t0 Tn

Figure 7.24: Response spectra - overview from Petersen [1996]

7.5.2 Exkursus: Kelvin-Voigt model In Section 7.1.1, we discuss the distinction between “quasi-static” and dynamic processes. To consider dynamic processes, the influence of inertia and damping forces are considered. However, also for processes, which vary very slowly in time, a static consideration of the problem does not lead to a full understanding of the process. This is for example the case for creep processes. To illustrate this, we derive the response of the Kelvin-Voigt model (spring and damper in parallel) to a step load.

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Example 7.5: Kelvin-Voigt model under step load

Consider the Kelvin-Voigt model governed by the differential equation (4.20), slightly reformulated to k f u˙ + u = c c where the stress is applied abruptly at t = 0 and then held constant, thus  f

f (t) = U (t) = 

0

0

for t ≥ 0 . for t < 0

In order to obtain the homogeneous solution, the following approach is given uh = uh0 e λt . Inserting into the ODE (f = 0) gives k λuh0 e λt + uh0 e λt = 0 c ! k uh0 e λt = 0 ⇔ λ+ c Thus, λ=−

k c

and k

uh = uh0 e − c t with uh0 being undetermined for now. As f (t) is constant for t ≥ 0, the particular solution up (t) is the constant function up (t) = up0 .

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Inserting into the ODE gives f0 k up0 = c c f0 ⇔ up0 = . k The full solution then reads k

u(t) = uh (t) + up (t) = uh0 e − c t +

f0 . k

The full solution now has to satisfy the initial condition, chosen here as u(0) = 0. Thus, u(0) = uh0 + ⇔ uh0 = −

f0 =0 k

f0 . k

Finally, we then obtain u(t) = −

 k f0 − k t f0 f0  e c + = 1 − e−ct . k k k

The resulting displacement solution is depicted in the following. f0 k

u

t Figure 7.25: Displacement response response of Kelvin-Voigt model to applied unit-step stress.

Due to the neglection of inertia forces in the equation of motion, this model does not give a oscillatory response.

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7.5.3 Frequency domain Reminder 7.2: Fourier transformation

The Fourier transformation is defined by Z∞

f˜(ω) =

f (t)e −iωt dt

−∞ ∞ 1 Z ˜ f (t) = f (ω)e iωt dω . 2π −∞

7.5.3.1 Response to harmonic excitation

For harmonic excitation f (t) = fb+ e iΩt + fb− e iΩt , the steady state response is wb = wb+ e iΩt + wb− e −iΩt = Ub (Ω)fb+ e iΩt + Ub (−Ω)fb− e iΩt

(7.99)

wb+ = Ub (Ω)fb+

(7.100)

wb− = Ub (−Ω)fb− .

(7.101)

Alternatively, we can also determine the response using the Duhamel integral, as w(t) =

Z∞

f (τ )h(t − τ ) dτ = −

−∞ Z

f (t − ρ)h(ρ) dρ =

∞

−∞

Z∞

f (t − ρ)h(ρ) dρ

where we substituted τ = t − ρ or ρ = t − τ and used the fact that that is acting since τ = −∞, it holds w(t) =

Z∞

(7.102)

−∞ dρ dτ

= −1. For an excitation

f (t − ρ)h(ρ) dρ

−∞

=

Z∞ 



fb+ e iΩ(t−ρ) + fb− e −iΩ(t−ρ) h(ρ) dρ

−∞

=

+e

fb

iΩt

Z∞ −∞

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h(ρ)e

−iΩρ

dρ + fb

−e

−iΩt

Z∞

h(ρ)e iΩρ dρ

−∞

7 Forced vibration of linear systems with single degree of freedom

124

We substitute τ = −ρ in the second part of the preceeding equation and obtain =

+e

fb

iΩt

Z∞

h(ρ)e

−iΩρ

dρ + fb

−e

−∞

Z∞

−iΩt

h(−τ )e −iΩτ dτ

−∞

= F {h(t)} fb+ e iΩt + F {h(−t)} fb− e −iΩt ˜ ˜ = h(Ω) fb+ e iΩt + h(−Ω) fb− e −iΩt

(7.103)

˜ h(Ω) is the Fourier transform of the unit impulse response function h(t). Comparing Eq. (7.103) with Eq. (7.100), we note that ˜ U (Ω) = h(Ω) =

Z∞

h(t)e−iΩt dt = F {h(t)}

−∞

Thus, the transfer function for harmonic excitation is equal to the Fourier transform of the unit impulse response function. 7.5.3.2 Response to aperiodic excitation

We consider the general, aperiodic load f (t) and its Fourier transform ∞ 1 Z ˜ f (t) = f (ω)e iωt dω 2π −∞

f˜(ω) =

Z∞

f (t)e −iωt dt .

−∞

The response of an SDOF system can then be written as ∞ 1 Z iωt w(ω)e ˜ dω . w(t) = 2π

(7.104)

−∞

This representation gives the response in the time domain as the inverse Fourier transform of the response in the frequency domain. Independently, we can also state the response as the Duhamel integral w(t) =

Z∞

f (t − τ )h(τ ) dτ

(7.105)

−∞

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We know replace f (t − τ ) in Eq. (7.105) by its Fourier transform w(t) =

Z∞

∞ 1 Z ˜  f (ω)e iω(t−τ ) dω  h(τ ) dτ 2π



−∞

1 = 2π



−∞

Z∞

 ∞ Z ˜  f (ω)



h(τ )e −iωτ dτ  e iωt dω

−∞

−∞

|

{z

˜ h(ω)

}

∞ 1 Z ˜ ˜ = f (ω)h(ω)e iωt dω . 2π

(7.106)

−∞

By comparing Eq. (7.104) with Eq. (7.106), we find that ˜ f˜(ω) . w(ω) ˜ = h(ω) with w(ω) ˜ : Fourier transform of the response f˜(ω) : Fourier transform of the excitation ˜ h(ω) : Frequency domain transfer function = U (ω) Thus, the system response in the frequency domain w(ω) ˜ can simply be calculated by mul˜ tiplying the transfer function h(ω) with the Fourier transform of the excitation f˜(ω). The system response in the time domain w(t) is given by the inverse Fourier transform of w(ω). ˜ w(t) = F −1 {w(ω)} ˜ The relations are further depicted in Fig. 7.26.

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126

f (t) m w(t) k

c

convolution w(t) = f (t)

R∞ −∞

f (τ )h(t − τ ) dτ w(t)

h(t) =

∗

f˜(ω)

w(ω) ˜

˜ h(ω) =

· multiplication ˜ w(ω) ˜ = h(ω) · f˜(ω)

Figure 7.26: Response of SDOF system to arbitrary excitation. The force signal f (t) is given. To calculate the response signal w(t), the problem can be considered in the time or frequency domain. In the time domain, the excitation signal is convoluted with the impulse response function h(t), and the resulting integral has to be solved. For a frequency domain calculation, the force signal f (t) is transformed into the frequency domain by a Fourier transformation, resulting in f˜(ω). The response in the frequency domain w(ω) ˜ ˜ is then found by multiplication of the force f˜(ω) with the systems’ transfer function h(ω) . The resulting response in the time domain is then given by an inverse Fourier transform of w(ω) ˜ . The two approaches yield the same result for the steady-state response.

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8 Free vibration of linear systems with multiple degrees of freedoms This section deals with the free vibration of systems with multiple degrees of freedom. We assume that the equation of motion is found by any of the methods discussed in chapter 4.

8.1 Preliminary remark: Maxwell-Betti theorem For a discrete, linear elastic system it holds w = Af with the flexibility matrix A, the displacement vector w, and the force vector f . We now consider a system, where two forces are applied one after the other, first at position k and secondly at position i. This case is called I. The reverse is called case II. The displacements in both cases are wi = aii fi + aik fk wk = aki fi + akk fk where aij is the (i,j) element of the flexibility matrix A. After applying the loads, the displacements are equal in both cases In a linear system, the work performed on the system by the forces has to be equal independently from the sequence of the loads. For case I the work is 1 1 WI = akk fk2 + aii fi2 + aki fi Fk . | {z } 2 2

(8.1)

wki

The last term corresponds to the work performed by load fk on the displacement wki = aki fi . For case II the work is 1 1 WII = aii fi2 + akk fk2 + aik fk fi , | {z } 2 2

(8.2)

wik

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where now the last term corresponds to the work performed by load fi on the displacement wik = aik fk . As the work performed in both cases is the same, we equate Eqs. (8.1) and (8.2) and obtain wki fk = wik fi .

(8.3)

It follows that the work performed by force fk on the displacement wki (displacement at position k due to fi ) is the same as the work performed by force fi on the displacement wik (displacement at position i due to fk ). Furthermore, from aki fi · fk = aik fk · fi

(8.4)

we obtain aki = aik . Here, aki corresponds to the displacement at position k due to a unit load 1 at position i and aik corresponds to the displacement at position i due to a unit load 1 at position k. Thus, we can state for the flexibility matrix AT = A. From this it follows that the transpose of the flexibility matrix is equal to the flexibility matrix. Thus, the matrix is symmetric. The stiffness matrix K is the inverse of the flexibility matrix and by this also symmetric, i.e. KT = K .

8.2 Solution of the homogeneous system of equations The general equation of motion for unforced MDOF systems with velocity proportional damping reads ¨ + Cw ˙ + Kw = 0. Mw

(8.5)

Even though the system is not excited by an external force, it can have non-trivial solutions w(t) 6= 0 responding to the initial conditions that are not in equilibrium. We first discuss the case of undamped MDOF systems and then generalize for the case of damped systems. In the following n denotes the number of DOFs.

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8.2.1 Undamped MDOF systems The derivation of the eigensolutions (f = 0, natural vibrations) of a MDOF is first given for systems without damping (C = 0). The system of equations thus reads ¨ + Kw = 0 Mw

(8.6)

We use an exponential approach to solve Eq. (8.6) w = φe iωt

(8.7)

¨ = −ω 2 φe iωt , w

(8.8)

and insert Eqs. (8.7) and (8.8) into Eq. (8.6) to obtain a homgoeneous linear system of equations representing a matrix eigenvalue problem. 



K − ω2M φ = 0

(8.9)

with λ = ω 2 . Nontrivial solutions for the displacements w are only possible if the determinant of the matrix (K − λM) is zero. The only unknown in Eq. (8.9) is the value λ = ω 2 (representing the square of the circular natural frequency in Eq. (8.7)). Thus, we find the circular natural frequencies of the system by det (K − λM) = 0.

(8.10)

Eq. (8.10) leads to an equation for λ = ω 2 with the order n (n = number of DOFs) and therefore has n solutions for the eigenvalues ω12 , . . . , ωn2 . The solutions of the system of equations for a vanishing determinant are called mode shapes or eigenvectors. The eigenvector φi related to λi = ωi2 is obtained by inserting ωi2 in eq. (8.9). 



K − ωi2 M φi = 0

(8.11)

For a vanishing determinant the above equation system becomes linearly dependent. Therefore, the eigenvectors φi (also called modes, mode shapes) are only determined up to a constant factor (hence also α φi is an eigenvector of the system). In general, various options can be chosen for the normalization of the eigenvectors: • The eigenvector is scaled, such that one vector component, e.g. the largest, is chosen arbitrarily. Usually this value is set equal to one. • The eigenvector is scaled, such that the L2 -norm is one, i.e., the eigenvector has the length one.

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• The eigenvector is scaled, such that the generalized mass matrix is the identity matrix (explained in the following), i.e., the norm of the eigenvectors with respect to the mass matrix is equal to one. A solution is finally given by the superposition of all the eigenvectors of the system weighted by time-dependent weighting functions. The solution w(t) to Eq. (8.6) can be written as w(t) =

n X





φi yˆi+ e iωi t + yˆi− e −iωi t =

i=1

n X

φi yi (t) = Φy(t) .

(8.12)

i=1

The solution can also be written in sine/cosine notation, as discussed in section 2.2. Then, w(t) =

n X

φi (yi,01 cos (ωi t) − yi,02 sin (ωi t)) = Φy(t) .

(8.13)

i=1

In order, to discuss the important characteristics of the eigenvalue solution, we write Eq. (8.9) for any two eigenvectors φi and φj with corresponding eigenvalues λi and λj Kφi = λi Mφi

(8.14)

Kφj = λj Mφj .

(8.15)

for all i,j = 1,2, . . . , n. We now premultiply on both sides of Eq. (8.14) by φTj and obtain φTj Kφi = λi φTj Mφi .

(8.16)

Considering the symmetry of the mass and (due to the law of Maxwell-Betti, see 8.1) the 

stiffness matrix MT = M and KT = K and using AT BC of Eq. (8.16) as

T

= CT BT A, we find the transpose

φTi Kφj = λi φTi Mφj .

(8.17)

Analogously Eq. (8.15) is premultiplied on both sides by φTi such that φTi Kφj = λj φTi Mφj .

(8.18)

If we substract Eq. (8.18) from Eq. (8.17), we obtain: 0 = (λi − λj ) φTi Mφj .

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From this we find the orthogonality relation φTi Mφj

=

 0

for i 6= j . for i = j

Mi

(8.19)

Eq. (8.19) shows that the eigenvectors are orthogonal with respect to the mass matrix. We call M∗ = ΦT MΦ the generalized mass matrix and Mi = φTi Mφi the generalized or modal mass of mode i. Here, Φ is the modal matrix that contains the eigenvectors column-wise. Since the eigenvectors are only determined up to a constant factor, for i = j the value Mi can take arbitrary values. If we now insert Eq. (8.19) into Eq. (8.17), we obtain φTi Kφj =

 0 K i

= Mi λi = Mi ωi2

for i 6= j . for i = j

(8.20)

We furthermore call K∗ = φTi Kφj the generalized stiffness matrix and Ki = φTi Kφi the generalized or modal stiffness of mode i. The eigenfrequencies ωi can be calculated from the generalized stiffnesses and masses as ωi2 =

Ki . Mi

(8.21)

As was discussed before, the eigenvectors need to be normalized. We can now choose the normalization of φi such that the generalized mass matrix is the identity matrix, i.e., φTi Mφj = δij

(8.22)

holds. This choice is also known as “unit modal mass”. It is Ai = 1 and thus we can write for Eq. (8.20) φTi Kφj = λi δij = ωi2 δij .

(8.23)

We obtain a compact notation of these n matrix equations by introducing the diagonal matrix 

λ1

[Λ] =   

..

. λn





  

= 

ω12





..

. ωn2

 , 

(8.24)

containing all the eigenvalues and the matrix Φ = [φ1 , φ2 , . . . , φn ] ,

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(8.25)

8 Free vibration of linear systems with multiple degrees of freedoms

132

containing all the eigenvectors. With this, we can write Eq. (8.22) and Eq. (8.23) as ΦT MΦ = I

(8.26)

ΦT KΦ = Λ .

(8.27)

Note that this only holds for mass normalized eigenvectors. Applying the derived properties of the solution, we can find a simple expression for the equation of motion (8.6) in terms of the modal coordinates y(t). The solution given in 8.12 is repeated her for convenience in the general form: w(t) = Φy(t) .

(8.28)

The time functions y(t) are time harmonic functions, that can be either expressed as complex exponential functions or sine/cosine functions. We now enter Eq. (8.28) into the equation of motion Eq. (8.6). It follows M Φ¨ y(t) + KΦy(t) = 0 .

(8.29)

By multiplying Eq. (8.29) with the transpose of the matrix of the eigenvectors ΦT , we obtain ΦT MΦ¨ y(t) + ΦT KΦy(t) = 0 .

(8.30)

If the eigenvectors are normalized with respect to the mass matrix, i.e. ΦT MΦ = I, after inserting Eqs. (8.22) and (8.23) it follows I¨ y(t) + Λy(t) = 0 .

(8.31)

As I and Λ are diagonal matrices the system of equations become decoupled and we obtain n equations y¨i (t) + ωi2 yi (t) = 0

(8.32)

for i = 1, . . . ,n. The decoupled equations of motion correspond to the equation of motion of the SDOF system.

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Example 8.1: Two degree of freedom system

In the general case, for n = 2, Eq. (8.30) reads φT1 Mφ1 φT2 Mφ1

"

φT1 Kφ1 φT1 Mφ2 ¨ (t) + T y φ2 Kφ1 φT2 Mφ2 #

"

φT1 Kφ2 y(t) = 0 . φT2 Kφ2 #

Through the orthogonality relations for M and K this leads to a decoupled system of equations m∗1 0 0 m∗2

"

#"

y¨1 k∗ 0 + 1 ∗ 0 k2 y¨2 #

"

#"

#

y1 =0 y2

with mi ∗ = φTi Mφi and ki ∗ = φTi Kφi . The above holds for any normalization of the eigenvectors. The response yi (t) can now be determined for every mode shape φi as the solution of the corresponding SDOF system: m∗1 y¨1 (t) + k1∗ y1 (t) = 0 m∗2 y¨2 (t) + k2∗ y2 (t) = 0.

8.2.2 Consideration of the initial conditions Furthermore the initial conditions of the MDOF system shall be given as w(t = 0) = w0

(8.33)

˙ = 0) = v(t = 0) = v0 w(t

(8.34)

Eqs. (8.33) and (8.34) are now represented by the initial conditions of the modal responses w(t = 0) = Φy(t = 0)

(8.35)

˙ = 0) = v(t = 0) = Φy(t ˙ = 0) w(t

(8.36)

The initial conditions w(t = 0) and w(t ˙ = 0) are projected on the modal basis. We multiply Eqs, (8.35) and (8.36) by ΦT M and obtain ΦT MΦy(0) = ΦT Mw(0)

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(8.37)

8 Free vibration of linear systems with multiple degrees of freedoms

134

˙ ˙ ΦT MΦy(0) = ΦT Mw(0) .

(8.38)

Considering the orthogonality relation (Eq. (8.22)) under the assumption of unit modal mass normalization it follows Iy(0) = ΦT Mw(0)

(8.39)

˙ ˙ Iy(0) = ΦT Mw(0) .

(8.40)

˙ Finally the values for y(0) and y(0) are y(0) = Tw(0)

(8.41)

˙ ˙ y(0) = Tw(0)

(8.42)

with T = ΦT M. Example 8.2: Two degree of freedom system (cont.)

The following MDOF system with two degrees of freedom w1 und w2 is considered. The homogeneous solution shall be determined. m1 w1 k1 m2

m1 = 80 kg m2 = 8 kg N k1 = 200 m N k2 = 125 m

w2 k2

For this purpose the eigenfrequencies and mode shapes need to be calculated in a first step. In a second step, the solution can be obtained for the decoupled system of equations. a) First, we calculate the eigenfrequencies ω1 and ω2 as well as the corresponding

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mode shapes φ1 and φ2 . From Eq. (8.6) and Eq. (8.8) it follows MΦ(−ω 2 )y(t) + KΦy(t) = 0 and thus 



−ω 2 M + K = 0.

A solution for the above system of equations exists if 



det −ω 2 M + K = 0 . With m1 0 M= 0 m2 "

"

#

k1 −k1 K= −k1 k1 + k2

#

it yields "

det

k1 − m1 ω 2 −k1 −k1 k1 + k2 − m2 ω 2

#!

= 640 ω 4 − 27.600 ω 2 + 25.000 = 0 .

Two solutions are obtained for ω 2 ω12 = 0.93 ω22 = 42.20 . The eigenfrequencies are ω1 = 0.96 ω2 = 6.50 . The corresponding eigenvectors can be determined from "

k1 − m1 ωi 2 −k1 −k1 k1 + k2 − m2 ωi2

#

φ1i 0 = φ2i 0 !

!

(8.43)

for i = 1,2. As the system of equations is underdetermined we solve the first

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equation for φ2i 



k1 − m1 ωi2 Φ1i − k1 Φ2i = 0

⇒ Φ2i =

(k1 − m1 ωi 2 )Φ1i k1

and choose Φ1i = 1, thus Φ21 =

k1 − m1 ω1 2 = 0.63 k1

Φ22 =

k1 − m1 ω2 2 = −15.88 . k1

The matrix of eigenvectors is then given as φ φ 1 1 Φ = [φ1 φ2 ] = 11 12 = . φ21 φ22 0.63 −15.88 "

#

"

#

Note that the eigenvectors are not normalized with respect to the mass matrix, i.e., ΦT MΦ 6= [I]. To obtain the mass normalized eigenvectors, we calculate the modal mass of each eigenvector m∗i = φTi Mφi . For the first eigenvector, we obtain m∗1

=

φT1 Mφ1

m∗2

=

φT2 Mφ2

h

i

"

= 1 0.63 h

80 0 0 8 i

= 1 −15.88

#"

1 = 83.17 0.63

80 0 0 8

"

#

#"

1 = 2 097.33 . −15.88 #

Thus, we obtain the mass normalized eigenvectors by dividing each eigenvector by the square-root of its modal mass m

φ φ . φi = √ i ∗ = q i T mi φi Mφi

The mass normalized eigenvectors are indicated by a left superscript m in this

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example. Thus, m

1 1 1 0.110 φ1 = √ ∗ φ1 = √ = 0.63 0.069 m1 83.17

m

1 1 1 0.022 = . φ2 = √ ∗ φ2 = √ −15.88 −0.347 m2 2 097.33

"

#

"

"

#

#

"

#

b) In the following we choose to use the mass normalized eigenvectors m φi . With the eigenfrequencies and eigenvectors determined, we can now solve the decoupled equation system 1 0 ω2 0 ¨ (t) + 1 y y(t) = 0 0 1 0 ω22

"

#

"

#

for each eigenform separately: y¨i (t) + ωi2 yi (t) = 0 .

(8.44)

The solution to the homogeneous differential equation for each yi (t) is given by yi (t) = yi1 cos (ωi t) − yi2 sin (ωi t) . For the given 2DOF system with eigenfrequencies ω1 and ω2 the solution reads y1 (t) = y11 cos (ω1 t) − y12 sin (ω1 t) y2 (t) = y21 cos (ω2 t) − y22 sin (ω2 t) . The four unknowns y11 , y12 , y21 , y22 can be determined from the initial conditions. Here, we choose w1 (0) = −1

w2 (0) = 1

w˙1 (0) = 0

w˙2 (0) = 0.

For that purpose the initial conditions need to be transformed to modal coordinates according to Eqs. (8.41) and (8.42) using the transformation matrix 0.110 0.069 T= Φ M= 0.022 −0.347 "

m

T

#"

80 0 8,772 0,552 = 0 8 1,747 −2,774 #

"

#

˙ Hence the initial conditions y(0) und y(0) relating to the eigenvectors are found

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by using Eqs. (8.41) and (8.42). "

#

−8,220 y(0) = Tw(0) = −4,521 0 . 0

" #

˙ ˙ y(0) = Tw(0) =

Note, that this transformation only holds, when the mass normalized eigenvectors are used. With the transformed initial conditions in the modal space, Eq. (8.44) can be solved for each mode shape separately for y(t). For the first mode shape, we determine the unknowns y11 und y12 by inserting the transformed initial conditions y(0) und y(0) ˙ into the solution of the homogeneous differential equation Eq. (5.9). y11 = −8.220 y12 = 0 . Analogously the coefficients C21 und C22 for the second mode shape are determined. y21 = −4.521 y22 = 0 . Thus the modal solutions read y1 (t) = y11 cos (ω1 t) + y12 sin (ω1 t) = −8.220 cos (0.962 t) y2 (t) = y21 cos (ω2 t) + y22 sin (ω2 t) = −4.521 cos (6.496 t) . The solution w(t) is obtained by w (t) 0.110 0.022 w(t) = 1 = m Φy(t) = w2 (t) 0.069 −0.347 "

#

"

#"

−8.220 cos (0.962 t) −4.521 cos (6.496 t)

#

−0.901 cos (0.962 t) − 0.099 cos (6.496 t) . = −0.568 cos (0.962 t) + 1.568 cos (6.496 t) "

#

We can easily identify the contributions of each mode shape to the displacement solution of the individual degree of freedom. The first summand in the first and second entry is the contribution of the first mode shape to the first and second degree of freedom, respectively. The second summand corresponds to the contributions of the second mode shape.

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c) Alternative solution The oscillation of the degrees of freedom w1 (t) und w2 (t) can also be directly given by Eq. (8.13). This procedure does not require the eigenvectors to be mass normalized. We still denote the unknowns as yij . w (t) Φ Φ12 w(t) = 1 = Φy(t) = 11 w2 (t) Φ21 Φ22 "

#

"

#"

y1 (t) y2 (t)

#

Φ (y cos (ω1 t) − y12 sin (ω1 t)) + Φ12 (y21 cos (ω2 t) − y22 sin (ω2 t)) = 11 11 . Φ21 (y11 cos (ω1 t) − y12 sin (ω1 t)) + Φ22 (y21 cos (ω2 t) − y22 sin (ω2 t)) "

#

The unknowns y11 , y12 , y21 , y22 are found using the initial conditions w(t = 0) ˙ = 0). The resulting equation systems read and w(t "

w1 (0) Φ11 −Φ12 = w2 (0) −Φ21 Φ22

"

w˙ 1 (0) Φ ω Φ12 ω2 = − 11 1 w˙ 2 (0) Φ21 ω1 Φ22 ω2

#

"

#

#"

y11 y21

"

#

#"

#

y12 . y22

Inserting the initial conditions w1 (0) = −1

w2 (0) = 1

w˙1 (0) = 0

w˙2 (0) = 0,

we can solve for the unknowns using eiter of the normalized eigenvectors. We again choose the mass normalized eigenvectors and obtain #−1 "

"

y11 0.110 −0.022 = y21 −0.069 −0.347

"

y12 0.110 · 0.69 0.022 · 42.2 =− y22 0.069 · 0.69 −0.347 · 42.2

#

#

"

"

#

−1 1

#"

Φ11 ωE1 Φ12 ωE2 Φ21 ωE1 Φ22 ωE2

#−1 " #

0 0 = . 0 0 " #

Since we used the mass normalized in both approaches we obtain the same results for the coefficients yij . Nevertheless, the second approach would also permit the use of otherwise scaled eigenvectors. Then the coefficients yij would differ, but the same result for w(t) would be achieved. d) The resulting amplitude is now depicted with respect to time for both degrees of freedom separately between t = 0 s (initial conditions) and t = 15 s.

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Figure 8.1: Solution for the 2DOF system. The left figure and right figure show the displacement solution for degree of freedom 1 and 2, respectively (—). Each figure gives the separate contributions from mode shape 1 (–·–) and mode shape 2 (- - -). It can be observed that the solution for DOF 1 is mainly governed by the contributions from mode shape 1, whereas the second DOF is predominantly governed by the second mode shape. The 2π 2π = 2π = 6.53 s and T2 = ω = 2π = 0.97 s. periods are T1 = ω 1 2 0.96 Hz

6.5 Hz

8.2.3 Damped MDOF systems Until now, we discussed the free vibration for the undamped MDOF system. In the following we investigate the influence of a damping matrix C 6= 0 on the solution strategy. Consider the equation of motion for the damped MDOF system without external loads: ¨ + Cw ˙ + Kw = 0. Mw

(8.45)

A major concern is the behavior of the damping matrix under the transformation to modal coordinates using the eigenvectors of the undamped problem φ, i.e., the form of the generalized damping matrix C∗ = ΦT CΦ.

(8.46)

For an arbitrary damping matrix C, in general, the generalized damping matrix C∗ is not diagonal, and thus the equations of motion do not decouple. Three strategies are usually introduced to cope with this problem. These are: • Solve the problem with the general damping matrix. This most general approach leads to a quadratic eigenvalue problem, which can be further augmented to a standard eigenvalue problem. The resulting eigenvectors and eigenvalues are complex. This approach has to be chosen, especially, if the spatial distribution of the damping differs

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significantly from the distribution of mass or stiffness (this holds especially for discrete damping elements like tuned mass dampers for example). • Assume a damping matrix that is diagonalizable. Whenever the generalized damping matrix C∗ = ΦT CΦ becomes diagonal, the equation system becomes decoupled and the single equations correspond to the equations of motion of the damped SDOF system. The simplest form of this kind of damping is called Rayleigh damping. • Project the damping matrix on each mode. With this approach, the damping matrix C is projected onto the modal space and only the diagonal terms are retained, i.e., only the terms (8.47)

ci = φTi Cφi

are used for the generalized damping matrix. All cross terms φTj Cφi for j 6= i are neglected. • Assume modal damping. With this approach, the generalized (modal) damping matrix C∗ is given directly. Each mode is assigned a damping value, thereby C∗ is directly diagonal. Often it is not possible to construct a damping matrix from a physical model. In this case it might be beneficial to assume damping values for the single modes directly. 8.2.3.1 The quadratic eigenvalue problem

In the case of a damped MODF system, the system of equations reads ¨ + Cw ˙ + Kw = 0. Mw

(8.48)

In the general case, the following ansatz holds w = φe λt ,

(8.49)

and thus the first two derivatives with respect to time are ˙ = λφe λt w

(8.50)

¨ = λ2 φe λt . w

(8.51)

We know insert Eq. (8.49) to Eq. (8.51) into Eq. (8.45) and obtain 



Mλ2 + Cλ + K φ = 0,

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(8.52)

8 Free vibration of linear systems with multiple degrees of freedoms

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which is commonly referred to as a quadratic eigenvalue problem. A nontrivial solution for φ is found for the case that the matrix (Mλ2 + Cλ + K) becomes singular and the determinant is zero 



det Mλ2 + Cλ + K = 0.

(8.53)

Eq. (8.53) is the characteristic equation of the problem. It is a polynomial with order 2n and gives 2n solutions for λ which are either all real or occur as complex conjugate pairs. Thus, we obtain 2n eigenvalues λ and corresponding eigenvectors φ. As for the undamped case, the eigenvectors are only determined up to an arbitrary factor. 8.2.3.2 The standard form of the damped eigenvalue problem

This derivation follows Humar [2012]. We introduce the a vector z of displacements and velocities (state space formulation) w z= ˙ w

!

(8.54)

and write the equations of motion (8.45) as ¨ + Cw ˙ + Kw = 0 Mw

(8.55)

˙ − Mw ˙ = 0. Mw

(8.56)

In matrix notation, this reads "

|

˙ w K 0 + ¨ w 0 −M

#" #

C M M 0 {z

:=A

}

"

|

#" #

{z

:=B

" #

w 0 = ˙ w 0

(8.57)

}

Hence, Eq. (8.57) is written as ˙ + Bz(t) = 0. Az(t)

(8.58)

A solution approach is given by z = φe λt . Inserted into Eq. (8.58) it reads (λA + B) φe λt = 0.

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(8.59)

8 Free vibration of linear systems with multiple degrees of freedoms

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Eq. (8.58) is a linear eigenvalue problem. This can further be transformed to a standard eigenvalue problem, by multiplying with the inverse of A, such that 



λI + A−1 B φe λt = 0.

(8.60)

Thereby Eq. (8.60) can be written as (λI − D) φe λt = 0.

(8.61)

with "

#

0 I D= . −1 −M K −M−1 C

(8.62)

We finally obtain the homogeneous equation system "

0 I λI − −M−1 K −M−1 C

#!

φ = 0.

(8.63)

This formulation is now a common eigenvalue problem. The resulting eigenvalues and eigenvectors are in general complex. Here, we get complex values for both the eigenvalues and the coordinates of the eigenvectors. Whereas vibrations in real mode shapes show for specific points in time for all nodes of the system zero crossings or maximum values at the same time, this is not the case for complex mode shapes. Since many available finite element codes cannot solve complex eigenvalue problems, in general, a direct integration using a time step method hast to be applied in this case (comp. Chapter 16).

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Example 8.3

The given 2DOF system is extended by one additional, viscous damper c2 , as depicted in the following figure. m1 m1 = 80 kg

w1 k1

m2 = 8 kg N k1 = 200 m N k2 = 125 m

c1

m2

c1 = 0

w2 k2

c2 = 0.6 Ns m

c2

The differential equation of the MDOF system with n degrees of freedom reads Mw(t) ¨ + Cw(t) ˙ + Kw(t) = 0.

(8.64)

For the given system the differential equation for the two degrees of freedom w1 and w2 are "

c1 −c1 k1 −k1 m1 0 ¨ ˙ w(t) + w(t) + w(t) = 0. −c1 c1 + c2 −k1 k1 + k2 0 m2 "

#

#

"

#

Again, the following initial conditions are given. w1 (0) = −1

w2 (0) = 1

w˙1 (0) = 0

w˙2 (0) = 0

We use the approach in Eq. (8.63), and introduce the vector of unknown state variables z, that includes displacements and velocities. w1 w  w   z= =  2 w˙1  w ˙ w˙2 



!

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Hence (8.64) can be rewritten as ˙ z(t) = Dz(t), or equivalently w˙ 1 " # w1 w  w˙  0 I  2  2  .  = −1 −1 w ¨1  −M K −M C w˙ 1  w˙ 2 w¨2 







With the approach z = φe λt the following homogeneous equation system is obtained "

"

##

0 I λI − −1 −M K −M−1 C

φ = 0.

The problem can be solved using MATLAB® via the polyeig command. The solution to the above equation system is given by the following four eigenvalues. λ1 = −3.61 · 10−2 + 6,50i λ2 = −3.61 · 10−2 − 6,50i λ3 = −1.43 · 10−3 + 0,96i λ4 = −1.43 · 10−3 − 0,96i . The eigenvalues are complex conjugate pairs. The corresponding eigenvectors can be calculated using Eq. (8.63). They are also complex conjugate pairs. The resulting eigenvectors from the polyeig command in MATLAB® are normalized such that they have unit-length. Here, we choose to give the eigenvectors such that the first entry is equal to one for better readability: 1,00  −15,88 + 0,12i    φ1 =    −0,04 + 6,50i  −0,11 − 103,15i 

1,00  −15,88 − 0,12i    φ2 =    −0,04 − 6,50i  −0,11 + 103,15i





1,00  0,63 + 0,69 · 10−3 i    φ3 =   −0,50 · 10−3 + 0,96i −0,97 · 10−3 + 0,61i 

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



1,00  0,63 − 0,69 · 10−3 i    φ4 =  . −0,50 · 10−3 − 0,96i −0,97 · 10−3 − 0,61i 



8 Free vibration of linear systems with multiple degrees of freedoms

146

The solution for the augmented state vector z is then written as the sum z = The unknowns zbj can be determined from the initial conditions:

bj φj e j=1 z

P2n

λj t

.

−1 zb1 Φ11 e λ1 0 + zb2 Φ12 e λ2 0 + zb3 Φ13 e λ3 0 + zb4 Φ14 e λ4 0 w1 (0)  1  zb Φ e λ1 0 + zb Φ e λ2 0 + zb Φ e λ3 0 + zb Φ e λ4 0  w (0)     1 21  2 22 3 23 4 24 2  = z(0) =   . =  0  zb1 Φ31 e λ1 0 + zb2 Φ32 e λ2 0 + zb3 Φ33 e λ3 0 + zb4 Φ34 e λ4 0  w˙1 (0) w˙2 (0) 0 zb1 Φ41 e λ1 0 + zb2 Φ42 e λ2 0 + zb3 Φ43 e λ3 0 + zb4 Φ44 e λ4 0 











The resulting vibration is finally depicted with respect to time for both degrees of freedom separately between t = 0 s (initial conditions) and t = 100 s. Note that due to the calculation approach the mode shape with higher eigenvalue is now the first mode shape as compared to the undamped case in Example 8.2.

Figure 8.2: Solution for the damped 2DOF system. The left figure and right figure show the displacement solution for degree of freedom 1 and 2, respectively (—). Each figure gives the separate contributions from mode shape 1 (–·–) and mode shape 2 (- - -). It can be observed that the solution for DOF 1 is mainly governed by the contributions from mode shape 2, whereas the second DOF is predominantly governed by the second mode shape for small values of t. Nevertheless, for larger t the response of mode shape 1 decays and the response is governed by mode shape 1. Due to the applied damping the response of mode shape 1 2π decays much more rapidly than for mode shape 2. The periods are T1 = ω = 2 2π 2π 2π = 0.97 s and T2 = ω1 = = 6.53 s. 6.5 rad Hz

0.96 rad Hz

8.2.3.3 Rayleigh damping

The equation of motion of an unloaded system with multiple degrees of freedom and viscous damping reads ¨ + Cw ˙ + Kw = 0. Mw

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(8.65)

8 Free vibration of linear systems with multiple degrees of freedoms

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We insert as ansatz the jth mode shape of the undamped MDOF system (modal ansatz), wj (t) = φj yj (t)

(8.66)

into Eq. (8.65) and obtain Mφj y¨j (t) + Cφj y˙ j (t) + Kφj yj (t) = 0.

(8.67)

Furthermore, we multiply with the transpose of the i-th eigenvector φi of the undamped system, from the left, φTi Mφj y¨j (t) + φTi Cφj y˙ j (t) + φTi Kφj yj (t) = 0. |

{z

δij

|

}

{z

ωj δij

(8.68)

}

For an arbitrary damping matrix the above equations are not decoupled anymore, since in general it holds φTi Cφj 6= Ci δij .

(8.69)

Only in the special case of C = αM + βK

(8.70)

the modal ansatz of the undamped system leads to decoupled equations in the damped case and thus a diagonalizable modal damping matrix: Mj y¨j + Cj y˙ j + Kj yj = 0

(8.71)

with Cj = αφTj Mφj + βφTj Kφj = αMj∗ + βωj2 Mj∗ .

(8.72)

In this case the damping is mass and/or stiffness proportional. It is called Rayleigh damping. The relationship between the Rayleigh damping coefficients α and β and the modal damping ratio Dj of the j-th eigenmode is given by (compare Tab. 5.1): 1 Cj Dj = q = = 2Mj ωj 2 2 Kj Mj Cj

α + βωj ωj

!

.

(8.73)

The stiffness proportional part increases for increasing eigenfrequencies ωj , whereas the mass proportional part decreases for increasing eigenfrequencies ωj , as it is proportional to its inverse

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1 . ωj

In the Rayleigh-damping model the free parameters α and β have to be chosen. A possibility is to choose desired damping values for two of the mode shapes φi and φj with eigenfrequencies ωi and ωj , respectively, say Di and Dj , and adjust the values for α and β such that 1 α + βωi Di = 2 ωi 

1 Dj = 2



α + βωj ωj

(8.74)

!

(8.75)

.

This leads to the following equation solution for the unknowns α and β. " #

α =2 β

"1 ωi 1 ωj

ωi ωj

#−1 "

#

Di . Dj

(8.76)

Dj

Dj α 2 ωj β ω 2 j

ωj Figure 8.3: Rayleigh damping

The decoupled equations of motion for systems with Rayleigh-damping can be solved with the methods already introduced for SDOF systems.

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8.2.3.4 Caughey damping model using additional orthogonality relations

According to Humar [2012], the following orthogonality conditions 

φTi M M−1 K

b

φj =

 0 λb i

i 6= j i=j

(8.77)

hold for all b ∈ Z. Thus, for any damping matrix C that is represented by the following linear combination C=

X



αb M M−1 K

b

(8.78)

b

the equations of motion become decoupled under the transformation into modal coordinates. This damping model is called Caughey damping model. The Rayleigh damping model is included as a special case in Eq. (8.78) for b = 0,1, i.e 

C = α0 M M−1 K

0

+ α1 MM−1 K = α0 M + α1 K

(8.79)

and α0 = α and α1 = β. With this model, damping ratios can be specified for more than two modes, as is the case for the Rayleigh damping model.

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9 Forced vibration of linear systems with multiple degree of freedoms Depending on the time structure of the load and the assumed damping model, different methods are applicable to determine the response of the forced multiple degree of freedom system. We introduce the direct solution, modal analysis and time step procedures hereinafter.

9.1 Direct solution for harmonic loads In case of an harmonic excitation, that is an excitation varying with a sine/cosine function in time, a direct solution for the response can be obtained by inversion of the dynamic stiffness matrix. This procedure is applicable for cases, where either, the solution is only needed for a single frequency, or a more general time dependence can be represented by the sum of harmonic functions (Fourier series, Fourier integral). Consider the general linear, damped MDOF system with the equation of motion. ¨ + Cw ˙ + Kw = f Mw

(9.1)

These are n coupled differential equation given in terms of the n unknown displacements w and their corresponding derivatives. Under the assumption of time harmonic excitation, we obtain f (t) = f 01 cos(Ωt) − f 02 sin(Ωt) = 



(9.2)

= f 0 cos Ωt + ϕf =

(9.3)

= bf + e iΩt + bf − e −iΩt .

(9.4)

Using the complex exponential approach in Eq. (9.4) simplifies the following derivations. For that reason, the same approach is made for the unknown displacements b + e iΩt + w b − e −iΩt . w(t) = w

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(9.5)

151

The first two derivatives with respect to time are then given by b + ) e iΩt + (−iΩw b − ) e −iΩt ˙ = (iΩw w 





(9.6)



b + e iΩt + −Ω2 w b − e −iΩt . ¨ = −Ω2 w w

(9.7)

Inserting Eqs. (9.5) to (9.7) and (9.4) into Eq. (9.1), we obtain 







b + e iΩt + −Ω2 w b − e −iΩt + C iΩw b + e iΩt − iΩw b − e −iΩt + M −Ω2 w 



b + e iΩt + w b − e −iΩt = f + e iΩt + f − e −iΩt . (9.8) +K w

The parts with positive and negative frequencies are then treated separately, thus 















b + e iΩt + C iΩw b + e iΩt + K w b + e iΩt = f + e iΩt M −Ω2 w 







b − e −iΩt + C −iΩw b − e −iΩt + K w b − e −iΩt = f − e −iΩt . M −Ω2 w

(9.9) (9.10)

Reorganizing the different terms leads to 

b + e iΩt = f + e iΩt −Ω2 M + iΩC + K w



(9.11)



b − e −iΩt = f − e −iΩt . −Ω2 M − iΩC + K w



(9.12)

The time functions e iΩt and e −iΩt can be cancelled from both sides in both the equations. As b + and w b − are complex conjugate, the solution only needs to be calculated for one of the w two. The solution for the complex coefficients is given by 

−1



−1

b + = −Ω2 M + iΩC + K w

b − = −Ω2 M − iΩC + K w

f+

(9.13)

b ∗+ . f− = w

(9.14)

The matrix −Ω2 M + iΩC + K is called the dynamic stiffness matrix Kdyn . Its inverse K−1 dyn = Rw is called the receptance matrix (flexibility matrix). If the solution is sought for different frequencies, the receptance matrix has to be calculated for each frequency separately. This can become very time consuming for large problems, especially because the dynamic stiffness matrix is complex. A different approach to the harmonic vibration problem is the application of the modal superposition.

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Example 9.1: Two degree of freedom system (cont)

We continue example 8.2 and apply a harmonic forces to the degrees of freedom. m1

f1 (t) m1 = 80 kg

w1 k1

m2 = 8 kg N k1 = 200 m N k2 = 125 m

c1 f2 (t)

m2

c1 = 0

w2 k2

c2 = 0.6 Ns m

c2

The loads p1 (t) and p2 (t) are now harmonic forces, such that f (t) f f f (t) = 1 = 1+ e iΩt + 1− e −iΩt = f + e iΩt + f − e −iΩt f2 (t) f2+ f2− !

!

!

We only consider the parts with positive frequency f + e iΩt since the parts with negative frequencies can be found as the complex conjugates of the former. In this example we choose f1+ = 1 and f2+ = 0. We consider the system without and with damping. Subsequently, the solution is depicted for both degrees of freedom. Real part, imaginary part, absolute value and phase are given.

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Figure 9.1: Displacement w1+ (—) and w2+ (–·–) for the 2DOF system under harmonic excitation, direct solution, undamped system.

Figure 9.2: Displacement w1+ (—) and w2+ (–·–) for the 2DOF system under harmonic excitation, direct solution, damped system.

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9.2 Modal analysis The modal analysis is based on the superposition of eigenmodes (eigenvectors) φi , that are scaled with time-dependent factors yi (t). It is only applicable for linear vibrations, since only then the principle of superposition is valid. For systems with a diagonalizable damping matrix, the displacement vector of a MDOF system is then written as w(t) =

n X

yi (t)φi = Φy(t)

(9.15)

i

with the matrix of eigenvectors Φ = [φ1 , φ2 , . . . , φn ]

(9.16)

obtained from the free vibration problem of the undamped system and the coefficients y(t) = [y1 (t), y2 (t), . . . , yn (t)]T .

(9.17)

The equation of motion for the MDOF system including external forces reads ¨ + Cw ˙ + Kw = f (t) . Mw

(9.18)

We assume that the eigenvectors and -values are known from a previous solution of the undamped, homogeneous problem (comp. section 8.2). Inserting Eq. (9.15) into Eq. (9.18), we obtain MΦ¨ y + CΦy˙ + KΦy = f .

(9.19)

Then, multiplying Eq. (9.19) with ΦT , it follows ΦT MΦ¨ y + ΦT CΦy˙ + ΦT KΦy = ΦT f

(9.20)

or written separately for each equation φTj M

n X i

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!

φi y¨i +

φTj C

n X i

!

φi y˙ i +

φTj K

n X

!

φi yi = φTj f .

(9.21)

i

9 Forced vibration of linear systems with multiple degree of freedoms

155

Reminder 9.1

From section 8.2 we know that the following relations hold ΦT MΦ = M∗ ΦT KΦ = K∗ = ΛM∗ where Λ is the diagonal matrix of squared eigenfrequencies. In the case of mass normalized eigenvectors this further reduces to ΦT MΦ = I ΦT KΦ = Λ

We furthermore require that the damping matrix is diagonalizable, i.e., C∗ is a diagonal matrix, which can, e.g., be obtained by Rayleigh damping or a modal damping approach. With this Eq. (9.20) reads ¨ + C∗ y˙ + K∗ y = ΦT f . M∗ y Thereby all mixed terms vanish from Eq. (9.20), and the equations of motion become decoupled. We can obtain a notation for the separate equations by writing Eq. (9.21) as φTj Mφj y¨j + φTj Cφj y˙ j + φTj [K] φj yj = φTj p . |

{z

Mj

}

| {z } Cj

|

{z

Kj

}

| {z } fj

Thus, n different and decoupled ordinary differential equations in the form of the SDOF differential equation are obtained. Mj y¨j (t) + Cj y¨j (t) + Kj yj (t) = fj . Therein are Mj = φTj Mφj

generalized (modal) mass

Cj = φTj Cφj = 2Dj ωj Mj

generalized (modal) damping

Kj = φTj Kφj = ωj2 Mj

generalized (modal) stiffness

pj = φTj f

generalized load

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Remember that this derivation of decoupled equations only holds under the assumption that the damping matrix is diagonalizable. In the special case of mass-normalized eigenvectors the differential equations simplify to y¨j + 2Dj ωj y˙ j + ωj2 yj = fj . Depending on the form of fj (t), the solution for each yj (t) can then be found by any of the methods for the forced SDOF  systemdiscussed in chapter 7. For example, in case of a harmonic excitation f (t) = f 0 cos Ωt + ϕf , all modal forces will also be harmonic. Then, the steady state (particular) solutions for single modes can be easily obtained by application of the amplification function V (η). In the case of transient loads, the response can be determined by the Duhamel integral. The physical solution w is finally obtained by inserting the solution y for in Eq. (9.15). Example 9.2: Two degree of freedom system (cont.)

We continue example 8.2 and apply two step loads to the two degrees of freedom, respectively. m1

f1 (t) w1

k1 m2

f2 (t)

m1 = 80 kg m2 = 8 kg N k1 = 200 m N k2 = 125 m

w2 k2

MΦ¨ y(t) + KΦy(t) = f (t) By left multiplication with ΦT we obtain ¨ (t) + Λy(t) = ΦT f (t) y

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The load is given by f1 (t) f U (t − t1 ) = 10 f2 (t) f20 U (t − t2 )

"

#

"

#

where U (t) is the unit-step function  1

U (t) = 

for t ≥ 0 . for t < 0

0

The mass normalized eigenvectors of the system that are derived in example 8.2 read m

0.110 φ1 = 0.069

m

0.022 φ2 = . −0.347

"

#

"

#

The generalized load thus results in Φ f (t) = Φ f (t) = m 11 Φ21 "

∗

m

T

m

Φ12 m Φ22

m

#T "

f1 (t) = f2 (t) #

Φ11 p1 (t) + m Φ21 f2 (t) . m Φ12 p1 (t) + m Φ22 f2 (t)

"

m

#

Since the equations of motion decouple under the transformation to the modal coordinates, the Duhamel-integral can be applied. As the decoupled equations correspond the the differential equation of the SDOF system, the impulse response function of the undamped SDOF system, given by hj (t) = q

1 kj mj

sin (ωj t) =

1 sin (ωj t) ωj

can be used in the integral equation directly: yj (t) =

Z∞ −∞

∞ 1 Z fj (τ )hj (t − τ ) dτ = fj (τ ) sin(t − τ ) dτ . ωj −∞

Thus: ∞ 1 Z y1 (t) = f1 (τ ) sin(ω1 (t − τ )) dτ = ω1 −∞

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∞ 1 Z m = ( Φ11 f1 (t) + m Φ21 f2 (t)) sin(ω1 (t − τ )) dτ = ω1 −∞

∞ 1 Z m = ( Φ11 f10 U (t − t1 ) + m Φ21 f20 U (t − t2 )) sin(ω1 (t − τ )) dτ = ω1 −∞

∞ 1 Z m = Φ11 f10 U (t − t1 ) sin(ω1 (t − τ )) dτ + . ω1 −∞

∞ 1 Z m + Φ21 f20 U (t − t2 ) sin(ω1 (t − τ )) dτ ω1 −∞

Since U (t − t1 ) = 0 and U (t − t2 ) = 0 for t < t1 and t < t2 , respectively, it holds ∞ ∞ 1 Z m 1 Z m Φ11 f10 sin(ω1 (t − τ )) dτ + Φ21 f20 sin(ω1 (t − τ )) dτ = y1 (t) = ω1 ω1 t1

= =

m

m

t2

1 Φ11 f10 − cos(ω1 (t − τ )) ω1 ω1

t



t1

+

m

1 Φ21 f20 − cos(ω1 (t − τ )) ω1 ω1 

t

=

t2

m Φ21 f20 Φ11 f10 [1 − cos(ω (t − t ))] + [1 − cos(ω1 (t − t2 ))] . 1 1 2 ω1 ω12

For the second modal coordinate y2 (t) we obtain analogously y2 (t) =

m

m Φ12 f10 Φ22 f20 [1 − cos(ω (t − t ))] + [1 − cos(ω2 (t − t2 ))] . 2 1 2 ω2 ω22

The solution in terms of the degrees of freedom w(t) is then given by y (t) w(t) = Φy(t) = Φ 1 = φ1 y1 (t) + φ2 y2 (t) . y2 (t) "

#

We calculate the response for f10 = 10, f20 = −5, t1 = 1 t2 = 3. The results are depicted in the following figure:

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Figure 9.3: Solution for the 2DOF system under two step loads. The left figure and right figure show the displacement solution for degree of freedom 1 and 2, respectively (—). Each figure gives the separate contributions from mode shape 1 (–·–) and mode shape 2 (- - -). The load application times are marked by vertical lines (|) It can be observed that the solution for both DOFs is mainly governed by the 2π = 2π = 6.53 s contributions from mode shape 1. The periods are T1 = ω 1 and T2 =

2π ω2

=

2π 6.5 Hz

0.96 Hz

= 0.97 s.

9.3 Time integration methods If the application of a diagonal damping matrix is not valid and the excitation cannot be split in single harmonic parts, the two methods presented above cannot be applied. Then, numerical time integration methods have to be applied. Different methods can be applied, such as difference methods, Crank-Nicholson, Houboldt-, Newmark, or generalized α methods. Some of these are presented in chapter 16. These procedures can become computationally very expensive, as the full system has to be solved in every time step.

9.4 Summary • Vibrations of system with n degrees of freedom under arbitrary, time dependent excitation can be described with the help of the modal analysis. Thereby the system is decoupled into n SDOF systems, which can be solved for separately and then superposed to obtain the complete solution. The resulting parameters of the single SDOF systems are called generalized quantities (generalized stiffness, mass, damping, and load). They can be interpreted by the kinetic and potential energy and the externally introduced work. • The number of degrees of freedom that describe the systems’ behavior is significantly smaller than the absolute number of DOFS n. This can be explained by the fact, that

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the stiffness term ωj2 Mj strongly increases for higher frequent modes. An exception to this is the case, where the system is excited at resonance. • For arbitrary damping, the generalized damping matrix is not diagonal in the modal coordinates. Only in the case, where the damping matrix is given as a specific combination of mass and stiffness matrix, diagonality is given. C=

X



αb M M−1 K

b

b

An often applied form of this damping model is the Rayleigh damping, which is achieved for b = {0,1}: CRayleigh = αM + βK The damping factors α and β of the Rayleigh damping are adopted to realistic damping, as far as the damping is distributed continuously over the system. In many cases, e.g. for single damper elements, an adoption via the Rayleigh damping is not possible. Then, a complex eigenvalue analysis can be performed. • With the help of the Fourier transformation, arbitrary loads, as presented for the SDOF system, can be transformed to harmonic loads, that then act on the different eigenmodes.

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10 Impedance models With the improvement of measurement capabilities, it has become common to characterize parts of a dynamic system by means of their dynamic stiffness (or impedance in general) and receptance (or admittance in general) matrix and to use a dynamic substructuring approach for the analysis of the dynamic behavior of the parts separately prior to the coupling. This reduces the computational time and improves the repeatability of numerical analysis. In vibroacoustics, the most used quantity to describe the vibrational behavior of a system or a subsystem is the impedance, which is defined as the ratio of the force spectrum to the resulting velocity spectrum at the loading point and describes how much a structural part resists to motions when subjected to a harmonic force. Impedance models provide an alternative simplified technique to FEM or analytical solutions for vibroacoustic predictions. Impedance modeling is a powerful tool for the simplified prediction of traffic-induced vibrations, or for the design with respect to acoustic requirements that require sophisticated regarding vibrations. They are also used in noise, vibration, and harshness control for cars and airplanes.

10.1 Preliminary definitions In the following we assume small amplitudes and linear systems. Time harmonic (sine/cosine) displacements w(t) result from harmonic forces f (t). The system response will be characterized by the velocity v(t) at the point of excitation. Furthermore, the force is defined in the same direction as the velocity (comp. Fig. 10.1a). A dynamic reaction force r(t) occurs in the system, which may result from different mechanisms. It is in equilibrium with the load f (t).

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f (t)

w,v r(t)

(a) Dynamic Equilibrium

(b) Impedance hammer

Figure 10.1: Reaction force under harmonic load

For the following considerations we choose the advantageous description using complex numbers f (t) = fb+ e iΩt + fb− e −iΩt 1 1 = (f01 + if02 )e iΩt + (f01 − if02 )e −iΩt = f01 cos (Ωt) − f02 sin (Ωt) 2 2 iΩt v(t) = w(t) ˙ = vb+ e + vb− e −iΩt

(10.1)

1 1 = (v01 + iv02 )e iΩt + (v01 − iv02 )e −iΩt = v01 cos (Ωt) − v02 sin (Ωt) 2 2

(10.2)

where Ω = 2π is the circular frequency and fb+ and fb− as well as vb+ and vb− are complex T conjugate to each other.

1 f 2 02

Im

fb+ Re

ϕf

1 v 2 02

Im −ϕf

1 f 2 01

1 f 2 01

Re

Im

Im vb+

ϕv

Re 1 v 2 01

− 12 f02

1 v 2 01

Re

−ϕv

fb− − 12 v02

vb−

Figure 10.2: Complex pointers fˆ+ and fˆ− as well as vˆ+ and vˆ−

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fmax vmax

f (t) v(t) t

T Figure 10.3: Physical quantities f (t) and v(t). One period of the harmonic oscillation is depicted, where f (t) and v(t) show an arbitrary phase shift ∆ϕ.

After applying an integration or differentiation to the velocity v(t), the vibration can also be expressed in terms of the displacement w(t) or the acceleration a(t) = w(t), ¨ respectively. 1 1 vb+ e iΩt + − vb− e −iΩt = wb+ e iΩt + wb− e −iΩt iΩ iΩ a(t) = w(t) ¨ = iΩvb+ e iΩt + (−iΩvb− ) e −iΩt = ab+ e iΩt + ab− e −iΩt 

w(t) =



(10.3) (10.4)

where the complex pointers are 1 1 vˆ+ = −i vˆ+ iΩ Ω 1 1 wˆ− = − vˆ− = i vˆ+ iΩ Ω a ˆ+ = iΩˆ v+ wˆ+ =

a ˆ− = −iΩˆ v− .

(10.5) (10.6) (10.7) (10.8)

The input power inserted at the the excitation point is given by P (t) = f (t)v(t).

(10.9)

Thus, the work within one period (one cycle of the load) is W = =

Z T 0

Z T 0

P (t) dt (fˆ+ e iΩt + fˆ− e −iΩt )(ˆ v+ e iΩt + vˆ− e −iΩt ) dt

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=

Z T 0

fb+ vb+ e 2iΩt + fb− vb− e −2iΩt + fb+ vb− e 0 + fb− vb+ e 0 dt .

(10.10)

The terms with e 2iΩt and e −2iΩt integrate to zero. Thus it follows 



W = fb+ vb− + fb− vb+ T.

(10.11)

We can also represent the complex numbers fb+ and vb+ as







fb+ = fb+ e iϕf , fb− = fb− e −iϕf

(10.12)

vb+ = |vb+ |e iϕv , vb− = |vb− |e −iϕv .

(10.13)

Inserting Eqs. (10.12) and (10.13) into Eq. (10.11), we obtain





  W = fb+ |vb− |e iϕF e −iϕv + fb− |vb+ |e −iϕF e iϕv T







= fb+ |vb+ | e i(ϕv −ϕF ) + e −i(ϕv −ϕF ) T. |

{z

=2 cos(ϕv −ϕF )







(10.14)

}

Here we use the fact that fb+ = fb− and |vb+ | = |vb− |. Finally, we obtain W =

Z T 0





P (t) dt = 2 fb+ |vb+ |T cos (ϕv − ϕF ) .

(10.15)

This result for the work W is depicted in Fig. 10.4 in dependency of ϕv −ϕF . We can distinguish between regions where the work performed is positive or negative W positive work π negative work

2π

3π

4π

ϕv − ϕF

Figure 10.4: Work performed by the force f (t).

π π ϕv − ϕF ∈ − , 2 2 

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→ positive work is performed on the system

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π 3π ϕv − ϕF ∈ , 2 2 



→ negative work is performed on the system

We can thus state: hIf the iphase shift between velocity v(t) and the force f (t) is within the range (ϕv − ϕF ) ∈ − π2 , π2 the force, which is in equilibrium with the reaction force in the system, carries out positive work. This work has got a maximum for ϕv = ϕF and it is equal to zero for ϕv − ϕF = ± π2 . Due to the requirement of a stationary harmonic vibration, which is defined in Eq. (10.2), this work has to hbe “consumed” by the system in each period. For i π 3π the phase shift in the range (ϕv − ϕF ) ∈ 2 , 2 , we observe negative work. Thus, an energy source would have to exist within the system.

10.2 Dynamic stiffness and impedance In statics the quotient of exciting force and displacement corresponds to the stiffness (“spring”). A dynamic equivalent is found for kb+ =

fb+ wb+

(10.16)

kb− =

fb− . wb−

(10.17)

Here, kˆ+ and kˆ− is the dynamic stiffnesses, also denoted by kdyn (Ω). The quotient of the force and the velocity at the excitation point is called impedance Z(Ω). Z(Ω) = zb+ =

fb+ vb+

Z(−Ω) = zb− =

(10.18)

fb− . vb−

(10.19)

It can be shown that both parts are complex conjugates of each other. F+∗ F− b z− = = ∗ = v− v+

F+ v+

!∗ ∗ = zb+ .

(10.20)

Impedance and dynamic stiffness are, in general, quantities that depend on the frequency and can be determined with the help of measurements. For this purpose we have to simultaneously measure force and velocity, displacement, or acceleration, respectively. Similarly, the quotient of force and acceleration can be computed, which is called dynamic mass (in German, “Reaktanz”). The inverse of the impedance is called mobility (in German, “Admittanz”); the inverse

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of the dynamic stiffness is called receptance (flexibility) (in German, “Nachgiebigkeit”); the inverse of the dynamic mass is called accelerance (in German, “Inertanz”).

10.3 Dynamic stiffnesses and impedances in passive systems In the following we only consider system, where forces insert mechanical work, that is dissipated by the (passive) system. The system can only dissipate energy, but not cannot generate energy, e.g., by an internal energy source. Thus, no work can be done on the exciting forceh by the i system, and the phase shift between velocity and force is in the range (ϕv − ϕF ) ∈ − π2 , π2 . In the following the reaction forces r(t) that are in equilibrium with the external load f (t) are discussed depending on the displacements at the excitation point (comp. Fig. 10.1). In contrast to statics it is important, which phase shifts occur between force and velocity. These relations are expressed with dynamic stiffnesses or impedances: f (t) = fb+ e iΩt + fb− e −iΩt = kb+ wb+ e iΩt + kb− wb− e −iΩt = zb+ vb+ e iΩt + zb− vb− e −iΩt .

(10.21)

10.3.1 Mass impedance According to Newton’s law, the inertial force of a single mass point is ¨ =f = mw

X

(10.22)

fn .

n

Here, f is the resultant of all forces f n acting on a point with a mass m. This equation can be rearranged as X

¨ = 0. f n − mw

(10.23)

n

This equation constitutes the “dynamic equilibrium” for an accelerated system, as the mass is accelerated. This equilibrium corresponds to the static equilibrium equation for masses ¨ moving at a constant speed except for the term −mw. X

fn = 0 .

(10.24)

n

¨ is directed in the opposite direction of the acceleration w. ¨ The reaction force −mw

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f (t) w,w, ˙ w¨

m m w(t) ¨

Figure 10.5: Mass excited by dynamic force.

In case of a harmonic force f (t) = fb+ e iΩt + fb− e −iΩt acting on the mass Eq. (10.23) results in f (t) = mw(t) ¨ .

(10.25)

Thus, with Eq. (10.4), it follows 



f (t) = fb+ e iΩt + fb− e −iΩt = m iΩvb+ e iΩt + (−iΩvb− ) e −iΩt .

(10.26)

Thus, fb+ and fb− are given by fb+ = imΩvb+

(10.27)

fb− = −imΩvb− .

(10.28)

The force with positive exponential undergoes a phase shift relative to the velocity due to the multiplication with the imaginary unit i, which is ϕf+ − ϕv+ = π2 corresponding to 90◦ . For the part with negative exponential the phase shift is ϕf− − ϕv− = − π2 due to multiplication with −i Thus, the force f (t) is “hurrying on ahead” of the velocity v(t). For this case, according to Eq. (10.14), no work is done on the system, and thus no energy is consumed. This can also be observed in Fig. 10.4. The dynamic stiffnesses in Eqs. (10.16) and (10.17) therefore read under consideration of Eqs. (10.3), (10.5), and (10.6), kb+ =

imΩvb+ fb+ = = −mΩ2 wb+ −i Ω1 vb+

(10.29)

kb− =

fb− −imΩvb− = = −mΩ2 . wb− i Ω1 vb−

(10.30)

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The impedances in Eqs. (10.18) and (10.19) under consideration of Eq. (10.3) follow to zb+ =

fb+ imΩvb+ = = iΩm vb+ vb+

(10.31)

zb− =

fb− −imΩvb− = = −iΩm. vb− vb−

(10.32)

10.3.2 Conservative forces 10.3.2.1 Potential

Forces performing work that can be explicitly characterized by the initial and the endpoint of the performance, have a potential Π , i.e., the work, which is done by the load on the system, is independent from the loading process and it only depends on the initial and the endpoint of the loading process. In case of forces that can be defined by means of a potential, the work due to a loading and subsequent relaxation to the initial state is equal to zero. An example of a force, which can be defined with the help of a potential, is the resulting force Fk of a linear elastic spring Fk = −kw w

(10.33)

where kw is the spring-stiffness. Another example is the resulting moment of a torsional spring Mγ = −kγ γ where kγ is the stiffness of the rotational spring. The minus sign indicates (comp. Fig. 10.1) that the spring force is acting against the displacement at the excitation point of the external load (the moment of the rotational spring is acting against the rotation) and is in equilibrium with the external load (with the external moment): fk + f = 0

(10.34)

−kw w + f = 0 .

(10.35)

In case of a longitudinal spring, the potential is defined as 1 Π = kw w2 . 2

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(10.36)

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The reaction force can be deduced from the potential f =−

δΠ = −kw w . δw

(10.37)

10.3.2.2 Spring impedance

fb+ e iΩt + fb− e−iΩt kw

wb+ e iΩt + wb− e −iΩt

Figure 10.6: Spring under harmonic force.

In case of an external harmonic force f (t) = fb+ e iΩt + fb− e −iΩt acting on a spring, the resulting force is obtained from the linear relation: f (t) = kw w(t) .

(10.38)

Thus, with Eq. (10.3), it follows f (t) =

+e

fb

iΩt

+ fb

−e

−iΩt

= kw



1 1 vb+ e iΩt + − vb− e −iΩt . iΩ iΩ 





(10.39)

Then fb+ and fb− are given by: kw vb+ Ω

(10.40)

kw vb+ . Ω

(10.41)

fb+ = −i fb− = i

This implies that the force with positive exponential undergoes a phase shift with respect to the velocity of ϕf+ − ϕv+ = − π2 corresponding to −90◦ . For the part with negative exponential the phase shift is ϕf− − ϕv− = π2 . That is, the force f (t) is “lagging behind” the velocity v(t). In this case, similar to the case where the force acts on a mass, no work W is done on the system (according to Eq. (10.14)) and no energy is dissipated. The values for kb+ and kb− under

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consideration of Eqs. (10.16) and (10.17) are given by kb+ =

−i kΩw vb+ fb+ = = kw wb+ −i Ω1 vb+

(10.42)

kb− =

i kw vb+ fb− = Ω1 = kw . wb− i Ω vb+

(10.43)

They are equal to the spring stiffness kw and thus positive and real numbers. Analogously, we obtain the impedances from Eqs. (10.18) and (10.19): zb+

−i kΩw vb+ fb+ kw = = = −i b b v+ v+ Ω

(10.44)

i kw vb− kw fb− = Ω =i . vb− vb− Ω

(10.45)

zb− =

10.3.3 Non-conservative forces 10.3.3.1 Damping impedance

For the viscous damper (Newton-Modell) a linear relation between force and velocity is given f (t) = cv(t) = cw(t) ˙

(10.46)

where c is the constant of the viscous damper. The damping force is acting oppositely to the velocity v(t). fb+ e iΩt + fb− e−iΩt

c

wb+ e iΩt + wb− e −iΩt Figure 10.7: Viscous damper under harmonic force.

For a stationary harmonic vibration it follows 



f (t) = fb+ e iΩt + fb− e −iΩt = c vb+ e iΩt + vb− e −iΩt .

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(10.47)

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Then, fb+ and fb− are given by fb+ = cvb+

(10.48)

fb− = cvb− .

(10.49)

For both force terms, no phase shift occurs between force and velocity. The values for kb+ and kb− under consideration of Eqs. (10.16) and (10.17) follow to fb+ cvb+ = iΩc = 1 wb+ vb iΩ +

(10.50)

cvb− fb− = 1 = −iΩc . wb− − iΩ vb−

(10.51)

kb+ = kb− =

Hence for viscous damping without a spring frequency proportional, purely imaginary dynamic stiffness is obtained. Analogously, we obtain the impedances from Eqs. (10.18) and (10.19): fb+ cvb+ = =c vb+ vb+

(10.52)

fb− cvb− = = c. vb− vb−

(10.53)

zb+ = zb− =

The impedance is equal to the viscous damping coefficient and thus purely real. 10.3.3.2 Loss factor

For passive systems the complex dynamic stiffness h ican only lead to a phase shift between π π velocity and force in the range (ϕF − ϕv ) ∈ − 2 , 2 . Except for the limit cases − π2 and π2 energy is dissipated. The forces are written in terms of their dynamic stiffnesses and the velocity. 1 = + −i vb+ + = Ω   1 fb− = kb− wb− = kb− i vb− . Ω 

fb

kb+ wb+

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kb



(10.54) (10.55)

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Thus, we obtain for the dynamic stiffness f+ v+ f01 + if02 = iΩ v01 + iv02 f01 v01 + f02 v02 + i (f02 v01 − f01 v02 ) = iΩ 2 2 v01 + v02 Ω = 2 [(f01 v02 − f02 v01 ) + i (f01 v01 + f02 v02 )] 2 v01 + v02

kb+ = iΩ

f01 v02 − f02 v01 f01 v01 + f02 v02 =Ω 1+i 2 2 v01 + v02 f01 v02 − f02 v01 "

#

(10.56)

and f− v− f01 − if02 = −iΩ v01 − iv02 f01 v01 + f02 v02 + i (f01 v02 − f02 v01 ) = −iΩ 2 2 v01 + v02 Ω =− 2 [(f02 v01 − f01 v02 ) + i (f01 v01 + f02 v02 )] 2 v01 + v02

kb− = −iΩ

f01 v01 + f02 v02 f01 v02 − f02 v01 =Ω 1−i . 2 2 v01 + v02 f01 v02 − f02 v01 "

#

(10.57)

The dynamic stiffness kdyn = kb+ is also termed as “complex spring”. It is often represented as follows 











kdyn = Re kb+ + i Im kb+ = Re kb+ (1 + iη) .

(10.58)

The parameter η=

Im(kb+ ) f01 v01 + f02 v02 = . f01 v02 − f02 v01 Re(kb+ )

(10.59)

is called loss factor (comp. Table 5.1). For coupled systems, the real part of the complex spring kdyn results from spring stiffnesses and masses. The imaginary part of kdyn is decisive for the energy dissipation. The loss factor η can be explained with the help of the model depicted

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in Fig. 10.8a. In a parallel arrangement of a spring with a stiffness k and a viscous damper with a damping coefficient c a complex spring stiffness with a frequency independent real part (resulting from the spring) as well as a frequency-dependent imaginary part (resulting from the damper) results from the sum of both the dynamic stiffnesses. The hysteresis shows the shape depicted in Fig. 10.8b. f fmax

fmax wmax

f (t) w(t)

w

c

kw

wmax

t

f 2π (a) System

(b) Hysteresis und time history Figure 10.8: Voigt-Kelvin-Modell, Petersen [1996].

According to Eq. (10.11), the work dissipated in one period is 



Wdiss = fb+ vb− + fb− vb+ T.

(10.60)

1 1 1 1 (f01 + if02 ) (v01 − iv02 ) + (f01 − if02 ) (v01 + iv02 ) T 2 2 2 2 1 = [f01 v01 − if01 v02 + if02 v01 + f01 v01 + if01 v02 − if02 v01 ] T 4 1 = (f01 v01 + f02 v02 ) T . 2

=





(10.61) (10.62) (10.63)

On the other hand, for a linear elastic spring with a spring stiffness kw = Re(kdyn ) = −f02 v01 Ω f01 vv02 (Eq. (10.56)), the energy Welast stored in the spring is for the maximum dis2 2 01 +v02 placement: 1 1 2 2 Welast = kw wmax = Re(kdyn )wmax . 2 2

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(10.64)

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Under consideration of Eq. (10.5) it follows wmax = 2|wb+ | =

v b+ 2

iΩ

=

2|vb+ | , Ω

(10.65)

and thus 1 f01 v02 − f02 v01 |vb+ |2 4 2 = Welast = Ω 2 Ω (2|vb+ |)2 =

1 (f01 v02 − f02 v01 ) . 2Ω

Hence the ratio Wdiss = Welast

(10.66)

1 2

Wdiss Welast

(10.67)

reads

(f01 v01 + f02 v02 ) T (f01 v01 + f02 v02 ) T f01 v01 + f02 v02 = T = 2π = 2πη f01 v02 − f02 v01 (f01 v02 − f02 v01 ) (f01 v02 − f02 v01 ) 2π

1 2Ω

|

{z

=η

(10.68)

}

where we identified η as in Eq. (10.59). Thus, the value 2πη gives the ratio of dissipated to restorable mechanical energy.

10.4 Parallel and serial connection The representation of mechanical quantities using complex numbers provides an elegant possibility to describe steady state vibrations of mechanical systems under harmonic loads. As discussed in Section 7.2.1.2, steady state implies that the transient effects due to initial conditions and activation operations already have decayed. Eqs. (10.32), (10.45), and (10.53) define the impedances of a single mass, a spring, and a viscous damper. For each of these elements the impedance describes the complex ratio of the force fˆ+ and the velocity vˆ+ at the excitation point. Assembling subsystems consisting of the already mentioned elements mass, spring, and damper in serial and parallel connections, we can deduce total impedances for complex systems (similar to the combination of springs in statics). It is useful to compute the impedance of coupled systems starting from the opposite side of the load.

10.4.1 Parallel connnection For subsystems that are subjected to the same displacement, and thus, the same velocity vˆ+ at the excitation point, the impedances are summed up because the total force fbtot+ results

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from the individual resistant forces of the subsystems. That is, the force has to “work against” the sum of the individual impedances. For a parallel connection composed of n subsystems, we obtain: fbtot+ = vb+ zb1+ + vb+ zb2+ + . . . + vb+ zbn+ = vb+ (zb1+ + zb2+ + . . . + zbn+ ) .

(10.69)

Thus, the impedance is given by zbtot+

fbtot+ = zb+1 + zb2+ + . . . + zbn+ . = vb+

(10.70)

10.4.2 Serial connnection In case of a serial connection of subsystems, the individual displacements are summed up, whereas the force remains the same in all the subsystems. Then, we obtain the following equation for the total impedance (serial connection of impedances): vbtot+ = vb1+ + vb2+ + ... + vbn+ =

fb+ fb+ + + ... + zb1+ zb2+

=

fb

1

1

fb+ zbn+ 1

!

.

(10.71)

Fb+ 1 = 1 1 1 . vbtot+ + bz2+ + ... + bzn+ b z1+

(10.72)

+

zb1+

+

zb2+

+ ... +

zbn+

Thus, the impedance is given by zbtot+ =

10.5 Arbitrary connections Using the idea of impedances, one can easily assess, e.g., the effect of additional masses installed in order to reduce vibrations (addition of a mass-impedance) or an elastic support (serial connection with a spring-impedance). Herafter, we renounce parts with negative frequencies, and only consider the part with positive frequencies as in, e.g., f (t) = fb+ e iΩt + fb− e −iΩt . The parts with negative frequencies can be obtained by complex conjugation, as the impedance for the negative frequency part is the complex conjugate of the part for positive frequencies.

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Example 10.1: SDOF system

The following SDOF system with mass m , spring stiffness k and viscous damping c is considered. m

k

fb+ e iΩt vb+ e iΩt

c

Here, the displacements and thus velocities of the three elements are equal. This corresponds to a parallel connection of mass, spring and damper. Hence, we obtain for the total impedance Z of the SDOF system z+ = zm+ + zc+ + zk+ = iΩm + c +

k . iΩ

The dynamic stiffness kdyn = wfb++ can be computed from the impedance by multiplication with iΩ considering Eq. (10.3): b

kdyn = −Ω2 m + iΩc + k. Thus, we immediately obtain the well-known equation of the dynamic stiffness of the harmonically forced SDOF system (cf. Section 7.2.1). The transfer function is given by the inverse relation U=

1 kdyn

=

−Ω2 m

1 . + iΩc + k

The impedance and transfer function are plotted over the excitation frequency in the following figure:

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(a) Impedance

(b) Transfer function

Figure 10.9: SDOF impedance model for different values of damping ratio, D = 0.05 (—), D = 0.2 (–·–), D = 0.7 (- - -).

Example 10.2: Inserting an elastic element

The following system is considered m fb+ e iΩt c

k

vb+ e iΩt

In this system the spring and the damper carry the same force. The displacements sum up. Therefore these two elements can be interpreted as a serial connection.

+e

fb k

vb+ e iΩt

c

iΩt

= ˆ

zc,k+ vb+ e iΩt

f+ e iΩt b

vb+ e

iΩt

Thus, the total impedance of these two subsystems at the interface to the mass is zc,k+ =

1 zk+

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1 +

1 zc+

=

iΩ k

1 +

1 c

.

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The endpoint of the subsystem spring-damper and the mass are subjected to the same displacement. Therefore, the connection of the spring-damper and mass subsystem is a parallel connection. m zc,k+ vb+ e iΩt

fb+ e iΩt vb+ e iΩt

Hence, for the computation of the total impedance we have to add the mass impedance as shown below: z+ = iΩm +

iΩ k

1 +

1 c

.

(a) Impedance

(b) Transfer function

Figure 10.10: SDOF impedance model for different values of damping ratio, D = 0.05 (—), D = 0.2 (–·–), D = 0.7 (- - -).

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Example 10.3: Undamped 2DOF system

The following undamped 2DOF system is considered. m1

m2 fb+ e iΩt k2

k1

vb+ e iΩt

We first represent the left subsystem by its impedance.

m1

zk1 ,m1 + vb+ e iΩt

fb+ e iΩt k1

=

fb+ e iΩt vb+ e iΩt

vb+ e iΩt

Therefore, we calculate the impedance for the parallel connection of the spring k1 and the mass m1 . zk1 ,m1 + = zm1 + + zk1 + = iΩm1 +

k1 . iΩ

In the next step, the spring k2 is added. This spring carries the same force as that of the first subsystem k1 , m1 , thus, the first subsystem and the second spring are connected serially. Hence, we obtain the impedance of the system k1 , m1 , and k2 as shown below. zk1 ,m1 + vb+ e

+e

iΩt

fb k2

vb+ e iΩt

iΩt

= ˆ

zk1 ,m1 ,k2 + vb+ e iΩt

fb+ e iΩt

vb+ e iΩt

It results zk1 ,m1 ,k2 + =

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1 . + k1 k2 iΩm1 + iΩ iΩ 1

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Finally, the mass m2 is coupled to the subsystem k1 , m1 and k2 . The displacement of the mass m2 is equal to the displacement at the endpoint of spring k2 . Thus, the mass m2 is coupled to the subsystem k1 , m1 and k2 in a parallel connection. m2 zk1 ,m1 ,k2 + vb+ e iΩt

fb+ e iΩt vb+ e iΩt

It follows z+ = zk1 ,m1 ,k2 ,m2 + =

1

1

+ k1 k1 2 iΩm1 + iΩ iΩ

+ iΩm2 .

(a) Impedance

(b) Transfer function

Figure 10.11: 2DOF impedance model.

10.6 Impedances for selected systems The representation of substructures through impedances or dynamic stiffnesses, respectively, enables the coupling to other substructures that can also be described via impedances. This way also structures defined on infinitely extended domains can be represented. In Tab. 10.1 selected impedances at the excitation points of point forces for single or two sided infinitely extended domains are given.

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Furthermore, results from measured impedances can be directly considered. For example, the dynamic stiffness of a foundation on an elastic isotropic halfspace can be coupled in serial or parallel to the dynamic stiffness of a structure (e.g., see Studer et al [2007]). rod

√ Z = A Eρ

beam

Z = 2ρAcB (1 + i)

beam

Z = 12 ρAcB (1 + i)

plate

√ Z = 8 B 0 ρh

plate

√ Z = 3.5 B 0 ρh

Table 10.1: Selected impedances for continuous systems, from Cremer et al [2005].

Here, A denotes the cross-section, ρ is the density, cB is the bending wave velocity, B 0 denotes the bending stiffness per length, h is the thickness of the plate, and E is the Young’s modulus.

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11 Solutions for selected continuous systems Until now, we discussed the dynamic characteristics of discrete systems, such as the MODF systems or continuous systems described by impedances. The systems’ description is reduced to the parameters of stiffness, damping and mass or in the are assumed to act as discrete elements of finite number. In example 4.9, this approach is used to model a three storey frame, where floors are assumed as discrete or lumped masses that are infinitely stiff and columns act as massless, but elastic springs. Through this, we reduce the governing equations to simple ordinary differential equations, that can be easily solved. However, this modeling approach is not always applicable and the continuous distribution of stiffness, damping and mass have to be accounted for. Furthermore, the analysis of continuous dynamic systems is helpful to gain a better understanding of physical phenomena. In general, only very simple models can be solved analytically. Especially for structures of complex shape, numeric methods have to be applied, that are based on the discretization of the system, such as the finite or boundary element method. In order to be able to interprete the behavior of continuous systems some analytical derivations are shown subsequently.

11.1 String, longitudinal and torsional rod 11.1.1 Equation of motion We derive the equations of motion for linear (one dimensional) elements with constant cross sectional values over the length analogously to the derivation of the equation of motion of a SDOF system (comp. Chapter 5). The formulation of the dynamic equilibrium of the differential element is given for rods subjected to longitudinal motion, under St. Venant torsion, as well as the taut string constrained to small sag. These formulations share the same structure. Fig. 11.1 illustrates the forces that act on the differential element with length dx. For the string in Fig. 11.1c we assume that the horizontal force H is constant over the length of the string.

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mT dx

px dx cu˙ dx

N

N + N 0 dx

u dx u µ¨ dx (a) EA-rod

cφ˙ dx MT

p dx

Hw0

cw˙ dx

MT + MT0 dx w H

φ Iφ φ¨ dx

H µw¨ dx Hw0 + Hw00 dx

dx

dx

(b) St.-Venant torsion

(c) String with small sag

Figure 11.1: Forces on differential elements: axial and torsional rod, string, with normal force N , Youngs modulus E , cross-sectional area A, mass per unit length µ = ρA, the horizontal displacement u, the torsional moment MT , the de St. Venant torsional stiffness GIT , the polar moment of inertia Iφ , the cross-sectional rotation φ, the horizontal tension force H (assumed constant) and the transversal displacement w.

The normal force follows to N = EAu0 , and the de St. Venant torsional moment to MT = GIT φ0 . Hence, the following identically structured equilibrium equations are found containing an elastic reaction force FK , a damper force FC and a inertia force FI together with external loading Fe . fK + fC + fI = fe

(11.1)

For the longitudinal vibration of the axial rod (Fig. 11.1a) it holds: −EAu00 + cu˙ + µ¨ u = px .

(11.2)

For the torsional vibration of the torsional rod (Fig. 11.1b) it holds: −GIT φ00 + cφ˙ + Iφ φ¨ = mT .

(11.3)

For the vibration of a string with small sag (neglecting displacement u, Fig. 11.1c) it holds: −Hw00 + cw˙ + µw¨ = p .

(11.4)

In the following we introduce y for any of the variables describing the deformation u, φ, or w. The stiffness related coefficients EA, GIT and H are denoted ke and the inertia related f. We now consider the free vibration of the above systems. coefficients µ and Iφ are denoted m

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11.1.2 Free vibration Neglecting all damping forces and external forces the partial differential equations read e 00 + m fy¨ = 0 −ky

(11.5)

f. We divide Eq. (11.5) by m f with the variable of motion y, the stiffness ke and the inertia m and obtain

−c2 y 00 + y¨ = 0

(11.6)

e k where c2 = m e . This equation is in general known as the classical one dimensional wave equation.

11.1.2.1 Solution via a separation approach

We use the separation approach y(x,t) = Y (x)f (t)

(11.7)

with the space function Y (x) and time function f (t). Inserting Eq. (11.7) into the differential equation (11.5) results in −c2 Y 00 (x)f (t) + Y (x)f¨(t) = 0 .

(11.8)

We divide Eq. (11.8) by Y (x)f (t), such that f¨(t) Y 00 (x) = c2 . f (t) Y (x)

(11.9)

Eq. (11.9) must hold at every instance in time and at every point along the system, i.e., both sides have to be equal to a constant, say λ = −ω 2 . f¨(t) Y 00 (x) = c2 = −ω 2 . f (t) Y (x)

(11.10)

Thus, we can write the following for both terms in Eq. (11.10) separately to obtain the two decoupled ordinary differential equations f¨(t) = −ω 2 f (t)

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(11.11)

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c2 Y 00 (x) = −ω 2 Y (x) .

(11.12)

These are eigenvalues for the continuous system. The above differential equation with respect to time is already known from the discussion on discrete SDOF systems. Additionally the second equation now introduces the continuous relations for the spatial domain. Under the assumption ω 2 > 0, we find the following solutions: f (t) = fb+ e iωt + fb− e −iωt

(11.13)

Y (x) = Yb+ e ikx + Yb− e −ikx .

(11.14)

with k2 =

ω2 . c2

(11.15)

Rewritten in sine and cosine terms, it follows f (t) = f01 cos (ωt) − f02 sin (ωt)

(11.16)

Y (x) = Y01 cos (kx) − Y02 sin (kx) .

(11.17)

The unknown coefficients f01 , f02 , Y01 and Y02 in Eqs. (11.16) and (11.17) can be found through the spatial boundary conditions (at both ends) and the temporal initial conditions (initial displacement and velocity). Possible spatial boundary conditions are depicted in Fig. 11.2. The above solution is the homogeneous solution of the continuous problem related to the given strucutres.

Fext x,Y (a) fixed, Y = 0 (Dirichlet)

x,Y 0

(b) free, SY = (Neumann)

Fext

x,Y (c) spring, SY (Robin)

0

= kY

Figure 11.2: Boundary conditions

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Example 11.1: Mode shapes of rod-shaped systems

We consider the following system according to Fig. 11.1 with length l and fixed boundaries at both ends x, Y (x)

ke = EA

l

f=µ m

The free vibrations of the depicted system are given by Eqs. (11.16). The boundary conditions are given as x=0:

Y =0

x=l:

Y = 0.

These conditions inserted into Eqs. (11.16) lead to the following equation system for the unknowns Y01 and Y02 "

1 0 cos (kl) − sin (kl)

#"

Y01 0 = . Y02 0 #

" #

(11.18)

This equation system only has a nontrivial solution (trivial would be Y01 = Y02 = 0), if the determinant of the coefficient matrix is zero, i.e., sin (kl) = 0 . Hence, the eigenvalues kj of the free vibration follow to kj =

jπ . l

Under consideration of Eq. (11.15) the radial eigenfrequencies ωj are jπ jπ ωj = kj c = c= l l

s

ke . f m

The eigenvectors of the system can then be obtained by inserting the result for kj into Eq. (11.18). From this we can deduce that Y01 = 0. As the equation system is

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underdetermined, Y02 6= 0 is arbitrary and thus jπ φj (x) = −Y02,j sin x l   ∞ X jπ x (f01,j cos (ωj t) − f02,j sin (ωj t)) . y(x,t) = sin l j=1 



The term −Y02,j can be neglected in the summation because the unknowns in the time function are also not yet determined. The functions φj (x) describe the mode shapes of the vibration with ωj as corresponding radial eigenfrequency. The system can vibrate freely in these mode shapes with the related frequencies. The unknowns f01,j and f01,j can be determined via the given temporal initial conditions y(x,t = 0) and y(x,t ˙ = 0). Note that the initial conditions are functions of x. In general the displacements and velocities for t = 0 can be written as y (x, t = 0) =

∞ X

sin

j=1 ∞ X



nπ x f01,j l 

(11.19)

nπ sin y˙ (x, t = 0) = − x ωj f02,j . l j=1 



(11.20)

Depending on the (spatial) boundary conditions different eigenvalues kj , radial eigenr e k frequencies ωj = kj m e and mode shapes φj are obtained. The following table shows selected cases. boundary condition

char. equation

eigenvalues kj

mode shape φj

fixed – fixed

sin (kl) = 0

jπ l

sin (kj x)

fixed – free

cos (kl) = 0

(2j − 1) 2lπ

sin (kj x)

free – free

sin (kl) = 0

jπ l

cos (kj x)

Table 11.1: Eigenvalues kj and mode shapes φj from Petersen [1996]

Consider the solutions to the Eqs. (11.13) and (11.14). The total solution w(x,t) is given as the product of the two separate solutions as w(t) = Y (x)f (t) = f+ Y+ e iωt e ikx + f− Y+ e −iωt e ikx + f+ Y− e iωt e−ikx + f− Y− e −iωt e −ikx

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= f+ Y+ e i(kx+ωt) + f− Y+ e i(kx−ωt) + f+ Y− e −i(kx−ωt) + f− Y− e −i(kx+ωt) .

(11.21)

Inserting ω = kc, we obtain h

i

h

}

|

i

w(t) = f+ Y+ e ik(x+ct) + f− Y− e −ik(x+ct) + f− Y+ e ik(x−ct) + f+ Y− e −ik(x−ct) . |

{z

wave travelling in negative x-direction

{z

wave travelling in positive x-direction

(11.22)

}

The arguments of the full solution x− = x + ct and x+ = x − ct describe waves traveling in the negative and positive x-direction, respectively, with the wave speed c=

ω = k

s

ke . f m

(11.23)

f, respectively, and is The wave speed c only depends on the stiffness and inertia terms ke and m thus a mere system parameter. It is independent of the frequency or the initial conditions.

We now represent Eq. (11.22) using sine/cosine functions w(t) = 2 [Re(f+ Y+ ) cos(kx + ωt) + Im(f+ Y+ ) sin(kx + ωt)] + + 2 [Re(f− Y+ ) cos(kx − ωt) + Im(f− Y+ ) sin(kx − ωt)] . (11.24) We then denote amplitudes of the solution that travel in negative x-direction (argument k(x + ωt)) by an uppercase −, e.g., w− , and the ones that travel in positive x-direction (argument k(x − ωt)) by an uppercase +, e.g., w+ : h

i

− − w(t) = w01 cos(kx + ωt) + w02 sin(kx + ωt) +

h

i

+ + + w01 cos(kx − ωt) + w02 sin(kx − ωt) . (11.25)

Applying Eqs. (2.18) and (2.19), we obtain i

h

+ + (cos(kx) cos(ωt) − sin(kx) sin(ωt)) + w02 (cos(kx) sin(ωt) + sin(kx) cos(ωt)) + w(t) = w01

h

i

− − + w01 (cos(kx) cos(ωt) + sin(kx) sin(ωt)) + w02 (cos(kx) sin(ωt) − sin(kx) cos(ωt)) . (11.26) − − + + Consider now the special case w02 = w02 and w01 = w01 . Eq. (11.26) then becomes





+ + w(t) = 2 cos(kx) w01 cos(ωt) + w02 sin(ωt)

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(11.27)

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This special case describes a standing wave with harmonically varying amplitude. 11.1.2.2 Direct solution via a wave approach

In the preceeding section, the solution to the differential equation (11.5), which reads −c2

∂ 2y ∂ 2y + 2 =0 ∂x2 ∂t

(11.28)

was derived. It can be interpreted as a sum of two waves that travel in opposite directions. We now discuss a different solution approch, that was introduced by d’Alembert. To this end, we introduce the new variables x+ = x − ct

(11.29)

x− = x + ct .

(11.30)

Applying the chain rule of differentiation we can write the partial differential operators ∂ as ∂x ∂x+ ∂ ∂x− ∂ ∂ ∂ ∂ = + =c + −c − + − ∂t ∂t ∂x ∂t ∂x ∂x ∂x + − ∂ ∂x ∂ ∂x ∂ ∂ ∂ = + = + −, + − + ∂x ∂x ∂x ∂x ∂x ∂x ∂x

∂ ∂t

and

(11.31) (11.32)

and furthermore ∂2 ∂2 ∂2 ∂2 2 = c − 2 + ∂t2 ∂x+ ∂x− ∂x− 2 ∂x+ 2 ∂2 = ∂x2

!

(11.33)

!

∂2 ∂2 ∂2 + 2 + . ∂x+ ∂x− ∂x− 2 ∂x+ 2

(11.34)

Eq. (11.28) can then be written as −c2

2 + − ∂ 2y ∂ 2y 2 ∂ y(x ,x ) + = −4c . ∂x2 ∂t2 ∂x+ ∂x−

(11.35)

A solution to Eq. (11.35) is given by y(x+ ,x− ) = f1 (x+ ) + f2 (x− ) = f1 (x − ct) + f2 (x + ct) .

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(11.36)

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This is the sum of two functions f1 and f2 , that move in opposite directions (f1 in the positive, f2 in the negative x-direction) with the velocity c. y(x,t) f1 (x,t = 0)

f1 (x,t = ∆t) x x+ = x − c∆t

c ∆t

Figure 11.3: Wave solution, f1 (x − ct), Eq. (11.36).

t t3

t2 y(x,t) t1

x x1

x2

x3

Figure 11.4: Wave traveling in the positive x-direction.

This approach solves for arbitrary functions and f1 and f2 Eq. (11.5): y¨ = c

2

d 2 f1 d 2 f2 + dx2+ dx2−

!

y 00 =

d 2 f1 d 2 f2 + dx2+ dx2−

(11.37)

The unknown functions f1 (x+ ) and f2 (x− ) have to satisfy displacement and velocity initial

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conditions respecting given boundary conditions. The initial conditions are given by y(x,t = 0) = y0 (x) = f1 (x) + f2 (x)

(11.38)

y(x,t ˙ = 0) = v0 (x) = f˙1 (x − ct) + f˙2 (x + ct) = −cf10 (x) + cf20 (x).

(11.39)

Eq. (11.39) is rewritten as 1 −f10 (x) + f20 (x) = v0 (x) c

(11.40)

Integrating Eq. (11.40) leads to −f1 (x) + f2 (x) =

1Z x v0 (ξ) dξ − f1 (0) + f2 (0) c 0

(11.41)

We now add Eq. (11.38) to Eq. (11.40) and obtain 1 Zx y0 (x) f1 (0) f2 (0) f2 (x) = v0 (ξ)dξ + − + . 2c 0 2 2 2

(11.42)

Analogously, we substract Eq. (11.40) from Eq. (11.38). Thus, it holds f1 (x) = −

y0 (x) f1 (0) f2 (0) 1 Zx v0 (ξ)dξ + + − . 2c 0 2 2 2

(11.43)

Inserting Eqs. (11.42) and (11.43) into Eq. (11.36), the displacement y(x,t) reads y(x,t) = f1 (x − ct) + f2 (x + ct) Z x−ct x+ct y0 (x − ct) + y0 (x + ct) 1 v0 (ξ) dξ = + v0 (ξ) dξ − 2 2c 0 0 y0 (x − ct) + y0 (x + ct) 1 Z x+ct = v0 (ξ)ξ . + 2 2c x−ct

=

Z



(11.44)

In cases, where the initial velocity is equal to zero, i.e., v0 (x) = 0, the initial displacement is split into two oppositely traveling wave packets with half of the initial amplitude.

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Example 11.2: Wave approach

The depicted system with given initial conditions shall be solved through a wave approach according to Eq. (11.36). y0 (x)

v0 (x) = 0 x

l

ymax

x

l

l

l

Figure 11.5: Initial conditions, e.g. deflection of string.

As the initial velocity is v0 (x) = 0, the displacement y(x,t) is given as y(x,t) =

y0 (x − ct) + y0 (x + ct) . 2

For t = 0, it holds y(x,0) =

y0 (x) + y0 (x) . 2

This solution corresponds to two oppositely traveling waves with the individual amplitudes ymax . In Fig. 11.6 both waves are depicted for different time instances t0 , t1 , t2 , 2 t3 .

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t = t0 so that ct0 = 0 ymax 2 ymax 2

x l

l ct0 = 0

t = t1 so that 0 < ct1 < l

ymax 2

ymax 2

x

l

l ct1

ct1

t = t2 so that ct2 = l

x

ymax 2

ymax 2

l

l

ct2

ct2

t = t3 so that ct3 = 2l x ymax 2

ymax 2

l ct3

l ct3

Figure 11.6: Oppositely travelling waves at times t0 , t1 , t2 , t3 .

Additionaly to the temporal initial conditions also the spatial boundary conditions have to be satisfied (Fig. 11.7).

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x = −l

y(−l,t) = 0

x = −l

y 0 (−l,t) = 0

x=l

x=l

y(l,t) = 0

y 0 (l,t) = 0

Figure 11.7: Spatial boundary conditions at the supports.

The boundary conditions at the bearings can be satisfied through oppositely traveling waves. The displacement at the respective side is given by the sum of the both waves. For the left side, the boundary condition is exemplarily given in Fig. 11.8. The illustration is valid for the time t = t0 = 0. Here, a wave is traveling to the left with wave speed c with known shape and amplitude. A second (reflected) wave with the same wave speed c but unknown amplitude AR , that travels to the right and is shifted to the left by 2l, is added. t = t0 so that ct0 = 0

AR fR (x + 2l − ct0 ) l

l ct0

Af (x + ct0 )

l

l ct0

Figure 11.8: Boundary condition at the left support

Fig. 11.9 illustrates the situation for a later point in time t = t1 .

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t = t1 so that 0 < ct1 < l

AR fR (x + 2l − ct1 )Af (x + ct1 ) l

l ct1

l

l ct1

Figure 11.9: Situation at the left support for time t = t1

The displacement at the boundary is given by the superposition of both waves y(x,t) = AR fR (x + 2l − ct) + Af (x + ct). The ratio of the amplitudes AR and A follows from the boundary conditions. For a support, that is fixed in x-direction at the location x = −l, the condition y(−l,t) = 0 results in y(−l,t) = AR fR (l − ct) + Af (−l + ct) = 0. Hence, fR (l − ct) = f (−l + ct), and thus, AR = −A. A boundary that is fixed in x-direction, is hence represented by two oppositely traveling, mirrored waves with opposite amplitudes. In the case of a free end at the position x = −l, the boundary condition of vanishing forces at the end, i.e., y 0 (−l,t) = 0, leads to the amplitude AR = A.

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11.2 Euler-Bernoulli beam 11.2.1 Equation of motion We derive the equation of motion of the Euler-Bernoulli beam via the equilibrium of forces on a differential element. p dx cwdx ˙ µwdx ¨ M + M 0 dx

M w Q

dx

Q + Q0 dx

Figure 11.10: Equilibrium on a differential element of the Euler -Bernoulli beam.

We assume the material properties to be independent of time. The dynamic equilibrium for the vibration of a beam with the bending stiffness EI and the mass per unit length µ, and under the assumption of Euler-Bernoulli theory (i.e., neglecting rotational inertia and shear deformation), reads −Q0 (x,t) + c(x)w(x,t) ˙ + µ(x)w(x,t) ¨ − p(x,t) = 0 .

(11.45)

We assume the material and cross-sectional properties to be time-independent. The shear force is Q(x,t) = (−EI(x)w00 (x,t))0 . Thus, we obtain 00

(EI(x)w00 (x,t)) + c(x)w(x,t) ˙ + µ(x)w(x,t) ¨ − p(x,t) = 0 .

(11.46)

11.2.2 Free vibration Neglecting the external loading as well as damping forces, and assuming a constant bending stiffness EI and mass density µ, the partial differential equation for the free vibration is obtained. EIw0000 (x,t) + µw(x,t) ¨ =0

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(11.47)

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11.2.2.1 Solution via a separation approach

The partial differential equation (11.47) of the free vibration can be solved by a separation approach with space function W (x) and time function f (t) 1 . w(x,t) = W (x)f (t) .

(11.48)

We insert Eq. (11.48) into Eq. (11.47) and obtain EIW 0000 (x)f (t) + µW (x)f¨(t) = 0 .

(11.49)

Dividing Eq. (11.49) by µ and W (x)f (t), it follows f¨(t) EI W 0000 (x) =− . f (t) µ W (x)

(11.50)

Eq. (11.50) must hold at every instance in time and space, i.e., both sides have to be equal to a constant, also called separation constant, say λ = −ω 2 . This at first not obvious choice will become apparent when discussing the solution characteristics. EI W 0000 (x) f¨(t) =− = −ω 2 . f (t) µ W (x)

(11.51)

Thus, we can write the following for both terms in Eq. (11.51) separately to obtain the two decoupled ordinary homogeneous differential equations f¨(t) + ω 2 f (t) = 0

(11.52)

W 0000 (x) − k 4 W (x) = 0

(11.53)

with k4 =

µ 2 ω . EI

(11.54)

We find the following solutions under the assumption ω 2 > 0. c e kx + K c e −kx + K c e ikx + K c e −ikx W (x) = K x1 x2 x3 x4 c e iωt + K c e −iωt . f (t) = K t1 t2

1

(11.55) (11.56)

The solution to the free vibration problem of the unloaded beam with the partial differential equation EIw0000 + µw ¨ = 0 is also called the homogeneous solution.

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The solution in time now corresponds to oscillating waves as the exponent of the exponential function is complex. Now, reconsider our choice for λ in Eq. (11.51). If the constant λ was assumed to be ω 2 for ω 2 > 0 we would have obtained solutions that correspond to decaying and growing exponential functions, which are not admissible in our problems. Eqs. (11.55) and (11.56) can alternatively be written as W (x) = Cx1 cosh(kx) + Cx2 sinh(kx) + Cx3 cos(kx) − Cx4 sin(kx) f (t) = Ct1 cos(ωt) − Ct2 sin(ωt) .

(11.57) (11.58)

Thus, the solution contains six unknown coefficients in Eqs. (11.55) and (11.56), or Eqs. (11.57) and (11.58), respectively. These can be found by inserting the spatial boundary conditions and the temporal initial conditions. Possible spatial boundary conditions are depicted in Fig. 11.11. hinged support:

clamped support: free end: dilational or rotational spring:

M = 0, w = 0

w = 0, φ = 0 M = 0, Q = 0 Q = kw or M = −kφ φ

Figure 11.11: Spatial boundary conditions.

Through the relations w(x,t) = W (x)f (t), w0 (x,t) = W 0 (x)f (t), M (x,t) = −EIW 00 (x)f (t), and Q(x,t) = −EIW 000 (x)f (t) all boundary conditions can be represented by derivatives of the displacement. At each boundary of a beam two boundary conditions can be found, i.e., at each boundary two equations for the unknown coefficients are given. This results in a 4 × 4 system of equations for the unknown coefficients Cxi . A solution is found by excluding the trivial solution and setting the determinant of the equation system to zero. As the solution for the time function f (t) is decoupled, the unknowns Cti are found through the initial displacement and velocities for t = 0.

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Example 11.3: Mode shapes of the Euler-Bernoulli beam

In order to find the unknowns of the solution to the Euler-Bernoulli beam vibration, the derivatives of the space function W (x) with respect to x and time function f (t) with respect to t are needed. Under consideration of Eqs. (11.57) and (11.58) it follows. W 0 (x) = k [Cx1 sinh(kx) + Cx2 cosh(kx) − Cx3 sin(kx) − Cx4 cos(kx)] W 00 (x) = k 2 [Cx1 cosh(kx) + Cx2 sinh(kx) − Cx3 cos(kx) + Cx4 sin(kx)] W 000 (x) = k 3 [Cx1 sinh(kx) + Cx2 cosh(kx) + Cx3 sin(kx) + Cx4 cos(kx)] f˙(t) = ω [−Ct1 sin(ωt) − Ct2 cos(ωt)] . The unknowns Cx1 to Cx4 are determined from the spatial boundary conditions. In the following this will be performed exemplarily for the hinged Euler-Bernoulli beam. µ = const., EI = const. w(x,t)

x

l

Figure 11.12: Hinged Euler-Bernoulli beam.

Since cos (ωt) and sin (ωt) change periodically, it follows, that w(x,t) and M (x,t = −EIW 00 (x,t) have to be equal to zero for all times t. Thus, x = 0 : W (x = 0) = 0

W 00 (x = 0) = 0

(11.59)

x = l : W (x = l) = 0

W 00 (x = l) = 0 .

(11.60)

These four equations lead to the following equation system for the unknowns Cx1 to Cx4 in matrix notation 1 0 1 0 Cx1 0   C  0 2 2 k 0 −k 0    x2      = .  cosh(kl) sinh(kl) cos(kl) − sin(kl)  Cx3  0 k 2 cosh(kl) k 2 sinh(kl) −k 2 cos(kl) k 2 sin(kl) Cx4 0 





 

From the first two lines, it immediately follows Cx1 = Cx3 = 0. Hence, we write the

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reduced equation system "

sinh(kl) − sin(kl) sinh(kl) sin(kl)

Cx2 0 = . Cx4 0

#

!

!

(11.61)

This homogeneous equation system only permits a non-trivial solution, if its determinant is zero, thus sinh(kl) − sin(kl) det = 2 sinh(kl) sin(kl) = 0 . sinh(kl) sin(kl) "

#

As sinh (k l) is only equal to zero for kl = 0, which is another trivial solutiona , the above characteristic equation for kl reduces to sin(kl) = 0 . Thus, we obtain the eigenvalues of the system kj l = jπ

(11.62)

for j ∈ N+ . The eigenvalue problem thus has an infinite number of solutions. Under consideration of Eq. (11.54), the eigenfrequencies for which non-trivial solutions are obtained, follows to s

ωj = kj2

EI jπ =± µ l 

2 s

EI . µ

(11.63)

These eigenfrequencies are associated with mode shapes φj . They follow from the solution of Eq. (11.61) after inserting the previously obtained values for k. φj (x) = Cx4,j sin



nπ x . l 

The beam vibrates in the mode shapes φj (x) with the corresponding eigenfrequencies ωj . The complete solution therefore follows as the addition of the individual solutions. wj (x,t) = φj (x)fj (t) w(x,t) =

∞  X j=1

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jπ jπ Ct1,j sin x cos (ωj t) − Ct2,j sin x sin (ωj t) l l 









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w(x,t) ˙ =

∞ X



ωj −Ct1,j sin



j=1

jπ jπ x sin (ωj t) − Ct2,j sin x cos (ωj t) l l 







The coefficients Ct1,j and Ct2,j then depend on the initial conditions. Depending on the spatial boundary conditions, different eigenvalues kj , eigenfrequencies ωj and mode shapes φj (x) result. Selected results are given in the following table. Table 11.2: Eigenfrequencies of the Euler-Bernoulli beam for different boundary conditions from Petersen [1996]

system

λj

Aj

Eigenform Φj

λj = j π

−

sin λj xl









λj =

2j+1 π 2

sinh(λj )−sin(λj ) cosh(λj )−cos(λj )

λj =

2j−1 π 2

sinh(λj )+sin(λj ) cosh(λj )+cos(λj )

4j+1 π 4

sinh(λj )−sin(λj ) cosh(λj )−cos(λj )

+Aj [cosh 

a



λj xl





x l



+



− cos λj





x l

]



sin λj xl − sinh λj xl + 







+Aj [cosh λj xl − cos λj xl ] 

λj =



sin λj xl − sinh λj







sin λj xl − sinh λj xl +     +Aj [cosh λj xl − cos λj xl ]

This trivial solution corresponds to the static limit case, for which another solution is valid.

11.2.2.2 Solution via a wave approach

The solutions given in Eqs. (11.55) and (??) can be interpreted by waves travelling in both directions and decaying near fields. c e −iωt results in functions c e iωt + K c e ikx and K c e −ikx with f (t) = K A multiplication of K t2 t1 x3 x4 ±ik(x− ω t) ±ik(x+ ω t) k k that depend on e or e . The argument x+ = x − ωk t is linked with a wave travelling in the positive x-direction, the argument x− = x + ωk t with a wave travelling in the negative x-direction with the velocity of propagation cB = ωk . Using Eq. (11.54), we find the bending wave velocity

ω cB = = k

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s 4

EI √ ω µ

s

i.e.

ω=

EI 2 k . µ

(11.64)

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In contrast to the wave propagation velocity from the solution of the differential equation of 2nd order (comp. Eq. (??), this so-called bending wave phase velocity only has the meaning of a propagation velocity for a harmonic (sinusoidal or cosinusoidal) waves. Waves, which are composed of different sine or cosine parts, distort themselves in the propagation, since portions with higher frequencies ω (or smaller wave numbers k), propagate faster than lower frequency components (with larger wavenumbers). Thereby the propagation on beams (differential equation of 4-th order) differs from the propagation the on rod-shaped systems described by the differential equation of 2nd order. The frequency dependence of the bending wave is called dispersion. It has special significance for the sound radiation of structures. For frequencies, for which the bending wave phase velocity cB takes values below the sound velocity of an adjacent fluid (air or water) cf luid, in the stationary case and with infinitely extended structures, the sound radiation is zero, because the wave pattern on the structure does not “couple” to a wave pattern in the adjacent fluid (see Fig. 11.13).

c Fluid λLuft

ϑ x

λB =

λLuft sin(ϑ)

z

Figure 11.13: Spuranpassung, Biegewellenlänge λB > Wellenlänge Fluid λf luid

The frequency, for which the bending wave phase velocity cB is equal to the sound wave velocity cf luid (speed of sound) of the adjacent fluid, is called coincidence frequency. The other parts of the solution in Eq. (11.55) and (??) that result from the multiplication of the c e kx and K c e −kx with f (t) = K c e iωt + K c e −iωt can be interpreted as harmonically terms K x1 x2 t1 t2 oscillating near fields. The arise at the boundaries of the beam and decay exponentially from there.

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Example 11.4: Semi-infinite beam for harmonic boundary moment

We consider the following semi-infinite Euler-Bernoulli beam with a harmonically varying moment at the boundary x = 0. M0 cos Ωt EI, µ x Figure 11.14: Beam under harmonic boundary moment

The general solution reads: c e kx+iωt + K c e −kx+iωt + K c e i(kx+ωt) + K c e −i(kx−ωt) + w(x,t) =K 4 1 2 3 c e kx−iωt + K c e −kx−iωt + K c e i(kx−ωt) + K c e −i(kx+ωt) +K 5 6 7 8

Since the domain of the beam extends to infinity in the positive x-direction, we can assume that no waves propagate from the infinite towards the origin, and the positive exponential near field cannot exist, since they would lead to infinitely high amplitudes for x → ∞. These conditions are also called the Sommerfeld radiation conditions. We furthermore insert the relationship ω = kcB , and obtain  

 

k(x+ic ik(x+c  B t) + K  B t) + K ce c e −k(x−icB t) + K c e c e −ik(x−cB t) + w(x,t) = K 1 2 3 4 

 

−ik(x+c k(x−ic  B t) + K (( B t) c(e( c e ik(x−cB t) + ( ce c e −k(x+icB t) + K + K K 8 5 6 7

((

where we skipped the above mentioned corresponding solution terms. The solution is c and K c ) and a thus given by a wave traveling in positive x-direction (parts with K 4 7 c and K c ). Considering the moment and displacement decaying near-field (parts with K 2 6 boundary condition at x = 0: EIw00 (0,t) = −M0 cos (Ωt) w(0,t) = 0

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the solution is found by w(x,t) =

  i M0 h  −kx cos k(x − c t) − e cos k c t . B B 2EI k 2

(11.65)

It is easy to show, that this solution satisfies the homogeneous PDE as well as the boundary conditions. This example considered a time harmonic boundary moment. If, for example, the boundary moment was modeled as a Dirac impulse, it could be represented as the superposition of single cosine terms by a Fourier transformation.

The dispersion of the beam wave leads to the fact, that vibrations with higher frequency propagate faster than the ones with lower frequency. If such a beam is excited with a Dirac impulse (composed of all frequency components with equal amplitude), at a finite distance from the excitation point the signal “shape” will be altered.

11.2.3 Displacement approaches - spectral elements based on analytic solutions (For information only, treated in “Ergänzungskurs Technische Mechanik”.) 11.2.3.1 Beams without loads and harmonic excitation at the boundaries

The free vibration of a system composed from individual beams or impedances at nodes can be determined from the solution of the individual elements (e.g., columns and bars) under the assumption of a harmonic motion. The procedure corresponds to the displacement approaches in statics. Every beam element is described by solution to Eqs. (11.57) and (11.58). W (x) = Φ (x) α

(11.66)

with Φ (x) = {cosh(kx), sinh(kx), cos(kx), sin(kx)}

(11.67)

and αT = {Cx1 , Cx2 , Cx3 , Cx4 }

(11.68)

Note that the sign of the sine function in Eq. (11.67) differs from the convention introduced in section 2.2. This is to keep consistency with literature on the topic. The unknown coefficients

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Cx1 to Cx4 can be determined via boundary and transition conditions. Corresponding to a finite element formulation, in a first step, the displacement field W (x) is represented through the ansatz functions Φ (x) and the nodal displacements we (comp. Fig. 11.15). wTe = {wi , φi , wk , φk }

(11.69)

The relationship between the coefficients α and the nodal displacements we is given by we = Aα

(11.70)

where we find through inserting 1 0 1 0   0 k 0 k    [A] =   cosh(kle ) sinh(kle ) cos(kle ) sin(kle )  k sinh(kle ) k cosh(kle ) −k sin(kle ) k cos(kle ) 



(11.71)

Hence we can express the unknwon coefficents α by the nodal displacements we through the inverse relation α = A−1 we

(11.72)

Inserting Eq. (11.72) into Eq. (11.66), we obtain W (x) = ΦA−1 we = Nwe

(11.73)

with the vector of shape functions N (the so called spectral ansatz). In order to obtain the element stiffness matrix the principle of virtual work is applied, analogously to section 12.1.1. Considering the appearing integrals that have to be solved under this approach, in the present case it is advantageous, to determine the nodal forces directly from the corresponding derivatives. With the vector "

w000 = −w00 and the

#

(11.74)

Matrix

EI 0 D= 0 EI "

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of stiffnesses

#

(11.75)

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the internal forces "

#

−Q σ= M

(11.76)

can be represented as σ = D

(11.77)

i

k

le

wk

wi +Mik +Vik

φi

λ4 =

µ ω2 4 l EI

+Mki

φk +Vki

Figure 11.15: Beam element, sectional forces/moments and degrees of freedom

The sectional forces at the boundaries of the beam qT = {Vik , Mik , Vki , Mki }

(11.78)

result from inserting the coordinates x = 0 and x = le , respectively and considering the definition of positive directions in Fig. 11.15 to 0 k 0 −k   −1 0 1 0  = EIk 2    α = Sα. −k sinh(kle ) −k cosh(kle ) −k sin(kle ) k cos(kle )  cosh(kle ) sinh(kle ) − cos(kle ) − sin(kle ) 

(

q=

EIi −EIk

)



(11.79) With Eq. (11.72) we obtain q = Sα = SA−1 we = Ke we ,

(11.80)

with the element stiffness matrix Ke . Introducing the auxiliary functions Fi (λ) introduced

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by Kolousek [1973], Ke reads  1 1 1 1 F4 (λ) − 2 F5 (λ) F3 (λ)   l2 F6 (λ) le le le   e   1 1     F (λ) F (λ) − F (λ) F (λ) 4 2 3 1  EI  le   le [Ke ] =  1 le − F (λ) − 1 F (λ) 1 F (λ) − 1 F (λ)    2 5 3 4 2 6   le l l l e e e     1 1 F3 (λ) F1 (λ) − F4 (λ) F2 (λ) le le 

(11.81)

with λ = kle . The auxiliary functions F1 (λ) to F6 (λ) read: F1 (λ) =

λ (sinh λ − sin λ) N1 (λ)

F2 (λ) =

λ (cosh λ sin λ − sinh λ cos λ) N1 (λ)

F3 (λ) =

λ2 (cosh λ − cos λ) N1 (λ)

F4 (λ) =

λ2 sinh λ sin λ N1 (λ)

F5 (λ) =

λ3 (sinh λ + sin λ) N1 (λ)

F6 (λ) =

λ3 (cosh λ sin λ + sinh λ cos λ) . N1 (λ)

with N1 (λ) = 1 − cosh λ cos λ Further derivation of element stiffness matrices can be, e.g., found in Petersen [1996]; Grundmann et al [1983].

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Example 11.5: Free vibration of a frame

We consider the following one-storey frame with flexible but longitudinally rigid bars. The system is depicted in Fig. 11.16 together with the identified degrees of freedom and the corresponding individual displacement curves for each degree of freedom. u

1 φ1

0

3 u=0

u=0 φ1 6= 0

2 φ2

φ2 = 0

φ1 = 0

u 6= 0 φ2 6= 0

φ1 = 0

φ2 = 0

Figure 11.16: System and degrees of freedom

Now, the three states 1 to 3 have to be superposed, such that all nodal forces are in equilibrium. a) Free body diagram at node 1 (only nodal moments are depicted)

M12 due to φ1 und φ2 M10 due to φ1 und u b) Free body diagram at node 2 (only nodal moments are depicted)

M21 due to φ1 , φ2 M23 due to φ2 und u

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c) Free body diagram of the bar (only nodal forces are depicted; the equilibrium would have to be dealt with separately for nodes 1 and 2 if the bar would not be modeled as longitudinally rigid) −µ12 l12 ω 2 u

V10 due to φ1 and u

V23 due to φ2 and u

The nodal forces are expressed in terms of λ or ω, respectively, via the auxiliary functions F (λ) (see Eq. (11.81)). The equilibrium conditions have to be expressed at the basis of those nodal forces. Then, we obtain a system of equations for the unknown nodal displacement values.

11.2.3.2 Loaded beam elements - particular solution

In the case of harmonically excited elements, the sectional forces and moments for a clamped beam can be calculated as follows: p(x,t)

Figure 11.17: Assessment of the sectional forces and moments

The equation of motion EIw0000 + µw¨ = p+ (x) e iΩt + p− (x) e −iΩt

(11.82)

is solved by the approach w (x) = W+ (x) e iΩt + W− (x) e iΩt

(11.83)

We now assume Im(p+ ) = 0 (then also Im(p− ) = 0), such that the load p(x,t) = p0 cos(Ωt) oscillates in phase and with constant amplitude for all x. This supposes that all motions, that result due to the nodal displacements and rotations, vary harmonically with the excitation frequency Ω. By that, the arguments λ of the functions Fj (λ) are known. Since the system

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is modeled without damping, the phase shift between the load and the displacement can only be 0 or π. Both phase shifts are covered by the cos function via the sign. Thus, with w(x,t) = W (x) cos (Ωt), the differential equation (11.82) reads EIW 0000 (x) − µΩ2 W (x) = p0

(11.84)

The partial differential equation is reduced to a ordinary differential equation in the variable x. The solution is composed of a homogeneous and a particular part. W = Wh + Wp

(11.85)

The homogeneous solution (part without load) is described by Eq. (11.73) For the particular solution to the right hand side p(x) = p0 = const., we use the approach Wp (x) = −

p0 p0 =− 2 µΩ EIk 4

(11.86)

The full solution Wh + Wp has to satisfy the boundary conditions wTe = {wi , φi , wk , φk } = 0. The corresponding system of equations reads 

1

0

1

         

0

k

0

cosh(k l)

sinh(k l)

cos(k l)

k sinh(k l) k cosh(k l) −k sin(k l)

 C   x1                   k  C    x2      sin(k l)      Cx3           k cos(k l)     

0

Cx4

=

 p0   µΩ2                    0      p0     2   µΩ            

(11.87)

0

The solution gives the constants Cx1 to Cx4 . From the second derivative of the displacement function, we find the moment M = −EIw00 under consideration of the sign convention and k = lλe as Mi0 = −Mk0 =

p0 l2 cosh λ cos λ − sinh λ sin λ λ2 cosh λ cos λ − 1

(11.88)

Analogously the nodal forces can be determined. Further loading conditions can be considered in a similar manner (see Grundmann et al [1983]; Petersen [1996]).

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11.3 Extended beam theories The dispersion equation (11.64) theoretically leads to arbitrarily high bending wave phase velocities. Physically the maximum propagation velocity in the elastic continuum is given by the compresssion wave velocity cp (comp. Continuum Mechanics, Soil Dynamics). This is not a contradiction, since for short wave lengths and corresponding high wave numbers k the assumptions of the Euler-Bernoulli beam theory are not valid anymore. Important assumptions herein are the neglection of rotatory inertia and shear deformation. In certain bounds the range of applicability can then be extended towards the Timoshenko beam theory. For shorter wave lengths, additional terms are then considered. Then, depending on these additional terms describing the rotatory inertia per unit length iθ (for the EulerBernoulli beam theory iθ = 0) and shear stiffness GAG (for the Euler-Bernoulli beam theory GAG → ∞), modified bending wave phase velocities are obtained (see Petersen [1996]; Cremer et al [2005]). Based on this—following Petersen [1996]; Cremer et al [2005]—the general relation between the wave number k und the frequency ω for a beam with uniform mass and stiffness distribution over its length, can be be written as H EIk 4 + 1+ GAG 



−

"

"

!

#

!

kφ EI kφ 1 + bett H + kwbett + kφbett k 2 + 1 + bett kwbett − GAG GAG GAG !

#

H EI kφ kw µ iθ 4 1+ iθ + µ k 2 + 1 + bett µ + bett iθ ω 2 + ω = 0 (11.89) GAG GAG GAG GAG GAG 



with the longitudinal force H (positive in tension), the rotational bedding kφbett and the translaµ iθ ω 4 , a solution tional bedding kwbett describing the attached subgrade. Neglecting the term GA G is directly found. s

ω=

EI µ

v u     kφbett kφbett kφbett kwbett kwbett u H u 1 + 1 + GA 2 + GA k 2 + EI k 2 + 1 + GA EI k EI k4 G G h G  i k2u t k k φ w iθ H EI bett bett iθ 2

1+

1+

GAG

µ

+

GAG

k +

GAG

+

(11.90)

GAG µ

Hence, for the single span beam the changes in the eigenfrequency compared to the EulerBernoulli beam theory can be easily estimated by inserting the wave numbers k = kn (comp. Eq. (11.62)). For the case of a beam hinged at both ends it follows kn =

nπ . l

(11.91)

Often, merely the influence of a normal force or a elastic bedding under the assumption

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GAG → ∞ is of interest. For this case we obtain from Eq. (11.90) s

ω=

v u H EI 2 u u 1 + EIk2 k t

k

φbett + EIk 2 + iθ 2 1+ µk

µ

kwbett EIk4

(11.92)

The eigenfrequencies of a single span beam subject to normal force, bedding, shear deformation and rotatory inertia are summarized in Tab. 11.3 (comp. Grundmann et al [1983]). EI, GA∗ , µ, iθ

H

k x

l

EI: bending stiffness GA∗ : shear stiffness GA κ µ: mass per unit length iθ : rotatory inertia per unit length i2 µ ω0 =

n2 π 2 l2

EI

GA

6= 0

∞

q

EI µ

µ

6= 0 6= 0

6= 0 6= 0 6= 0 6= 0

∞

iθ

6= 0

0 0

H

k

radial eigenfrequncy

0

0

q

0

0

q

6= 0

0

ω0 n2 π 2 l2 ω0 2 2 1+i2 n 2π E G l

1+i2

ω0

q

1+

κ

H PK

with PK =

n2 π 2 l2

EI

s

0

0

6= 0

0

6= 0

6= 0

∞

6= 0

0

0

6= 0 6= 0 6= 0 6= 0

0

nπ H lq µ 6 0 ω0 1 + = 0

0

k l4 EI n4 π 4

!

ω0 q

1 + i2

n2 π 2 l2

(1 +

E G

κ)

for

µ2 i2 κ ω4 ∼ =0 GA

Table 11.3: Radial eigenfrequncies of the single span beam with normal force, bedding, shear deformation and rotatory inertia.

In Kreutz [2013], the beam theory is extended for higher cross-section deformation. There— depending on the frequency—dispersion diagrams for different wave numbers are generated. Fig. 11.18 illustrates the results for a beam with rectangular cross-section. The limits of the corresponding theories can be identified compared to the finite element calculations.

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10.00 8.00

FT FEM compression torsion (St.-Venant)

kx [ rad ] m

6.00

y-bending (Euler-Bernoulli) z-bending (Euler-Bernoulli)

4.00

y-bending (Timoshenko) z-bending (Timoshenko)

2.00 0.00

y-bending 2 (Timoshenko) z-bending 2 (Timoshenko)

0

4000

8000

12000

16000

20000

] ω[ rad s Figure 11.18: Dispersion diagram for different wave numbers for a rectangular cross-section (from Kreutz [2013])

11.4 Thin plates Analogously to beam structures, the dispersion equations can also be determined for the vibration of thin plates. The differential equations of a elastically supported plate according to Kirchhoff theory in Cartesian coordinates with the normal forces Hx and Hy in x- and ydirection, respectively, per unit length (in [N/m]) and the spring stiffness per unit area kw0 bedd (in [N/m3 ]) is given as B

0

∂ 4w ∂ 4w ∂ 4w +2 2 2 + ∂x4 ∂x ∂y ∂y 4

!

+

2 ∂ 2w w 0∂ w 0 + Hy 2 + kwbedd w + ρh 2 = p(x,y,t) ∂x2 ∂y ∂t

∂ Hx0

2

(11.93)

3

Eh with the plate stiffness B 0 = 12(1−ν 2 ) , the plate thickness h, the Poisson ratio ν, the displacement perpenicular to the central plate plane w = w(x,y,t), and the density ρ. Under consideration of shear deformations and rotatory inertia according to the Mindlin theory, Eq. (11.93) can be extended (comp. Cremer et al [2005]).

B

0

∂ 4w ∂ 4w ∂ 4w +2 2 2 + ∂x4 ∂x ∂y ∂y 4

!

B0ρ h3 − +ρ G 12

!

∂ 4w ∂ 4w + ∂x2 ∂t2 ∂y 2 ∂t2 B0 = p− Gh

!

h3 ρ ∂ 4 w +ρ = 12 G ∂t4

∂ 2p ∂ 2p + ∂x2 ∂y 2

!

+

ρI ∂ 2 p (11.94) Gh ∂t2

Here, influences resulting from the normal forces and an elastic bedding are neglected. The solutions for the free vibration problem (p = 0) can be obtained by applying an exponential

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approach to Eqs. (11.93) and (11.94) similar to the procedure for the beam in Eqs. 11.55 and 11.56. Hence, it follows from Eq. (11.93) with the wave numbers kx and ky in x- and y-direction, respectively, and Hx0 = Hy0 = H 0 

B 0 kx2 + ky2

2





+ H 0 kx2 + ky2 + kw0 bett − ρhω 2 = 0.

(11.95)

Introducing the wave number kr2 = kx2 + ky2 , it follows B 0 kr4 + H 0 kr2 + kw0 bett − ρhω 2 = 0.

(11.96)

The resulting dispersion relation thus corresponds to the relations for the beam in Eq. (11.92).   The wave number kr can be interpreted with a wave, inclined by the angle α = arcsin kkxr (comp. Fig. 11.19) in the x-y-plane.

λr λy α λx =

λr sin(α)

x

y Figure 11.19: Einfallswinkel der Biegewelle

Further relations are covered in Cremer et al [2005]. Tab. 11.4 shows the radial eigenfrequencies for rectangular plates under various boundary conditions.

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2π ωi = βi 2 a ν: Poissons ratio µ: mass per unit length hinged

s

E d3 12 (1 − ν 2 ) µ

β1 = β1,1 β2 = min(β1,2 , β2,1 ) γ=

a b

clamping

b

β1,1 = 1.57 (1 + γ 2 ) β2,1 = 6.28 (1 + 0.25 γ 2 )

a

β1,2 = 1.57 (1 + 4 γ 2 )

b a

√ β1,1 = 1.57 1 + 2.5 γ 2 + 5.14 γ 4 √ β2,1 = 6.28 1 + 0.625 γ 2 + 0.321 γ 4 √ β1,2 = 1.57 1 + 9.32 γ 2 + 39.06 γ 4

b a

√ β1,1 = 1.57 5.14 + 2.92 γ 2 + 2.44 γ 4 √ β2,1 = 9.82 1 + 0.266 γ 2 + 0.0625 γ 4 √ β1,2 = 1.57 5.14 + 10.86 γ 2 + 25.63 γ 4

b a

√ β1,1 = 1.57 1 + 2.33 γ 2 + 2.44 γ 4 √ β2,1 = 6.28 1 + 0.582 γ 2 + 0.152 γ 4 √ β1,2 = 1.57 1 + 8.69 γ 2 + 25.63 γ 4

b a

√ β1,1 = 1.57 2.44 + 2.72 γ 2 + 2.44 γ 4 √ β2,1 = 7.95 1 + 0.395 γ 2 + 0.095 γ 4 √ β1,2 = 1.57 2.44 + 10.12 γ 2 + 25.63 γ 4

b a

√ β1,1 = 1.57 5.14 + 3.13 γ 2 + 5.14 γ 4 √ β2,1 = 9.82 1 + 0.298 γ 2 + 0.132 γ 4 √ β1,2 = 1.57 5.14 + 11.65 γ 2 + 39.06 γ 4

Table 11.4: Eigenfrequencies of rectangular plates (Kirchhoff, Eq. (11.93)), without elastic bedding and without horizontal forces) from Grundmann et al [1983]

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11.5 Eigenfrequencies and mode shapes of selected systems In the following eigenfrequencies and mode shapes for two selected examples are given. Further examples can be found in, e.g., Müller and Möser [2004]; Cremer et al [2005]. Circular ring

(from Grundmann et al [1983]) EA: axial stiffness EI: bending stiffness µ: mass per unit length r: radius w: deflection in radial direction

w r

ϕ

mode shape φj j≥2

wj cos(jϕ)

j=0

wj

radial eigenfrequency ωj j (j 2 −1) 1 q EI √ 2 µ j 2 +1 r q 1 EA r µ

(from Müller and Möser [2004])

Cavity filled with air or fluid

V : volume c: compresssional wave velocity in the fluid lx , ly , lz : dimensions in the spatial directions lz z

ly

y

x

boundary

pressure mode shape φj

reverberant cos absorbent

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lx

sin



j1 π x lx





j1 π x lx



cos



j2 π y ly



sin



j2 π y ly



cos



sin



radial eigenfrequency ωj

j3 π z lz

j3 π z lz





πc πc

r

 j1 2 lx

+



 j2 2 ly

+



 j3 2 lz

r

 j1 2 lx

+



 j2 2 ly

+



 j3 2 lz

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11.6 Modal analysis for loaded Euler-Bernoulli beam The modal superposition method can be applied to continuous systems analogously to the application for MDOF systems (comp. Section 9.2). We present the method exemplarily for an Euler-Bernoulli beam hereinafter.

11.6.1 Euler-Bernoulli beam under step load For every time t the vibration of a continuous system can be described by a superposition of mode shapes φj (x) (analogously to the eigenvectors φj for discrete systems) multiplied with time dependent factors yj (t), often called modal coordinates. w(x, t) =

∞ X

φj (x)yj (t) = φ1 (x)y1 (t) + φ2 (x)y2 (t) + . . . + φn (x)yn (t) + . . .

(11.97)

j=1

φ2 (x)

φ1 (x)

φ1,1 φ

φ3 (x)

...

1,2 φ1,3

Analogously to the approach for MDOF systems (comp. Chapter 8, Eqs. 8.19 and 8.20), orthogonality conditions for the mode shapes are given. For an undamped Euler-Bernoulli beam, we find: Zl

φi (x)µ(x)φj (x) dx = mj δij

(11.98)

φ00i (x)EI(x)φ00j (x) dx = kj δij

(11.99)

0

Zl 0

A more formal introduction concerning properties of mode shapes for different differential equations can be found, e.g. in Humar [2012].

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p(x) ms

Iθ

Ps

MS EI(x), µ(x)

kϕ

kw x

x1

x2

x3

l

x4

x5

x6

l For arbitrary systems the generalized quantities then are: mj =

Zl

2

µ(x)φ2j (x) dx + Ms φ2j (x1 ) + Iθ φ0j (x2 )

(11.100)

0

kj =

ωj2 mj

=

Zl

2

2

EI(x)φ00j (x) dx + kw φ2j (x3 ) + kϕ φ0j (x4 )

(11.101)

0

pj =

Zl

p(x)Φj (x) dx + Ps φj (x5 ) + Ms φ0j (x6 )

(11.102)

0

Remark

If the eigenfrequencies of a system are not known, the relation in Eq. (11.101) can be used for estimating the eigenfrequency ωj . For this a mode shape φj (x) is inserted into Eqs. (11.100) and (11.101) and the generalized quantities are evaluated. Then the Rayleigh quotient can be determined (here neglecting rotational inertia and attached springs): Rl

ω2 ≤

0

EI(x)φ2 (x)00 dx Rl 0

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µ(x)φ2 (x) dx

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With the generalized quantities the differential equations can be decoupled, which leads to the known form of the SDOF system mj y¨j + kj yj = pj The time dependent functions yj (t) are generalized or modal coordinates that give the contribution of each mode shape. The accuracy of the solution in Eq. (11.97) improves with the number of considered mode shapes (eigenvectors). In practice, the representation in Eq. (11.97) has to be truncated. For this reason, only the contributions of n mode shapes are considered. w(x, t) ≈

n X

φj (x)yj (t) = φ1 (x)y1 (t) + φ2 (x)y2 (t) + . . . + φn (x)yn (t)

j=1

Then sequentially modal contributions can be added until the solution has converged. Theoretically the exact is obtained by adding all terms (infinitely many), however, considering the Bernoulli-theory. In the case of continuous systems, e.g. beams, it has to be verified that the underlying theory (Euler-Bernoulli- hypothesis) is still valid. Often, a low number of mode shapes is sufficient to obtain an accurate solution. Frequently, only considering the first mode shape is sufficient to describe low frequency vibrations accurately enough. Depending on the form of the load, the different methods for solving a SDOF system can be applied (comp. Chapter 7.4). Example 11.6

F (t)

F

F l 2

l 2

t

x w The depicted single span beam is excited by the point force p(x,t), modeled by a step function at t = 0. We want to determine the displacement and internal forces response, whereby we neglect any damping influences. First, we derive the eigenfrequencies and -vectors. The homogeneous equation of motion

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of the Euler-Bernoulli beam with constant bending stiffness reads EIw0000 + µw¨ = 0 with can be solved by a separation approach w(t) = W (x)f (t) (comp. Section 11.1.2.1) that leads to the following general solution (comp. Eq. (11.57)): W (x) = Cx1 cosh(kx) + Cx2 sinh(kx) + Cx3 cos(kx) − Cx4 sin(kx) f (t) = Ct1 cos(ωt) − Ct2 sin(ωt) For the single span beam, the mode shapes (eigenfunctions) are derived in Ex. 11.3. They read x φj (x) = sin jπ l 



for j ∈ N+ . We can check that the boundary conditions are satisfied: x = 0 : φj (0) = 0,

φ00j (0) = 0

x = l : φj (l) = 0,

φ00j (l) = 0

The corresponding eigenfrequencies are given in Eq. (11.63), repeated here: ωj2

π = j l 

4

EI = j 4 ω12 . µ

This can be solved for ω 2 , which are the eigenfrequencies given in Eq. (11.63). We can now derive the generalized quantities Mj , Kj , pj of the non-homogeneous problem by: mj =

Zl

µ(x)φ2j (x) dx

=µ

0

kj = ωj2 Mj = j 4

Zl 0

x 1 j·π dx = µl l 2





4

π EI 2l3 

pj = F φj (xp ) = F sin j · π

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sin

2

xp l





= F sin j

π 2



11 Solutions for selected continuous systems

221

For the generalized load pj we find π F sin j 2 



=

   F   

0 −F

for j = 1, 5, 9, . . . for j = 2, 4, 6, . . . for j = 3, 7, 11, . . .

This leads to decoupled differential equations that correspond to the differential equation of the SDOF system: mj y¨j + kj yj = pj or y¨j + ωj2 yj =

pj mj

The solution to the non-homogeneous differential equation consists of the sum of the homogeneous and particular solution. The general solution for the j-th mode shape reads: yj,h = yj,h01 cos(ωj t) − yj,h02 sin(ωj t) The following heuristic approach is made, since the load term is a constant function yj,p = yj,p0 Inserted into the equation of motion, it reads ωj2 yj,p0 =

pj Mj

The particular solution finally reads yj,p0 =

pj 2 ωj mj

=

pj kj

Thus, the full solution is given by yj (t) = yj,01 cos(ωj t) − yj,02 sin(ωj t) +

pj kj

y˙ j (t) = −ωj (yj,01 sin(ωj t) + yj,02 cos(ωj t))

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We assume a system at rest at time t = 0. Thus the initial conditions are given by: w(x, 0) = 0 w(x, ˙ 0) = 0 The first condition leads to ∞ X

w(x, 0) =

φj (x)yj (t = 0) = 0

j=1

Thus, all yj (t = 0) have to be zero. yj,01 +

pj =0 kj

yj,01 = −

→

pj kj

The initial condition with respect to velocity results in w(x, ˙ 0) =

∞ X

φj (x)y˙ j (t = 0) = 0

j=1

Thus, all y˙ j (t = 0) have to be zero. This leads to −ωj yj,02 = 0

yj,02 = 0

→

Then, the response of the j-th mode shape follows to pj π yj (t) = (1 − cos(ωj t)) = F sin j kj 2 



2l3 (1 − cos(ωj t)) j 4 π 4 EI

The solution in the physical space is found by superposition: w(x,t) =

∞ X

φj (x)yj (t)

j=1 ∞ X

π x F sin j = sin jπ l 2 j=1 =

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2l3 (1 − cos(ωj t)) j 4 π 4 EI

2F l3 x 1 x sin π (1 − cos(ω t)) − sin 3π (1 − cos(ω3 t))+ 1 π 4 EI l 34 l    1 x + 4 sin 5π (1 − cos(ω5 t)) − . . . 5 l 









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223

The internal forces can be derived from the derivatives of w(x,t), as EI is constant, M (x, t) = −EIw00 (x, t) Q(x, t) = −EIw000 (x, t) or through integration using the equilibrium of forces, where the inertia forces are considered as external loads pI (x, t) = −mw(x, ¨ t), with 2F l3 x ω2 x w(x, ¨ t) = 4 ω12 sin π cos(ω1 t) − 43 sin 3π cos(ω3 t)+ π EI l 3 l "









x ω52 cos(ω5 t) − . . . + 4 sin 5π 5 l 

=



#

2F x x sin π cos(ω1 t) − sin 3π cos(ω3 t)+ ml l l    x cos(ω5 t) − . . . + sin 5π l 









The internal forces follow from the derivatives to: 2F l x x 1 M (x, t) = 2 sin π (1 − cos(ω1 t)) − 2 sin 3π (1 − cos(ω3 t))+ π l 3 l    x 1 + 2 sin 5π (1 − cos(ω5 t)) − . . . 5 l      2F x 1 x Q(x, t) = cos π (1 − cos(ω1 t)) − cos 3π (1 − cos(ω3 t))+ π l 3 l    1 x + cos 5π (1 − cos(ω5 t)) − . . . 5 l 









or through integration 2F x x sin π cos(ω1 t) − sin 3π cos(ω3 t)+ l l l    x + sin 5π cos(ω5 t) − . . . l      F 2F x 1 x Q(x, t) = − cos π cos(ω1 t) − cos 3π cos(ω3 t)+ 2 π l 3 l

pI (x, t) = −

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11 Solutions for selected continuous systems

224

1 x + cos 5(π ) cos(ω5 t) − . . . 5 l      F x 2F l x 1 x M (x, t) = − 2 sin π cos(ω1 t) − 2 sin 3π cos(ω3 t)+ 2 π l 3 l    1 x + 2 sin 5π cos(ω5 t) − . . . 5 l 

table where, due to symmetry, we only consider the range 0 ≤ x < 2l . The following      shows the displacement w 2l , t0 , the moment M 2l , t0 and shear force Q 2l , t0 for the time t0 = ωπ1 relative to the corresponding static value and for a different number of considered modes. j w wst M Mst

1 2 · 96 = 1,971 π4 16 = 1,621 π2

3 1,971 + 0,024 = 1,995 1,621 + 0,180 = 1,801

5 1,995 + 0,003 = 1,998 1,801 + 0,065 = 1,866

M Mst

1+

1,811 + 0,090 = 1,901

Q Qst

8 = 2,546 π

Q Qst

1+

8 = 1,811 π2

4 π

= 2,273

limit 2,0

int. forces derived by

2,0

derivative EIw00

1,901 + 0,032 = 1,933

2,0

2,546 − 0,848 = 1,698

1,698 + 0,509 = 2,207

equilibrium pI = −µw¨

2,0

derivative EIw000

2,273 − 0,424 = 1,849

1,849 + 0,254 = 2,103

2,0

equilibrium pI = −µw¨

The results show that under consideration of a few modes only, already a good approximation of the response is obtained. Note though that this is even valid for the applied load. The generalized load usually decays with increasing frequency, if, e.g. in the case of an earthquake, the load acts evenly distributed on the system. An exception are single loads, for which the generalized load stays more or less equal for increasing j.

+

-

+

+

+

-

+

+ +

Furthermore, a Fourier transform reveals that the amplitude of the force in the frequency

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domain f˜(ω) decays with ∼ ω1 . For this reason, the modal forces pj for higher j decay rapidly for the applied step load. Furthermore we note that the convergence of the internal forces is slower than for the displacement. This is due to the fact that the higher frequency mode shapes exhibit larger bending terms, whereas they give small contributions to the displacement. The accuracy of the internal forces is better when they are determined by the equilibrium method, where the inertia forces are assumed as external loading, rather when determined by the derivatives of the displacements. This leads to the conclusion that dynamic internal forces for structures with known mode shapes should be calculated preferably by the method of external inertia forces. load

moment

tranvsversal forces

F Fl 4 F 2

=

2F l

−

− 2F l π2

2F π

−

−

−

2F l

+ . . . limit:

+ . . .

2F l 9π 2

Fl 2

2F 3π

+ . .

F

The presented method can also be applied to different structures, e.g. plates or shells. Nevertheless the modal analysis is restricted to linear systems, as here the superposition principle is valid.

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11.7 Summary For the elements in the technical bending theory different differential equations apply. Therein, in general, the beam differential equation as differential equation of fourth order has to be distinguished from the differential equations of the torsional, shear or longitudinal rod and the string, those being of second order. Whereas no unique wave velocity results for the beam differential equation due to dispersion of the beam wave, for the differential equations of second order frequency-independent wave velocities can be derived.

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12 Approximate solutions of continuous systems 12.1 Discretization of continuous systems Systems with distributed elasticities and masses have an infinite number of degrees of freedom. In some rare cases, differential equations are found, that can be solved analytically (see Chapter 11). In general this is not possible, and the motion of such systems is approximated by appropriate shape functions. The task to be solved is then described by a finite number of discrete unknowns. Often the unknowns are then the nodal displacements and/or rotations. The system is thus transformed into a discrete system, and corresponds to a MDOF system. For the finite element method, the field of the displacement vector uT (x,y,z,t) = [u(x,y,z,t), v(x,y,z,t), w(x,y,z,t)]

(12.1)

is approximated through spatial shape functions φj (x,y,z). Thereby an approximate solution e (x,y,z,t) is obtained. The problem is hence reduced to a determination of the unknown u time-dependent coefficients αj (t). u(x,y,z,t) ≈

e (x,y,z,t) u

=

n X

αj (t)φj (x,y,z)

(12.2)

j=1

Herafter, the notation e· is dropped for the sake of clarity. The applied shape function have to be able to describe the significant properties of the solution, especially the wavelengths, typically described by the frequency dependent wave numbers k. Therefore knowledge about the analytic solutions is necessary. The single shape functions can extend over large or infinite domains of the system or be limited to finite elements. Especially in dynamics, global shape functions that are directly obtained from analytic solutions (comp. Chapter 11) or that are based on analytic solutions for idealized boundary conditions (e.g. Component Mode Synthesis Buchschmid [2011]; Buchschmid et al [2010]), can be very efficient.

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Example 12.1: Shape functions Euler-Bernoulli beam

We consider a planar problem and uniaxial bending. The displacement w(x) of a beam element perpendicular to the beam axis over the element length le is described by shape functions w (x) = Φ (x) α

(12.3)

with the row vector of the shape functions Φ (x) and the column vector of coefficients α. The displacements in longitudinal direction directly follow from the Euler-Bernoulli beam theroy. The spatial displacement field can thereby be completely described by w(x). In the following, a cubic approach h

Φ (x) = 1, x, x2 , x3

i

with the four coefficients αT = [C0 , C1 , C2 , C3 ] is exemplarily chosen. Practically the coefficients are expressed in terms of the unknown nodal displacements and roations of the beam element wTe = [wi , φi , wk , φk ] with φ =

dw . dx

i

k

le

wk

wi +Mik +Vik

φi

λ4 =

µ ω2 4 l EI

+Mki

φk +Vki

Figure 12.1: Nodal forces and displacement degrees of freedom

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Thus, it reads we = Aα with 1 0  A= 1 0

0 0 0 1 0 0   le le2 le3  1 2le 3le2 



For the coefficients α it holds α = A−1 we with 1 0  1  0 = − 32 − 2  l le 

A−1

2 le3

e

1 le2

0 0 3 le2 − l23 e

0 0   1 − le   

1 le2

Using Eq. (12.3), the displacements along the beam axis can be expressed through the nodal displacements we . w (x) = Φ (x) A−1 we = N (x) we With ξ =

x le

h

(12.4)

we obtain 





N (ξ) = 1 − 3ξ 2 + 2ξ 3 , le ξ − 2ξ 2 + ξ 3 , 3ξ 2 − 2ξ 3 , le −ξ 2 + ξ 3

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i

(12.5)

12 Approximate solutions of continuous systems

230

1

ξ

1 N1 (ξ)

ξ

N2 (ξ)

1 1

N3 (ξ)

1

1

ξ

1 N4 (ξ)

1

ξ

Figure 12.2: Functions for unit displacements/rotations - cubic approach

12.1.1 The principle of virtual work In the dynamic equilibrium, the sum of internal and external virtual work δW of a finite element is given by δW = −

Z

δε σ dΩ − T

(Ω)

Z (Ω)

δu ρ¨ u dΩ + T

Z

δu p dΩ + T

(Ω)

Z

δuT t dΓ = 0

(12.6)

(Γ)

with the domain of the structure Ω, the boundary of the structure Γ, the density ρ, the stresses in the interior of the domain σ, the corresponding vector of stresses on the boundary t, and the vector of external forces p. The vector field δu describes a virtual displacement field, δε the corresponding virtual strains. Reminder 12.1

For the derivation of the equations of motion for a linear elastic isotropic material (Lamé equation, refer to the lecture Continuum Mechanics) the relation between the displacements u, v, and w (in x-, y-, and z-direction) and the resulting strains for linearized systems (small deformations) are: ∂u ∂x ∂v εy = ∂y ∂w εz = ∂z

εx =

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∂v ∂u + ∂x ∂y ∂w ∂v γyz = + ∂y ∂z ∂u ∂w γzx = + ∂z ∂x γxy =

12 Approximate solutions of continuous systems

231

with the strains εk and the shear deformations γmn . Furthermore, for a linear elastic isotropic material the stress-strain relationship is given by σx = λ(εx + εy + εz ) + 2Gεx

τxy = τyx = Gγxy

σy = λ(εx + εy + εz ) + 2Gεy

τyz = τzy = Gγyz

σz = λ(εx + εy + εz ) + 2Gεz

τzx = τxz = Gγzx

The relations given in Eq. (??) can be applied to the n shape functions in Eq. (12.2) (small deformations are assumed, that allow linearization, Bathe [2007]). Hence the strains result in

ε(x,y,z) =

n X

Bj (x,y,z)αj = Bα

(12.7)

j=1

with εT = [εx , εy , εz , γxy , γyz , γzx ]

(12.8)

Thereof n different virtual displacements and the corresponding virtual strains can be derived. δuj (x,y,z) = φj (x,y,z)δαj ,

δεj (x,y,z) = Bj (x,y,z)δαj

(12.9)

The acceleration field is found analogously: u ¨ (x,y,z) =

n X

φj (x,y,z)¨ αj = Φ α ¨

(12.10)

j=1

The stresses can be found under consideration of Eq. (??). Thus, σ(x,y,z) = Eε(x,y,z) = E

n X

Bj (x,y,z)αj = EBα

(12.11)

j=1

with σ T = [σx , σy , σz , τxy , τyz , τzx ]

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(12.12)

12 Approximate solutions of continuous systems

232

and λ + 2G λ λ  λ + 2G λ  λ   λ λ λ + 2G [E] =   0 0 0    0 0 0 0 0 0 

0 0 0 G 0 0

0 0 0 0 G 0

0 0   0  0   0 G 

(12.13)

¨ , φj and σ are column vectors that depend on x, y, and z. α is the The vector fields Bj , ε, u vector of the coefficients αj . In the matrices B and Φ these are contained column-wise. It results a system of equations with n equations, in which every equation corresponds to one single virtual displacement state. Thereby Eq. (12.6) results in δW = −δα

T

Z

B EB dΩ α − δα T

(Ω)

T

Z

¨ ρΦT Φ dΩ α+

(Ω)

+ δα

T

Z

Φ p dΩ + δα T

T

(Ω)

Z

ΦT t dΩ = 0 (12.14)

(Γ)

Example 12.2: Virtual work at the Euler-Bernoulli beam

This example is a continuation of Ex. 12.1. First, the strains in the Euler-Bernoulli beam for uni-axial bending without normal force are approximated by: εx = −zw00

(12.15)

The virtual work in Eq. (12.6) can then be written as:

δW = −

Zle Z 0 (A)

σx δεx dA dx −

Zle Z 0 (A)

ρwδw ¨ dA dx +

Zle

p(x,t)δw dx + Vik δw(x = 0)

0

+ Vki δw(x = le ) + Mik δw0 (x = 0) + Mki δw0 (x = l) = 0 (12.16) with the cross sectional are A and the outer load p(x,t). With Eq. (12.15), the inner

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233

virtual work linked with the stresses follows to: Zle Z

σx δεx dA dx =

Zle Z

Ez w δw dA dx = 2

00

00

EIy w δw dx = 00

00

0

0 (A)

0 (A)

Zle

Zle

M δw00 dx .

0

Further the work related to the d’ Alembert inertia forces reads: Zle Z

ρwδw ¨ dA dx =

0 (A)

Zle

µwδw ¨ dx .

0

The nodal forces are gathered in the vector qT = [Vik , Mik , Vki , Mki ] (comp. Fig. 12.3). node Vki Mki beam element x Mik Vik

w

node Figure 12.3: Nodal forces for the beam element

For determining the inner virtual work using the shape functions in both, the actual and virtual displacement fields, the second derivative of the displacement in Eq. (12.5) has to be determined. B (x) = −

d2 1 N (x) = 2 [6 − 12ξ, le (4 − 6ξ) , −6 + 12ξ, le (2 − 6ξ)] 2 dx le

d2 −w (x) = − 2 N (x) we = B (x) we dx 00

The bending stiffness D = EI is assumed constant over the element length. The d’Alembert inertia forces result in ¨e pi (x,t) = µw(x) ¨ = µN (x) w

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For the application of the principle of virtual work, here, the shape functions are also used for the virtual displacements. Thus, it follows δw (x) = N (x) δwe ,

−δw00 (x) = B (x) δwe

Hence we can write for the work expression in Eq. (12.16) (also comp. Eq. (12.14)):

δW = −

Zle

EIy w (x) δw (x) dx − 00

{z

|

0

00

}

pi (x) δw (x) dx + |

0

T δwT e B (x)[D]B(x)we

Zle

Zle

{z

}

T ¨e δwT e N (x)µN(x)w

p (x) δw (x) dx + Vik δwi + Mik δφi + Vki δwk + Mki δφk = 0 (12.18) |

| {z }

0

T δwT e N (x)

{z

}

δwT eq

Out of this result the stiffness- and mass-matrix: 

δW =

T

δwe 

−

Zle

BT (x) DB (x) dx we + q−

0

|

{z

}

Ke

−

Zle

¨e + µNT (x) N (x) dx w



p (x) NT (x) dx  = 0 (12.19)

0

0

|

Zle

{z

Me

}

|

{z p

}

Thus, the equation of motion for a single beam element is given in terms of the nodal displacements and rotations as ¨ e + Ke w e = q + p Me w For cubic shape functions applied here, the mass- and stiffness-matrices Me and Ke for an element of length le with constant stiffness EI and mass distribution µ are 12 6l −12 6l 2 EI  6l 4l −6l 2l2   Ke = 3    , l −12 −6l 12 −6l 6l 2l2 −6l 4l2 



156 22l 54 −13l 2 µl  22l 4l 13l −3l2    Me =   13l 156 −22l 420  54 −13l −3l2 −22l 4l2 



Me is termed as consistent mass matrix of the beam element. Depending on the ratio of wave length and element length, the mass is often concentrated at the nodes (lumped

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mass) and the inertia forces and moments are applied as single forces and moments. In the case of a viscous damping distributed like the mass µ, the damping matrix cw˙ (x) is treated like mw¨ (x) (comp. Eqs. (12.17) and (12.18)): ˙e pD (x) = cw(x) ˙ = c N (x) w and Ce =

Zl

c NT (x) N (x) dx .

0

This leads to the equation of motion of the beam element with distributed damping. Further reference is given in section 8.2.3.3. ¨ e + Ce w ˙ e + Ke w e = q + p Me w

(12.20)

12.1.2 Lumped mass matrix and static condensation In general, the dynamic analysis of a consistent mass system considerably requires more computational effort than a lumped mass description. At the same time, the accuracy is often only slightly improved. Therefore, it can be advantageous to substitute the general consistent mass system by a lumped mass system. In the lumped mass description, the masses are concentrated in the nodes i.e., for the node i. 1 1 Mi = mi−1 · li−1 + mi+1 · li+1 2 2 where the subscripts i−1 and i+1 are related to the mass or length of the neighboring elements, respectively. The rotatory inertia is neglected in this case. Accordingly, the mass matrix is simply a diagonal matrix, with zero elements in the position referring to the rotational degrees of freedom (The damping matrix can be brought into a corresponding form.). If the nodal rotations do not lead to contributions to the mass and damping matrix, it appears advantageous to eliminate them by a procedure called static condensation. To derive this, the unknowns are rearranged as follows: h

i

wT = [w1 , w2 , . . . , wn , φ1 , φ2 , . . . , φn ] = wTT , wTφ .

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The equations of motion are then written as "

Kww Kwφ Kφw Kφφ

#"

#

"

#"

wT Mw 0 + wφ 0 0

¨T w Cw 0 + ¨φ w 0 0 #

"

#"

˙T w p = T ˙φ w 0 #

"

#

with the nodal forces pT . The second row describes the static equilibrium without inertia forces and gives a dependence between wφ and wT : Kφw wT + Kφφ wφ = 0

→ wφ = −K−1 φφ Kφw wT

If this is substituted into the first row, the equation of motion with the order reduced by half ¨ T + Cw w ˙ T + Kw wT = pT Mw w is obtained, with the modified stiffness matrix Kw = Kww − Kwφ K−1 φφ Kφw

12.1.3 Principle of Hamilton Alternatively the equations of motion can be determined from the principle of Hamilton (comp. 4.1.3.1). The term extends the approach with the work of the non-conservative forces (external forces, damping forces, general forces, that cannot be described by a potential, comp. section ??).

δ

Zt2

Ekin − Epot dt +

t1

Zt2

δWnc dt = 0

(12.21)

t1

The procedure is analogous to the example 4.10.

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13 Random vibrations 13.1 Introduction Most of the excitation processes in structural dynamics are of random nature, respectively have random components. However, many of them are modeled by deterministic loads. With the help of the probability theory we can deduce properties of the stochastic response under the knowledge of the random properties of the excitation process. Thereby an uncertain quantity, e.g. an excitation force or displacement, is modeled as a random variable. Using the theory of probability certain characteristics of a random response can then be determined, which allow to assess probabilities of the underlying events, e.g. probability of exceeding a certain threshold. This procedure is outlined briefly in the following. Further insights can be found in [Wirsching et al 2006; Roberts and Spanos 2003; Petersen 1996]. [Li and Chen 2009] gives an overview over models for stochastic processes for typical excitations. The derivation in this chapter follow Lutes and Sarkani [2004]. Excitation and response time histories that are random in nature can be modeled as a stochastic (or random) processes. A stochastic process can be understood as a set of random variables. Consider for example some process x(t), such as an excitation or response that evolves over time. At every instance of time t the outcome of the process is uncertain and x(t) is modeled as a random variable X(t). The whole stochastic process is then denoted by {X(t)}1 . The parameter t can be viewed as the index parameter for the process. Usually this index set is continuous. A stochastic process can also be interpreted in terms of all possible realizations. Note though that this concept only holds in the limit for an infinite number of observations. Nevertheless, through a sufficiently large number of realizations (observations) of the process a good understanding of the likelihood of certain outcomes can be obtained. The collection of realizations, also called time histories, of a stochastic process is then called ensemble. From m observations discrete histograms can be derived, from which the relative frequency of occurence nmk emerges. The histogram describes how often the process {X(t)} at time ti takes a certain value x. For this, m time histories xi (t) have to be evaluated. In the limit ∆x → 0 1

This follows the common practice to denote a set

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and m → ∞ the histogram corresponds to the probability density function fX(t) (x). In most applications however, it is not feasible to collect a sufficient number of process time histories in order to obtain an accurate description of the process {X(t)}

xm (t)

xm (t1 )=am

xm (t2 )=bm

t x2 (t) x1 (t)

x2 (t1 )=a2

x2 (t2 )=b2

t x1 (t1 )=a1

Ensemble

x1 (t2 )=b1

t t = t1

τ t = t2

Figure 13.1: Stochastic process with m realizations

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x1 (t)

realization t

histogram at time ti

∆x

(for the total ensemble)

x2 (t) ensemble

∆x t

nk m

xm (t) x ∆x t

∆x ti

tj Figure 13.2: Histogramm X(ti )

13.2 Description of random variables 13.2.1 Single random variable A real single random variable is a function, X : S → R that maps events in S to intervals in R. In the scope of this lecture we are usually concerned with continuous random variables. Through random variables we can identify the probability of occurence of certain events. To this end, we describe the probability Pr of occurrence of the event {X ≤ x}, i.e., the event that the random variable X takes a value smaller or equal than x, by the cumulative distribution function (CDF) FX in terms of the outcome x. Thus, FX (x) = Pr (X ≤ x) .

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13 Random vibrations

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For continuous random variables we define the probability density function (PDF) fX (x) by fX (x) dx = Pr (x < X ≤ x + dx) .

(13.2)

It gives the probability that the outcome of X takes a value in the interval x < X ≤ x + dx (comp. Fig. 13.3). We note, that the PDF of a random variable itself does not give a probability, the probability in the above equation is given for the infinitesimally small interval dx. From differentiation of the CDF, we find the PDF and vice versa, from the integration of the PDF we find the CDF, fX (x) = FX (x) =

dFX (x) dx Zx

(13.3)

fX (x) dx

(13.4)

−∞

Especially it holds Z∞

fX (x) dx = FX (∞) = 1.

(13.5)

−∞

fX (x) fX (x) dx = Pr(x < X < x + dx) µX

dx

x

Figure 13.3: Probability density function fX (x) of the random variable X

Information on the random variable X can also be given in terms of its moments. To this end, we define the expected value of any function g (X) of the random variable X as E [g (X)] =

Z∞

g (x) fX (x) dx

(13.6)

−∞

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We define the n-th moment of random variable X as follows: E [X ] = n

Z∞

xn fX (x) dx

(13.7)

−∞

For the case n = 1, the first order moment is obtained. It is called the mean or expected value of X: µX = E [X] =

Z∞

xfX (x) dx

(13.8)

−∞

For n = 2, the second order moment, also termed mean square of X, is obtained. Furthermore, the n-th central moment of X is defined as Z∞

E [(X − µX ) ] = n

(x − µX )n fX (x) dx

(13.9)

−∞

It holds E [(X − µX )] = 0

(13.10)

For the case n = 2, we obtain the variance of X. h

Var [X] = E (X − µX )

2

i

=

Z∞

(x − µX )2 fX (x) dx

(13.11)

−∞

It gives a characterization of the deviation of the outcomes of X around the mean value µX . The standard deviation σX is then defined as σX =

q

Var [X]

(13.12)

Most commonly the normal distribution (Gaussian distribution) is used as a probability distribution. Its PDF is given by fX (x) =

1 √

σX 2π

e

−

(x−µX )2 2σ 2 X

(13.13)

This distribution is fully characterized by two parameters, which can be derived from its first two moments, i.e., mean value µX and standard deviation σX .

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σ2 2σ2 1

fX (x)

fX (x) = probability density function 2

2σ1

x

µX = mean value σ = standard deviation µX1 = µX2 σ1 < σ2

µX σ 1 Figure 13.4: Normal distribution

13.2.2 Two random variables Two random variables X and Y are described by the joint PDF fXY (x,y): fXY = fXY (x,y) y fXY (x,y)

y

x x

Now, we have two mean values and variances, each for the single random variables X and Y . The mean values are given by µX = E [X] =

Z∞ Z∞

xfXY (x,y) dx dy

(13.14)

−∞ −∞

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Z∞ Z∞

µY = E [Y ] =

yfXY (x, y) dx dy

(13.15)

−∞ −∞

and the variances by: h

2 Var [X] = σX = E (X − µX )2

h

i

(13.16)

i

Var [Y ] = σY2 = E (Y − µY )2 .

(13.17)

Analogously to Section 13.2.1, we furthermore introduce the joint moments of X and Y of order (nx + ny ) E [X Y nx

ny

Z∞

]=

xnx y ny fXY (x,y) dx dy

(13.18)

−∞

and the joint central moments of X and Y of order (nx + ny ) E [(X − µX )

nx

(Y − µY ) ] = ny

Z∞

(x − µX )nx (y − µY )ny fXY (x,y) dx dy

(13.19)

−∞

Of most interest are lowest order joint moments, i.e., nx = ny = 1. For this case the joint central moment is called covariance, given by Cov [X,Y ] =

2 σXY

= E [(X − µX ) (Y − µY )] =

Z∞ Z∞

(x − µX ) (y − µY ) fXY (x,y) dx dy

−∞ −∞

Furthermore the correlation coefficient ρXY is defined as ρXY =

Cov [X,Y ]

q

2 σXY = σX σY Var [X] Var [Y ]

q

The covariance and the correlation coefficient can be interpreted as a measure of linear dependence between X and Y .

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ρXY > 0

ρXY < 0

ρXY = 1

ρXY ≈ 0

ρXY ≈ 0

ρXY ≈ 0

For the case that X and Y are completely independent, covariance and correlation coefficient vanish. Example 13.1

Consider the two random variables U and F , with mean value µF and standard deviation σF , and where additionally it holds U = kF + u0 . k and u0 are deterministic. This relation describes, e.g. the random displacement response of a linear system (spring) subject to a random static load. Thus, F and U are perfectly linearly dependent. The expected value of U is given by µU = kµF + u0 Then, we find for the variance σU2 σF2 U = E [(F − µF )(kF + u0 − c1 µF − u0 )] = E [(F − µF )k(F − µF )] h

= k E (F − µF )2

i

= kσF2 and the covariance σF2 U h

σU2 = E (kF + u0 − kµF − u0 )2 h

= k 2 E (F − µF )2

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13 Random vibrations

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= k 2 σF2 The correlation coefficient then reads ρF U =

σF2 U kσF2 = = ±1 σF σU σF |k| σF

Because spring stiffness are greater than zero here, the case −1 can be omitted.

13.3 Description of stochastic processes As noted in Section 13.1, a stochastic process is a set of random variables. It can thus be seen as a generalization of random vectors, where the random vector consists of infinitely many random variables. In the following we introduce common descriptions of a stochastic process.

13.3.1 Probability distribution In order to have a complete probabilistic description of a stochastic process, the joint probability of all single random variables {X(t1 ), X(t2 ), . . . , X(tn )} needs to be known. This can be given by the joint probability density function fX(t1 )X(t2 )···X(tn ) (x1 , x2 , . . . , xn )

(13.20)

In general, the information available on a random process is at most its first or second order marginal distribution, e.g. fX(t1 ) (x1 )

(13.21)

fX(t1 )X(t2 ) (x1 , x2 )

(13.22)

The marginal distributions can be easily derived from the more general description in Eq. (13.20). Thereby fX(t1 ) (x1 ) describes the probability density for the random variable X(t1 ) at a single time instance only, and fX(t1 )X(t2 ) (x1 , x2 ) describes the joint distribution of any two random variables X(t1 ) and X(t2 ) at two time instances. An equivalent description can also be given in terms of the joint cumulative distribution function. Often, descriptions of stochastic processes are not given in terms of the joint probability, but in terms of their moment functions.

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13.3.2 Moment functions The mean or expected value of a stochastic process is defined by µX (t) = E [X(t)] =

Z ∞ −∞

xfX(t) (x) dx .

(13.23)

Note that now the mean value is a function of time. The autocorrelation function of the stochastic process {X(t)} is defined as φXX (t1 , t2 ) = E [X(t1 )X(t2 )] =

Z ∞ Z ∞ −∞

−∞

x1 x2 fX(t1 )X(t2 ) (x1 ,x2 ) dx1 dx2

(13.24)

Correspondingly, we also define the autocovariance function of {X(t)} KXX (t1 , t2 ) = E [[X(t1 ) − µX (t1 )] [X(t2 ) − µX (t2 )]]

(13.25)

Mean, autocorrelation and autocovariance are related by KXX (t1 , t2 ) = φXX (t1 , t2 ) − µX (t1 )µX (t2 )

(13.26)

Mean-square and variance function are obtained, by evaluating autocorrelation and autocovariance function, respectively, at the same time t = t1 = t2 : E X 2 (t) = E [X(t)X(t)] = φXX (t, t)

h

i

(13.27)

2 σX (t) = KXX (t,t)

(13.28)

Similarly, higher order moments can be defined, which is not further illustrated in the scope of this lecture. It is noted that a full probabilistic description requires full knowledge about all moments of the process. In this case the information contained in the full PDF of the process and all the moment functions is the same. Example 13.2: Cosine function with random amplitude and phase

Consider the following stochastic process X(t) = X0 cos (ωt + Φ) where X0 and Φ are random variables and ω is the deterministic frequency. Furthermore X0 and Φ are assumed to be independent and Φ is assumed to be uniformly distributed

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over the interval [0,2π]. The mean of the process is given by µX (t) = E [X0 cos (ωt + Φ)] Due to the independence we can write µX (t) = E [X0 ] E [cos (ωt + Φ)] = µX0

Z2π

cos (ωt + ϕ) fΦ (ϕ) dϕ

0

The integral over the cosine function equals zero, thus it is µX (t) = 0. This holds for any µX0 < ∞. Furthermore, since the mean value is zero, autocorrelation φXX and covariance function KXX coincide. We write φXX (t1 ,t2 ) = E [(X0 cos (ωt1 + Φ)) (X0 cos (ωt2 + Φ))] Again, due to the independence of X0 and Φ it follows h

i

φXX (t1 ,t2 ) = E X02 E [cos (ωt1 + Φ) cos (ωt2 + Φ)] We introduce the identity cos(x) cos(y) = second part of the preceding equation as E [cos (ωt1 + Φ) cos (ωt2 + Φ)] =

1 2

(cos(x + y) + cos(x − y)) and rewrite the

1 E [cos (ω(t1 + t2 ) + 2Φ) + cos (ω(t1 − t2 ))] 2 2π

1 1Z (cos (ω(t1 + t2 ) + 2ϕ) + cos (ω(t1 − t2 ))) dϕ = 2 2π 0

=

1 cos (ω(t1 − t2 )) 2

Then, the autocorrelation function reads φXX (t1 ,t2 ) =

E [X02 ] E [X02 ] cos (ω(t1 − t2 )) = cos (ωτ ) 2 2

with τ = t1 − t2 . Fig. 13.5 depicts the autocorrelation function in terms of the time difference τ and Fig. 13.6 shows five different realizations of the process, where additionally it is assumed that the amplitude X0 follows a standard normal distribution. Subsequently we will see that a process, for which the autocorrelation only depends on the time shift between two points, is called stationary. Thus, the stochastic process introduced in this example is stationary.

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φXX

x(t)

E[X02 ] 2

−3π −π −2π

4π ωt

3π ωτ

π 2π

−

E[X02 ] 2

Figure 13.5: Autocorrelation function φXX for the above process in terms of time shift.

Figure 13.6: Five realizations of the stochastic process with X0 following the standard normal distribution.

13.3.3 Stationarity of stochastic processes Stationarity of a stochastic process describes the property that some description of the stochastic process is unchanged by an arbitrary shift along the time axis. A stochastic process {X(t)} is called mean value stationary, if µX (t + τ ) = µX (t).

(13.29)

For this to hold, it must be µX (t) = µX ,

(13.30)

i.e., the mean value of the process is constant over time. The process {X(t)} is called secondmoment stationary if the second-moment function, i.e., the autocorrelation function φXX is invariant under a time shift: φXX (t1 + r, t2 + r) = φXX (t1 , t2 )

(13.31)

This results in the fact that the autocorrelation function only depends on one time argument, which is the time difference τ = t1 − t2 , or t1 = t2 + τ . Thus, RXX (τ ) = φXX (t + τ, t) = E [X(t + τ )X(t)] .

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Here, we set t2 = t. Furthermore, the process {X(t)} is called covariant stationary if the second central moment function, i.e., the covariance function KXX is invariant under a time shift: GXX (τ ) = KXX (t + τ, t) = E [(X(t + τ ) − µX (t + τ )) (X(t) − µX (t))] .

(13.33)

A process that is both mean and second-moment stationary can be shown to also be covariant stationary. Processes with this type of stationarity are also termed weakly stationary or widesense stationary. Strict stationarity implies stationarity in the probability density function. For a covariant stationary process the variance of the process is found by evaluation of the covariance function for the time shift τ = 0 (Eqs. ??,): h

i

Var [X] = E (X(t) − µX (t))2 = KXX (t, t) = GXX (0)

xm (t)

xm (t1 )=am

(13.34)

xm (t2 )=bm

t x2 (t) x1 (t)

x2 (t1 )=a2

x2 (t2 )=b2

t x1 (t1 )=a1

Ensemble

x1 (t2 )=b1

t t = t1

τ t = t2

13.3.4 Ergodicity of stochastic processes Ergodicity describes the property of a stochastic process that a mathematical expectation is equally described by a time average over an infinitely long time history, i.e., one realization of the process. Analogously to the definition of stationarity, different types of ergodicity exist. The process {X(t)} is called ergodic in mean if it is mean-value stationary and it holds T 2 1 Z x(t) dt µX ≡ lim T →∞ T T

(13.35)

−2

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Figure 13.7: from [Petersen 2000, p.1220, Fig. 34]

Furthermore, the process is called ergodic in second moment if it is second-moment stationary and it holds T 2 1 Z RXX (τ ) ≡ lim x(t + τ )x(t) dt T →∞ T T

(13.36)

−2

Also the process is called ergodic in covariance if it is covariant stationary and it holds T

1 T →∞ T

GXX (τ ) ≡ lim

Z2

[x(t + τ ) − µX ] [x(t) − µX ] dt

(13.37)

− T2

For this case, we find the variance of the process by T

2 Var [X(t)] = σX = GXX (0) ≡ lim

T →∞

1 T

Z2

[x(t) − µX ]2 dt

(13.38)

− T2

13.3.5 The Gaussian process The Gaussian process is a stochastic process {X(t)} for which every finite subset of random variables that constitute the process {X(t1 ), X(t2 ), . . . , X(tn )} has a multivariate normal distribution. A special property of the Gaussian process is that it is fully defined by its mean vector and covariance matrix. The PDF of the collection X = {X(t1 ), X(t2 ), . . . , X(tn )} for any n can then be written as 1

1 exp − (x − µX )T Σ−1 fX (x) = q XX (x − µX ) 2 (2π)d det (ΣXX )

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(13.39)

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where µX is the vector containing the mean values of all X(ti ) and ΣXX is the covariance matrix. Many physical phenomena are the result of the superposition of a large number of stochastic processes, e.g. wind turbulences result from the superposition of different vortices or the excitation due to waves results from the superposition of an infinite number of different surface waves with different directions, phases and amplitudes. The central limit theorem [Maybeck 1979] states (in short) that the sum of independent and identically distributed random variables can be described by a normal distribution. An important property is that a Gaussian load process [Bendat and Piersol 2010; Runtemund 2013] results in a Gaussian response process for linear systems. Then by the knowledge of mean and covariance function, the stochastic response process is completely defined.

13.3.6 Frequency domain analysis of stochastic processes Subsequently we are interested in a frequency domain analysis of stochastic processes. In order to obtain a frequency domain representation we can apply the Fourier transformation. For convenience we repeat the definition of the Fourier transformation, given in Eq. (3.34) and Eq. (3.35). Z∞

f˜(ω) =

f (t)e −iωt dt

(13.40)

−∞ ∞ 1 Z ˜ f (ω)e iωt dω f (t) = 2π

(13.41)

−∞

We now apply the Fourier transformation to the process {X(t)} to define the new process ˜ {X(ω)}. ˜ X(ω) =

Z∞

X(t)e −iωt dt

(13.42)

−∞

˜ The mean value of the Fourier transformed process {X(ω)} is given by ˜ µX˜ (ω) = E X(ω) = h

i

 ∞ Z  E −∞



X(t)e

−iωt

dt

=

Z∞ −∞

˜X (ω) E [X(t)] e −iωt dt = µ |

{z

µX (t)

(13.43)

}

We note that the mean value of the Fourier transformed process is simply the Fourier transform of the mean value function of {X(t)}. This only holds, if the Fourier transformation

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of µX (t) exist, which is not the case for stationary signals, because then the mean value is a constant and thus, not absolute-value integrable. We will consider the analysis for stationary signals hereafter. Furthermore, we consider the second moment function of the process in the frequency domain. To this end, we modify the definition of the autocorrelation function, because the process in the frequency domain is now in general complex: ˜ 1 )X ˜ ∗ (ω2 ) φX˜ X˜ (ω1 ,ω2 ) = E X(ω h

i

=

 ∞ Z E

 ∞ Z X(t1 )e −iω1 t1 dt1 

−∞

−∞

=

 ∞ ∞ Z Z E

∗ 

X(t2 )e −iω2 t2 dt2   

X(t1 )X(t2 )e −iω1 t1 e iω2 t2 dt1 dt2 

−∞ −∞

Z∞

=

Z∞

E [X(t1 )X(t2 )] e −iω1 t1 e iω2 t2 dt1 dt2

−∞ −∞

Z∞ Z∞

=

φXX (t1 ,t2 )e −i(ω1 t1 −ω2 t2 ) dt1 dt2

(13.44)

−∞ −∞

Analogously, we find the expression for the covariance function Z∞ Z∞

KX˜ X˜ (ω1 ,ω2 ) =

KXX (t1 ,t2 )e −i(ω1 t1 −ω2 t2 ) dt1 dt2

(13.45)

−∞ −∞

This expression will also not exist, if the process is not second-moment stationary. In order to circumvent the problem of undefined Fourier transforms for stationary signals, we present an alternative procedure that applies a slight modification to the considered signals. To this end, we consider a truncated version of the process {X(t)}, denoted by {XT (t)}: T XT (t) = X(t) H t + 2 





T −H t− 2 



(13.46)

where H(t) is the Heaviside function. This modified process coincides with the original process in the limit T → ∞. For the truncated stochastic process, the covariance function reads T

KX˜T X˜T (ω1 ,ω2 ) =

− T2

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T

Z2 Z2

GXX (t1 − t2 )e −i(ω1 t1 −ω2 t2 ) dt1 dt2

(13.47)

− T2

13 Random vibrations

253

Here we made use of the knowledge that the covariance of the stationary process depends only on the time difference τ = t1 − t2 and is given by GXX . According to Lutes and Sarkani [2004], Eq. (13.47) can be reformulated, which yields an expression for the covariance function KX˜T X˜T (ω1 ,ω2 ) that is bounded for all T including the limit T → ∞, but ω1 6= ω2 . It can be further derived that for the case ω = ω1 = ω2 and in the limit T → ∞ the following holds lim KX˜T X˜T (ω,ω) = T

T →∞

Z∞

GXX (τ )e −iωτ dτ

(13.48)

−∞

This expression grows proportionally with T for large T . To obtain an expression that exists in the limit T → ∞, we define a new function through dividing KX˜T X˜T (ω,ω) by T : 1 KX˜T X˜T (ω,ω) T →∞ T

SXX (ω) = lim

(13.49)

The function SXX (ω) is called autospectral density function or power spectral density or just spectral density. By inserting Eq. (13.48) into Eq. (13.49), we find: Z∞

SXX (ω) =

GXX (τ )e −iωτ dτ

(13.50)

−∞

Thus, the power spectral density is given by the Fourier transform of the autocovariance function of the process in the time domain. For the zero mean process Z(t) = X(t) − µX (µZ = 0), covariance function and autocorrelation function coincide and the spectral density is equivalently given by Z∞

SZZ (ω) =

RZZ (τ )e −iωτ dτ .

(13.51)

−∞

The inverse relation to Eq. (13.50) is given by ∞ 1 Z GXX (τ ) = SXX (ω)e iωτ dω 2π

(13.52)

−∞

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sinusoidal

sinusoidal with

stochastic

superimposed

process with narrow band

random fluctuations

with broad band

Figure 13.8: Comparison of different processes from [Petersen 2000, p. 1228, Fig. 43])

Example 13.3: White noise process

A special case of a stochastic process is the white noise process. The white noise process {X(t)} is the weakly stationary process with zero mean µX = 0, and the following autocorrelation function 2 RXX (τ ) = GXX (τ ) = σX δ(τ )

where δ(τ ) is the Dirac delta function that is equal to zero for all τ 6= 0. We can

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calculate the spectral density of this process by evaluating Eq. (13.50): SXX (ω) =

Z∞

GXX (τ )e

−iωτ

−∞

dτ =

Z∞ −∞

2 δ(τ )e −iωτ σX

dτ =

2 σX

Z∞

δ(τ )e −iωτ dτ

−∞

By application of the sampling property of the Dirac delta function, f (a), we obtain

R∞

−∞

δ(x−a)f (x) dx =

2 −iω0 2 SXX (ω) = σX e = σX ,

which is constant throughout the entire frequency range and has an amplitude that corresponds to the variance of the stochastic process. It should be noted that this process is unphysical as it has an infinite energy content. Nevertheless it is often used as an approximation in practical applications.

Figure 13.9: Autocorreation function and spectral density for the white noise process [Petersen 2000, p. 1228, Fig. 43])

13.3.7 Stochastic response of linear dynamic system In the preceding section we derived expressions to describe the first two moment functions of stochastic processes. Consider now the deterministic linear problem, where usually an excitation is given from which we want to calculate the response of a system. In order to do so, we describe the relation between excitation and response by differential equations. As was discussed, e.g. in Chapter 7 for the linear SDOF system, the deterministic system response in the frequency domain can be described by its transfer function H(ω), x˜(ω) = H(ω)f˜(ω).

(13.53)

This relationship can also be stated for the stochastic excitation and response processes {F˜ (t)} ˜ and {X(t)}: ˜ F˜ (ω) ˜ X(ω) = h(ω)

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(13.54)

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For a known excitation process, we now want to derive the first two moment functions of the response process. Taking the expectation of Eq. (13.54) gives ˜ F˜ (ω) = h(ω) ˜ ˜ ˜ µX˜ (ω) = E X(ω) = E h(ω) E F˜ (ω) = h(ω)µ F˜ h

i

h

i

h

i

(13.55)

For the case of covariant stationary processes, we now derive the spectral density function for the response process. For this we first rewrite Eq. (13.49):   i 1 1 h ˜ ˜ T − µ ˜ (ω) ∗ KX˜T X˜T (ω,ω) = lim E XT − µX˜ (ω) X X T →∞ T T →∞ T

SXX (ω) = lim

(13.56)

Rephrasing the multiplication with the complex conjugate as the square of absolute values, we find  2  1 ˜ SXX (ω) = lim E XT − µX˜ (ω) T →∞ T

(13.57)

Inserting Eqs. (13.54) and (13.55) into Eq. (13.57), we obtain  2  1 ˜ F˜ (ω) − h(ω)µ ˜ E h(ω) F˜ T →∞ T  2 2  1 ˜ ˜ = h(ω) lim E F (ω) − µF˜ T →∞ T | {z }

SXX (ω) = lim

KF˜ F˜ (ω,ω)

|

{z

SF F (ω)

}

˜ 2 = h(ω) SF F (ω)



(13.58)

The result shows that the spectral density of the response process is obtained by multiplication of the excitation spectral density by the square of the absolute value of the transfer function. From this, further response characteristics can be obtained, e.g. the response variance by consideration of Eq. (13.52): ∞ ∞ 1 Z 1 Z iω0 Var [X] = GXX (τ = 0) = SXX (ω)e dω = |H(ω)|2 SF F (ω) dω 2π 2π −∞

(13.59)

−∞

If the stochastic load process is a Gaussian process then the response process is also Gaussian and fully characterized by the mean and covariance function derived in this chapter. The following figure exemplarily shows the work flow for a SDOF system. Here we see how the system “filters” the significant contributions of the load via its transfer function.

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13 Random vibrations

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F (t) m k

c

time domain w(t)

F (t) σF

convolution

t T

1 w(t)= m·ω D

R(τ )= lim

T →∞

R

F (t)·F (t+τ )dt

−T 2

Fˆ F (f )=

R

F (t) sin[ωD (t−τ )]e−δ(t−τ ) dτ

R(τ )e−i2πτ f dτ

+∞

F (t)e−i2πtf dt

SF (f )

2= σw

−∞

SF (f )= lim

R

Sw (f )df

−∞

|Fˆ F (f )|2 T

T →∞

−∞

multiplication ˆ (f )|2 Sw (f )=SF (f )·|U 2 ˆ U (f )

Sw (f )

1 f

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T

+∞

+∞

SF (f )=

RT

σw

t

frequency domain

T

R2

w(t)

fe

f

fe

f

13 Random vibrations

258

14 Aeroelastic vibrations (wind) Different types of wind excited vibrations are distinguished: • By gusts induced vibrations (vibrations in wind direction) • Vibrations due to vortex shedding (vibrations transversely to the wind direction) • Self excited vibrations (stability problems) • Galloping (vibrations transversely to the wind direction) • Divergence and fluttering (vibrations transversely to the wind direction) • Interference phenomena (interaction phenomena between different structures) First the wind effects will be discussed phenomenologically. Then current standards are addressed.

14.1 General information 14.1.1 Loads The basic value of all flow induced forces is the velocity pressure q (also termed dynamic pressure): ρ q = v2 2

(14.1)

kg and the wind velocity v. The force values W are obtained with the density of air ρ = 1,25 m 3 from the velocity pressure through multiplication with the aerodynamic coefficients. These coefficients are found in the literature or can be determined from the following relation. cf =

W qA

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(14.2)

259

where W is the measured resulting force of all flow induced pressure, pull and friction forces, acting on the structure, and A is the loaded area. Analogously the moment coefficient is found.

cM =

M qAl

(14.3)

14.1.2 Description of wind 14.1.2.1 Wind velocity

The effect of different wind velocities is given through the Beaufort scale. Beaufort number

Description of wind

0 1 2 3 4

calm light air light breeze gentle breeze moderate breeze

5

fresh breeze

6 7 8

strong breeze near gale gale

9 10

strong gale storm

11 12

violent storm hurricane

Land conditions Smoke rises vertically Smoke rises almost vertically almost felt on face moves leaves and small twigs raises dust and loos paper, moves small branches moves larger branches, inconvenience felt whistling heard whole trees in motion, twigs break of trees, impedes pedestrians slight structural damage trees uprooted, considerable structural damage widespread damage devastation

Average wind speed in m/s 0 - 0,4 0,4 - 1,8 1,8 - 3,6 3,6 - 5,6 5,6 - 7,9 7,9 - 10,4 10,4 - 13,1 13,1 - 16,0 16,0 - 19,1 19,1 - 22,1 22,1 - 26,0 26,0 - 29,7 > 29,7

14.1.2.2 Temporal and spatial structure of wind velocity measurements

Fig. 14.1 depicts typical measurements of the wind speed at three different heights of a mast.

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260

u [m/s]

153 m

64 m

12 m

30 25 20 15 10 1

2

3

4

5

t [min.]

Figure 14.1: Measured wind velocities at three different heights [Holmes 2007, p. 56]

From this it can be deduced, e.g. [Holmes 2007, p. 55-56]: • The average wind velocity increases with the height up to a limit value. • Turbulences/gusts occur at all heights. • Gusts are broadband, i.e., slow changes are superimposed with highly frequent vibrations. • Low frequency vibrations occur similarly at all heights. • The gustiness over smooth terrain (sea, flat land) is smaller than for rugged terrain (cities) und decreases for increasing height [Stathopoulos and Baniotopoulos 2007, p. 3-4].

14.1.3 Temporal structure of the wind velocity 14.1.3.1 Power spectral density

[Van der Hoven 1957] investigated the frequency content of the wind and the weather system in the atmospheric boundary layer on the basis of differently long measurements of the wind velocity in the height of 80 m at the tower of the Brookhaven National Laboratory, New York. Fig. 14.2 shows a typical wind spectrum, that is gained by assembling the different time scales. A logarithmic frequency scale is chosen to cover the large frequency domain. In order to obtain a measure of energy per frequency by the area under the depicted power spectral density, the y-axis is scaled via the following relation: σv2

=

Z ∞ 0

Gv (f ) df =

Z ∞ 0

f Gv (f ) d(ln f )

(14.4)

with the one sided power spectral density function Gv (f ) = 2S(f ). The wind spectrum can be split into the “mesometeorological” and the “micrometeorological” range, with periods of over and below one hour, respectively. In between lies the “spectral gap” with low energy input (periods of under a day to around ten minutes).

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mean wind velocity

Anual peak

f S(f)

(speculativ not measured)

turbulence

Synoptic peak (passage of weather systems)

Turbulent peak Diurnal peak (day-night)

Spectral gap

log f -4

frequency 1/h 10 period in h 10000

-3

10 1000

1 year

-2

10 100

10 10

-1

1 1

4 days 0,5 days mesometeorological

10 0,1

100 0,01

5 min. 1 min.

1000 0,001

5 sec.

micrometeorological

Figure 14.2: Frequenzspektrum der horizontalen Windgeschwindigkeit (1957), Messungen in Brookhaven, New York [Burton et al 2001, p. 12], [Petersen 2000, p. 596]

In wind engineering, in particular the higher frequency parts in the “micrometeorological” range are of interest. These can be represented as stochastic processes with zero mean value on the basis of the power spectral density [Burton et al 2001, p. 11ff]. 14.1.3.2 Turbulence intensity

The fluctuating wind velocity v is described by the turbulence intensity, the power spectral density and the correlation lengths. The turbulence intensity is a measure for the standard deviation σ in the “micrometeorological” range. s

σV =

sZ ∞ 1ZT v(x,y,z,t)2 dt = Gv (f ) df T 0 0

(14.5)

with the averaging time T , which results in the turbulence intensity IV =

σV . µV (z)

(14.6)

14.1.4 Spatial structure of the wind velocity 14.1.4.1 Atmospheric boundary layer - wind profile

Due to the “spectral gap”, for sufficiently long averaging times, e.g., T = [10,60] min, the average value µV (z,t) becomes time invariant (mean value stationary), and thus only depends

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262

on the height z. The mean wind velocity increases with height up to a surface independent limit value. This is due to the friction on the ground surface. The “gradient wind velocity” thus obtained results from air pressure differences in the atmosphere, that are linked to the different solar heating of the earth [Mendis et al 2007]. The sphere of influence of the surface friction is called atmospheric boundary layer. The height of this layer zg (gradient height) depends on the surface roughness. The dependence of the average wind velocity on the height can be either represented through a logarithmic relation or through an empirical exponential approach. The first one can be analytically derived from boundary layer considerations (z.B. [Holmes 2007, p. 56ff.]). The exponential approach results in µV (z) = µV (zref )

z

!α

(14.7)

zref

with the reference height zref = 10 m. The exponent α [Balendra 1993] depends on the surface roughness. It is related to the roughness length z0 of the logarithmic approach α = ln(zref/z0 )−1 [Holmes 2007, p. 58-59]. The dimensionless “surface drag coefficient” κ relates the velocity at √ the ground v∗ to the average wind speed at the height of 10 m via v∗ = κµV (10). ground conditions very flat ground (snow, desert) open ground (grasland, few trees) suburban areas (buildings 3 - 5 m) dense cities (buildings 10 - 30 m)

roughness length z0 [m] 0.001 - 0.005 0.01 - 0.05 0.1 - 0.5 1-5

surface drag coefficient κ [-] 0.002 - 0.003 0.003 - 0.006 0.0075 - 0.02 0.03 - 0.3

Table 14.1: Surfaces, roughness lengths and surface drag coefficient [Holmes 2007, p. 58]

The distribution of the temporal mean value of the velocity is thus described, e.g. by the following relation: z [m] I

100

III

50 10 5 20

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40

60

v

h i m s

14 Aeroelastic vibrations (wind)

263

v(z) = v10m



z 10

α

≥ 0.9v10m

Here, v10m is the 10 minute time average, that is reached or exceeded in the statistical average in 50 years and occurs at 10 m height. Fig. 14.3 shows the the decay for different surface roughnesses. z[m]

% 100

α=0.4 k=0.05

400

89

α=0.28 k=0.015

77

200

61 42

% 100 α=0.22

90 76 59

k=0.009

100 94

% α=0.16 k=0.005

82 67

% 100 91

79

Figure 14.3: aus Ruscheweyh, Dynamische Windwirkung an Bauwerken Vol. Abb. 2.3

2 (1982), S. 16,

14.1.4.2 Turbulences

Turbulences are caused by differently large vortices, that are transported with the mean wind velocity and showing a time structure with higher frequencies. Geoffrey I. Taylor (1939) proposed, that–under certain conditions–the statistical composition of the turbulences can be understood as a “frozen pattern”, that is transported with the mean wind velocity (Taylor’s frozen-in turbulence hypothesis [Holm 2005]). It is based on the assumption, that the changes of the vortices are slow compared to the wind velocity. On this basis, the size (wave length λ) of the vortices is related to the excitation frequency f and the mean wind velocity µV (z): λ = µV (z)/f [m].

(14.8)

14.2 Wind induced vibrations Along with the static wind force, various dynamic effects have to be considered.

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14.2.1 Gust induced vibrations Due to included gusts within the flow, the wind velocities are distributed unevenly over the height. It can then be composed of two parts. One part is the static part vm (z) that is constant over time according to Fig. 14.4 and a superimposed, randomly variable part v(z,t). z vges (z,t) = vm (z) + v(z,t) vm (z)

v Figure 14.4: Static and dynamic part of the wind velocity

This distribution can be used for slender buildings. For a structural element with sufficiently low width in the flow1 , the corresponding wind force per unit length is: ρ 2 q(z, t) = cf vges (z,t) 2 ρ 2 = cf (vm (z) + 2vm (z)v(z, t) + v 2 (z, t)) 2 ρ 2 ρ = cf vm (z) + cf ρvm (z)v(z,t) + cf v 2 (z, t) 2 2

(14.9)

with cf ρ2 vm (z)2

static term

cf ρvm (z) v(z,t)

linear dynamic term

cf ρ2 v(z,t)2

non-linear dynamic term (is usually neglected)

The frequency content of the excitation is given by the power spectral density S(f ) or S(T ) of the wind. Fig. 14.2 exemplarily shows such a power spectral density. For the “mesometeorological” range (comp. Fig. 14.2) there are different approaches. Some of those are summarized in Tab. 14.2. The differences are exemplarily depicted in Fig. 14.5. 1

Sufficiently small in the sense, that the same wind speed can be assumed over the whole width.

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2

Non−dimensional wind velocity spectra fG(f)/σ , z = 10 [m] Kolmogorov (z) Kolmogorov (L) Kareem Simiu Harris Davenport Kaimal von Karman ESDU

−1

10

−2

10

−2

−1

10

0

10

1

10

2

10

10

2

Wind velocity spectra G(f) in [(m/s) /Hz], z = 10 [m] 700 600 500 400 300 200 100 0

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

f [Hz]

Figure 14.5: Comparison of different models: z = 10 m, v¯10 = 20 m s , z0 = 0.05 (dimensionless), κ = 0.006 (open grassland, grassland); above: log-log illustration of the dimensionless turbulence spectra; below: Comparison for low frequencies [Runtemund 2013]

Exemplarily the approach of Davenport is explained. The dynamic additional velocities for the gusts are here given by Sv (f ) = I(z)2 v 2 (z)

2 ξ2 , 3 (1 + ξ 2 )4/3

(14.10)

with the frequency f in Hz, the average wind velocity v in m s , the dimensionless frequency

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266

application

limitations

Kolmogorov limit

basis

Kaimal

off-shore structures; land based structures with low surface roughness

z < 200 [m]

fulfilled

empirical, strong wind conditions, flat homogeneous terrain

Kareem

off-shore structures

von Kármán

atmospheric turbulences (e.g. airplane constructions), wind tunnel tests

ESDU

adjustable parameters

z ≥ 200 [m]

fulfilled

empirical

theoretically justified

empirical modification of the von Kármán spectrum with adjustable parameters

Simiu

off-shore Strukturen; landgestützte Strukturen mit geringer Geländerauigkeit

Davenport

land based structures

Harris

land based structures

z < 50 [m]

underestimated

empirisch

empirical, mean values based on different measurements on land at difunderestimatedferent heights overestimated

f > 0.01 [Hz]

Table 14.2: Comparison of different approaches from [Runtemund 2013]

coefficient ξ=

1200 f, v10m

(14.11)

and the dimensionless turbulence intensity z I (z) = 6k 10 2



−2α

.

(14.12)

Here z is the height in m and k is the roughness coefficient. It ranges from k = 0,05 in cities to k = 0,005 in open land.

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Sv (f )

f [Hz] 10

10

−1

−2

1

10

1

Figure 14.6: Example for Davenport spectrum.

The power spectral density of the dynamic load, neglecting the non-linear term in Eq. (14.9) is given by: ρ Sq (f ) = cf 2vm,z 2 

2

Sv (f )

(14.13)

For a small structure, with equally distributed wind pressure, the vibration response can be determined from this excitation (Eq. (13.58)). ˜ 2 Sw (f ) = h(f ) Sq (f )



(14.14)

Under the assumption of a stationary Gauß process, by the above a probabilistic description of the response is given. It allows to develop design concepts, e.g. for a gust factor. For this, a target probability for the case, that during an intended usage period a specified response value is not exceeded, is specified. For linearly shaped or extensive structures it further needs to be considered, that pressures are unevenly distributed over length- and cross-dimension of the structure. With the help of a modal superposition, design calculation concepts of practical applictions, e.g., cantilever systems, have been derived.

14.2.2 Vortex induced vibrations Due to vortices, periodic forces result at structures that form obstacles in the flow, which are initially at rest and loaded by a flow. These forces predominantly act transverse to the flow direction. They can induce forced vibrations. For coincidence of the vortex frequency and the eigenfrequency of the structures, the risk of resonance exists. This is often the case for chimneys, masts, ropes, and pipes. There is also the risk of resonant deformations, as far as mode shapes of the structures, that are connected to ovalization, are excited in resonance, e.g. ovalization of not sufficiently stiff cylinders.

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Figure 14.7: Vortices, from [Petersen 2000]

14.2.2.1 Occurence of vortices

In order to describe the character of the flow and the occurrence of vortices, the Reynolds number is introduced: Re =

vd η

which is dimensionless, with

d

m s diameter of the structure in m

η = 15 · 10−6 m2 /s

kinematic visosity in air,

wind velocity in

v

The following cases can be distinguished: • Subcritical:

40 < Re < 3,5 · 105

laminar flow in the boundary layer, before the cross section maximum, i.e., widest possible cut through the cross section, orthognoal to the flow direction, stable vortex trail ⇒ large transverse forces • Supercritical:

3,5 · 105 < Re < 7 · 106

turbulent flow in the boundary layer, vortex shedding due to shedding and re-application on the cross-section far behind the maximum of the cross-section, disordered wake flow with irregular vortices and low correlation of the lateral pressures ⇒ low transverse forces

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Re < 5 laminare Strömung keine Wirbel

5 < Re < 40 kleine symmetrische Wirbel keine Ablösung

40 < Re < 3,5 · 105 Stabile unsymmetrische Wirbel mit periodischer Ablösung (unterkritisch)

Re > 3,5 · 105 hochturbulente Wirbel mit unregelmäßiger Ablösung, für Re > 3,5 · 105 − 7 · 106 nahezu periodischer Nachlauf (über-/transkritisch)

Figure 14.8: Different states of vortex shedding

• Transcritical:

7 · 106 < Re

Vortex shedding again close to the cross-section maximum, but again periodic vortices in the wake flow ⇒ again increasing transverse forces Below a Reynolds number of 40, no significant vortex shedding occurs. Fig. 14.8 shows states of vortex shedding, from the laminar to the transcritical flow. 14.2.2.2 Frequency of vortices

The frequency f of the alternating force resulting from the vortices acting on the structure is described by the dimensionless Strouhal number S: fd v Sv → f= . d S=

(14.15) (14.16)

Thereby f corresponds to number of vortices created per second at one side of the obstacle.

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14.2.2.3 Forces related to vortices

The force in transverse direction can be determined with the help of a lateral aerodynamic coefficient cs . It results to ρ F (t) = v 2 cs dg(t) 2

(14.17)

The lateral aerodynamic coefficient cs can, e.g. be determined in wind tunnel testing. The function g(t) is a periodic function with amplitude 1 and the fundamental excitation frequency f. 14.2.2.4 Assessment of vortex induced vibrations

Through measurements the dependency of the Strouhal and the Reynolds number can be determined. For cylindrical structures, the Strouhal number takes values of around 0.2 and is almost independent from the Reynolds number. St, clat irregular shedding

0,6

clat

0,4

St 0,2 103

102

104

subcritical

106

105

107

Re transcritical

supercritical

Typically, the Strouhal number and the lateral aerodynamic coefficient cs are illustrated in simplified diagrams. The following figure shows the relation exemplarily for a cylindrical profile. S,cs 0.7

cs S

0.2

Re 10

4

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10

5

10

6

10

7

14 Aeroelastic vibrations (wind)

271

By setting the excitation frequency f equal to a specific natural frequency fn , the critical wind velocity can be determined. vcrit =

dfn S

The corresponding amplitude of the transverse force per unit length of a line structure follows to: ρ p k = v 2 cs d 2 Through the resonance, the “static” response is amplified by the factor (comp. Tab. 5.1): V =

π δ

The risk of resonace exists, whenever the critical wind velocity falls within the range of possible wind velocities. If this is the case, the resulting response has to be determined and the structural reliablity has to investigated with respect to the response. If the structure loaded by wind is not held at rest, i.e., vibrations of the structure occur, a so called lock-in effect can be of significance. Here, an influence of the periodic motion of the structure on the shedding frequency is observed. If a structure is excited by a wind velocity, for which correspondence between the shedding frequency of the vortices and system eigenfrequency (resonance) exists, and the wind velocity changes, the resonance state is not lost. The lock-in effect causes (as long as the change in wind velocity is not too large) that the shedding frequency of the vortices for larger vibration amplitudes is fixed at the previous frequency of the structure, i.e., at the eigenfrequency of the structure. Thereby, the excitation frequency and thus the resonance effect is maintained independently of wind velocities within a certain range. Constructive measures can be applied, e.g., the arrangement of spirals. These disturb the regularity of the the shedding and simultaneously lead to an increase of the force coefficient in the flow direction.

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a)

fe region of synchronization

vcrit Wind speed v

Related vibration amplitude y/d

S·v d

amplification function

b)

vortex excitation

Λ = 0,01

Reduced wind speed vr

Figure 14.9: vgl: Bachmann, Hugo, Vibration Problems in Structures (1997), in: Vibrations in acrosswind direction induced by vortex-shedding, S. 203, Abb. H.10

wire helix

scruton helix disturbance shrouds stripes

rings

14.2.3 Galloping Galloping describes self excited vibrations. It only occurs, when the system is already in motion due to inevitable flows. Thereby flow induced forces are caused that can stimulate the motion further, until a non-linear increasing damping is able to dissipate the energy introduced within on period of motion. This is then typically linked with large amplitudes. For circular cross-sections due to the full symmetry, such an excitation will not occur. Exceptions from this are asymmetries that occur, e.g. through ice cover or the flow of water in case of rain. The situation is depicted here for the example of a half circle cross section. In a coordinate system attached to the moving structure the flow is acting at the angle α.

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d

vwind w˙

Fh

α

Faero (t) flow force

Fv vres

vres

w˙

α vwind

mw¨ kw cw˙

closer streamlines: higher velocities, thereby lower pressures

The force component in vertical direction depends on the relative flow angle α. It is determined, e.g. in wind tunnel testing and given in terms of a aerodynamic coefficient cs (α) that depends on the flow angle α: ρ 2 dcs (α) Faero (α) = qdcs (α) = vwind 2

(14.18)

The aerodynamic coefficient cs (α) is shown for two typical cross-section in the following figure. Therein it is assumed that for α = 0 no transversal force Faero occurs. cs

cs dcs | dα α=0

>0

α

α dcs | dα α=0

0, stimulation occurs occurs from a critical wind velocity on. α=0



dcs ρ c≥ vcrit d dα α=0 2

→

vcrit =

2c dcs dα

(14.24) ρd

α=0

Galloping can occur for different cross-sections.

turbulent

Figure 14.11: Profiles at which Galloping can occur

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In [DIN 1055-4 2005; EN 2010] galloping vibrations are treated in appendix E.

14.2.4 Divergence Divergence is not a vibration problem. This phenomenon is described by “static tilting” that occurs, whenever the torsional moment, induced by the flow on the structure, linked with the rotation of the cross-section exceeds the reaction moment of the structure. In this case a stability loss can occur. Usually, the critical wind speed for static divergence in civil engineering applications is very high. It is higher than the velocities that are usually considered in the design process.

14.2.5 Flutter Fluttering of bridges is a stability problem, just as galloping. Small deformations of the system (often caused by vortex shedding) are a precondition for the stimulation.

w α

The relevant wind forces result from flow induced deformations due to bending and torsional motion of a bridge. Here, the flutter phenomenon is depicted in the following figure, showing the cross-section of a bridge vibrating simultaneously in a torsional and bending mode hape (two degrees of freedom). If the phase of the two degree of freedoms are coupled in such a way that the integral of the work done on the self-excited, vibrating system over one period is larger than zero, an amplification of the vibration occurs.

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Figure 14.12: Flutter, [Ruscheweyh 1982, Vol. 2 (1982), S. 118, Abb. 4.67]

In the above illustration translatory and rotatory motions are in phase. In the first quarter of the motion the wind does positive work on the system, as the direction of force and motion coincide. In the second quarter of the motion force and motion are counter-directed, such that the force does negative work on the system. Over the whole period, the energy in the system is thus not increased. In the lower illustration the translatory and the rotatory motions are phase shifted, such that for every instance in time the wind direction corresponds to the direction of motion and positive work is done. Further information can be found in Klöppel / Thiele 2 . Investigations based on potential theory, as for aircrafts, are hardly accurate for typical bridge cross-sections. Therefore, usually wind tunnel testing is necessary. Instead of a model of a miniature bridge, only a bridge segment, supported by springs (displacement and rotation), is investigated. In order to obtain valid results, it is necessary that the vibration of the bridge girder is not significantly affected by the motion of other structural parts, e.g. the pylons. Necessary data, such as the mass, the rotatory inerty, the bending and torsonial frequencies and the detailed cross-section design are prerequisites for the investigation. Compared to the experimental values from Klöppel / Thiele significant differences in the critical wind velocity can result from details in the constructive detailing, such as edge smoothing and railing perforation, etc.. Therefore the data given in these publications are especially useful for a predesign. As long as the risk of flutter cannot be excluded, wind tunnel test have to be performed. 2

K. Klöppel/F. Thiele, Modellversuche im Windkanal zur Bemessung von Brücken gegen die Gefahr winderregter Schwingungen, 1967, Der Stahlbau, 32, 353-365

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14.2.6 Interference effects Vortices behind structures can induce excitations to the structures behind those buildings. The details can only be determined in wind tunnel tests. For a series or parallel arrangement, the proof has to be done that the following interference effects do not occur: • interference effects for transverse vibrations due to vortices, • interference galloping, • classical galloping for coupled cylinders.

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15 Earthquakes 15.1 General information and terms Earthquakes result from fractures in the lithosphere (earth crust). The thick, brittle plates with a thickness ranging from a few to 200 km float on the comparably “liquid” asthenosphere. The plates are in constant movement, which is termed continental drift. The fracture zone of the lithosphere is called hypocenter. The projected point on the surface is called epicenter, as shown in Fig. 15.1. Often the hypocenter is covered by a sediment layer that is many kilometers thick. Caused by the fracture movement of the crust, compression waves (P-waves), shear waves (S-waves) and Rayleigh waves (R-waves) occur. These propagate with different propagation velocities (cP > cS > cR ) and are responsible for the movement of structures on the earth surface. The epicenter depth can be estimated from the arrival time difference between the P - and the S-waves. The waves can be attenuated or even amplified as they propagates through the soil toward the structures. In particluar, the properties of top 30 meters of the soil can strongly influence the amplitude and the frequency content of the seismic excitation. This was evident during the earthquake in Mexico City in 1985. Even though the hypocenter was 400 km far away from Mexico City, enormous damages occurred. The city is built upon a 30 − 40 m thick, very soft clay layer. The amplification of the earthquake acceleration due to the presence of the soft clay layer especially in the range of around 2 Hz corresponded very well with the eigenfrequency of the buildings with 8-12 floors. On the stiff soil outside of the inner city, almost no damages occurred

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epicentral distance ∆

depth of focus h

epicenter intensity I0

building intensity I

sediment layer fault

hypocentral distance s

hypocenter (focus) depth of focus isoseismal curves I0

epicenter earth surface I=I0 −1 I=I0 −2 I=I0 −3

Figure 15.1: Terminology of an earthquake, from [Flesch 1993, p. 203, Fig. 7.1]

The magnitude and intensity are used to characterize the strength of an earthquake.

15.1.1 Magnitude The energy release of an earthquake can be described by the magnitude M on the Richter scale. The largest magnitude, that was ever measured, was 9,5 (Validivia, Chile). The magnitude is described by the maximum displacement smax that is measured 100km away from the epicenter. An empirical relation between the energy released in an earthquake and the magnitude is given by lg(E) = 11.8 + 1.5M The difference ∆M = 1 then corresponds approximately to a factor 31 in the energy release. Here, the energy is given in the non SI-conform unit [E] = 10−7 J = 2,8 · 10−14 kWh. An earthquake of magnitude 9 thus has an energy release of around 6 · 1011 kWh. Considering the average energy consumption of 50.000 kWh per person and year in highly civilized countries, this means, that with the energy of such an earthquake around 10 million people could be supplied for one year. Furthermore, a relation between the magnitude and the length of the fracture zone (in km) can be empirically given as M = 5.65 + 0.98 lg(l0 ).

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Then, M = 9 corresponds to a length of l0 > 1000 km.

Earthquake/year

Fig. 15.2 shows the frequency of occurrence of earthquakes with a certain magnitudes per year. 103 102 101 100 10−1

a b

10−2 −4 max M ≈8.75 10 −3 10 10−4 max M ≈6.25 10−5 10−6 2 3 4 5 6 7 8 9 10 Magnitude M

Figure 15.2: Frequency of occurrence of magnitudes for the whole earth (a) and the Rhine valley (b) from [Flesch 1993, p. 203, Fig. 7.2]

15.1.2 Intensity Earthquakes can classified according to their consequences on nature, humans and structures (similar to the Beaufort scale for wind). This is done by the modified Mercalli-scale (MM scale) or the Medvedev, Sponheuer, Karnik scale (MSK scale). An earthquake thus has only one magnitude but different intensities from place to place. The intensity is a measure for the perception and damaging effect of earthquakes. The relations between magnitude M , intensity I and soil acceleration a can be approximately described empirically. For Germany the relation between the horizontal soil acceleration a at the surface of the earth and the intensity I can be described as follows lg a = 2 −

8 − I cm 2.4 s2

There are a number of empirical relations between the magnitude M and the epicenter intensity I0 , e.g., for Austria: M = 0.7 I0 − 0.1

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and Switzerland M = 0.67I0 + 2.3 log (focal depth in km) − 2.0

15.1.3 Calculation procedures For the dynamic investigation of a structure subjected to an earthquake excitation, the soil acceleration is applied as root point excitation. This typically shows an arbitrary time history, which is referred to as accelerogram.

Figure 15.3: Exemplary accelerogram given in terms of the acceleration relative to the gravitational acceleration on earth g .

Horizontal and vertical soil accelerations occur simultaneously. Usually, the horizontal movements are more important, as building structures are typically designed for comparatively low horizontal forces. The maximum vertical accelerations can be assumed to the in the order of 0,3 - 1,0 times the maximum horizontal acceleration. In the Swiss code SIA 160 a ratio of 2/3 is given, the German code DIN 4149 gives the ratio 0.7. Essential for the demand, are the acceleration, the frequency spectrum of the accelerogram and the duration of the loading. Exemplarily, a few numbers are given from Bachmann [2013] Maximum values of the horizontal acceleration: · · · ·

Friaul, 1976, 15 km distance from epicenter Mexico, 1985, on rock Mexico, 1985, in the city (soft soil layers) Nortridge, 1994, Southern California

0,37 0,04 0,17 1,82

g g g g

Frequency ranges of the soil acceleration:

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· · · ·

stiff soil, rock medium stiff soil soft soil Mexico, very soft soil

3 - 10 Hz 2 - 8 Hz 0,5 - 2 Hz 0,3 - 0,5 Hz

Duration: ·M=6-7 · Mexiko 1985

5 - 20 s 80 s

Different calculation methods can be used depending on the task to investigate the seismic response of structures • Approaches based on accelerograms: – Frequency-domain analysis (only linear) – Time-domain analysis (time history method; also non-linear) Using a statistically relevant number of accelerograms, several nonlinear transient analysis are carried out and the maxima of the forces and displacements are evaluated with averaging methods. The Eurocode 8 suggests a minimum number of analysis equal to 8. • Linear static methods – Equivalent static design – Response spectrum analysis (RSA) • Nonlinear methods: – Nonlinear static analysis – Response history analysis (RHA) (via modal analysis or direct integration methods) – Capacity design based the hierarchy of strengths

15.2 Response spectrum analysis (RSA) 15.2.1 Introduction In practical applications often the response spectrum analysis is applied in the earthquake design of structures. This approach is only valid for linear system, nevertheless it is also often applied for non-linear problems.

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w

fg (t) wg

wr

The linear equation of motion for the root-point excited SDOF system with the relative displacement wr and the soil displacement wg is given by: m(w¨g (t) + w¨r (t)) + c w˙r (t) + k wr (t) = 0

(15.1)

or, equivalently m w¨r (t) + c w˙r (t) + k wr (t) = −m w¨g (t) . |

{z

Fg (t)

(15.2)

}

The right hand side can be seen as seismic excitation Fg (t). The system response can now be determined numerically using the Duhamel-integral wr (t) = −

Zt

mw¨g (τ )h(t − τ ) dτ ,

0

or with other time integration methods. After dividing Eq. (15.2) by the mass m, it can be observed that the system q properties can be completely described using the definition of the natural frequency (ω = k/m and δ = c/2m) and the damping. w¨r (t) + 2δ w˙ r (t) + ωE2 wr (t) = −w¨g (t) With this approach, the maximum relative response values for different systems can be obtained in dependence of the natural frequency ωE and the damping ratio ς = D = δ/ωE for a given accelerogram w¨g (t). wr,max (ωE , ς) = Swr (ωE , ς) For this, the Duhamel integral (or equivalent time integration methods) is applied to evaluate the maximum response of the SDOF system for different natural frequencies and damping ratios, as shown in Fig. 15.4.

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Swr (ωE,i ,ςi )

w¨g Figure 15.4: SDOF system with different natural frequencies and damping ratios for the response spectrum analysis

Under the assumption of harmonic responses, w¨r,max occurs at the same time as wr,max = Swr 1 , that is when w˙ r = 0, and the following holds: w¨r,max = −ωE2 Swr and |w¨r,max | = Sa (ωE , ς) = ωE2 Swr (ωE , ς) or |w¨r,max | = Sa (TE , ς) =



2π TE

2

Swr (TE , ς)

The latter is also called “pseudo absolute acceleration”. The word “pseudo” refers to the fact that the used formulas are actually only valid for linear harmonic vibrations, but are applied here for non-harmonic vibrations introducing an approximation. It is also important to notice, that only the absolute values of the maximum accelerations are accounted for, therefore the information about the phase is lost. The following figures shows response spectra for absolute and pseudo-absolute acceleration, relative and pseudo-relative velocity and relative displacement. Absolute and pseudo-absolute acceleration differ only slightly, whereas relative and pseudo-relative velocity show larger differences.

1

This assumptions is an approximation and holds for linear and harmonic, i.e., sinusoidal vibrations.

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pseudo acc. acc.

m/s2

pseudo relative vel. relative vel.

m/s

relative displ.

cm

Figure 15.5: Response spectra for absolute and pseudo-absolute acceleration, relative and pseudorelative velocity and relative displacement

For SDOF systems, the response spectra can be directly used. Under certain conditions on geomtery and material properties, buildings can be idealized as SDOF system. For more complicated system–when applying the linear elastic response spectra–the modal superposition analysis can be used. A direct superposition of the modal responses would imply, that the maxima of all modal vibrations occur at the same time, thus leading to unrealistically high displacement and internal forces. Therefore the maximum value of quantities such as displacements and forces are calculated as the root of the square sum: Smax =

r

X

(15.3)

Si2

with Smax

Resultant

Si

Resultant corresponding to the -thi eigenform

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In reality loads from earthquakes lead to plastic deformations in structures, which dissipate energy when limited to a certain controlled value. Therefore the application of the linear RSA is a very conservative method. For a more realistic analyis, the response spectrum is reduced to a factor that accounts for the structure ductility.2 . This app visualizes the response spectrum method according to DIN 4149. It calculates the maximum displacement of a building with three floors under earthquake excitation. http://www.bm.bgu.tum.de/lehre/interactive-apps/ response-spectrum-analysis/

15.2.2 Design codes Relevant codes are: • Eurocode 8 - Seismic Design of Buildings • DIN 4149 (April 2005) - Building in German seimisc areas • KTA - Guidelines for nuclear power plants DIN 4149 allows calculations based on a SDOF system and the response spectrum method using a linear elastic structural model and the design spectrum for the design of building constructions, if the building has a regular structure in the layout and elevation and a soil dependent, maximum fundamental period between 0,8s and 2s (DIN 4149, chapter 6.2.2). The subsurface conditions are specified in DIN 4149, chapter 5.2. The design codes made available design response spectra for seismic design purposes. The design response spectra are simplified functions that envelope a large number of response spectra for recorded seismic input motions. For each specific site, the probabilistic evaluation of the seismic history, also called Probabilistic Seismic Hazard Analysis (PSHA), provides a statistically significant number of response spectra for events with a certain probability of occurrence in 50 year. From this response spectra population, an enveloping function is estimated and the design spectra is obtained. Therefore, for each probability of occurrence, a design spectrum can be derived. The smaller the probability of occurrence, the stronger is the event for which the structure is design. The probability of occurrence can be expressed also in terms of return period of the event. According to the DIN 4149, the probability of occurrence of the design earthquake for residential buildings is 10 % within 50 years. This corresponds to a return period of 475 years. 2

In DIN 4149 the non-linear effects of the building material, load-bearing system and constructive details are considered by a coefficient q.

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The elastic response spectrum from DIN 4149 for the linear calculation under consideration of the hysteretic damping has the following form3 : Se (T ) ag · γI · S ·

B

β0 q

C D

A

ag · γI · S

TB

TC

TD

T

Figure 15.6: Design response spectrum according to the DIN 4149.

"

0 < T ≤ TB :

T Se (T ) = ag · γI · S · 1 + TB

TB < T ≤ TC :

Se (T ) = ag · γI · S ·

TC < T ≤ TD : TD ≤ T :

!#

β0 −1 q

β0 q   β0 TC Se (T ) = ag · γI · S · · q T β0 TC · TD · Se (T ) = ag · γI · S · q T2

with

Se (T )

ordinates of the design spectrum as a function of the structure natural period T

ag

design ground acceleration on type A ground 0 ≤ ag ≤ 0,8 sm2 (which depends on the design seismic zone)

γI

importance factor of the building 0,8 ≤ γI ≤ 1,4 (high value for buildings of importance for the general safety, such as hospitals or emergency centers)

S 3

soil factor 0,75 ≤ S ≤ 1,5

In DIN 4149 two expressions for damping effects are chosen. Here, we restrict to the presentation of the behavior factor.

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TB ,TC ,TD corner periods (depends on the soil class) β0

damping correction factor with a reference value of β0 = 2,5 für 5 % viscous damping

q

behavior factor (according to DIN 4149, 8 to 12, depends on the regularity of the building, on the structural type (frame/walls), on the material behavior): Concrete Steel Timber

1,0 ≤ q ≤ 3,0

1,0 ≤ q ≤ 8,0 1,0 ≤ q ≤ 4,0

The energy dissipated due to the formation of plastic hinges cannot be represented with the linear approach. As an approximation the factor q is used to describes the amplitude reduction due to favorable dissipative effects. With sufficient structural ductility, the design forces are smaller than in the linear-elastic case (which is for q = 1) due to the plastic behavior. The seismic excitation Fb is estimated according to the spectral value Sd (T ) of the design response spectrum, depending on the natural period and structural damping: Fb = Sd (T ) · M · λ with M

toal mass of the building

λ

0,85 ≤ λ ≤ 1

The factor λ is introduced to consider the difference between effective modal mass and building mass.

15.2.3 Constructive notes (short excerpt from DIN 4149 and EC 8) • Structural forms with stable box-behavior • Simple geamotries, uniformity, symmetry, redundancy • Resolution of angled structural parts in cuboid

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• No large lumped masses on the extremity of slender structures • Even distribution of stiffening elements • Prevention of torsional vibrations → center of mass near center of rigidity • Connections and joints carefully done, including ring anchors • Mindestauflagertiefen • No discontinuities in stiffness and mass distribution • Joints sufficiently wide, i. Joint width> 1.5 of the max. displ.; at least 2cm • Foundation should not be placed at different depths using different foundation elements A sufficient ductility needs to be ensured. Particular attention needs to be paid to avoiding concrete compressive fracture and to keeping the foundation in the elastic range.

15.3 Capacity design 15.3.1 Basics For a given earthquake, a structure can be designed differently. A purely elastic design of structures for stronger earthquakes can lead to over-conservative constructions. Alternatively, also an elastic-plastic design with lower structural resistance but planned plastic deformations, is possible. This assumes, that damages are acceptable–without collapse–and requires very careful planing and constructive design of the building. The essential prerequisite for the reliability of the utilization of elastic-plastic reaction of the structure is a sufficiently high structural resistance against horizontal forces and a sufficient ductility. For general orientation the following rule holds: “Quality” of earthquake behavior ≈ load resistance · ductility Qualitatively “equal” possibilities are shown in the following sketch:

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Resistance

high resistance low ductility, no plastic deformation medium resistance A1 medium ductility low resistance A2 high ductility, large A3 plastic deformation Displacement (global displacement)

Force

Ductility ] Fracture

(..)y

(..)u

µ=

(..)u (..)y

Displacement

The ductility gives the ratio of the total elastic-plastic deformation to the elastic deformation at the beginning of the yield process. It can be introduced for different values, e.g. displacement ductility (global ductility), strain ductility, rotational ductility, or curvature ductility (local ductility). Due to the narrow local limitation of the plasticized region, the curvature ductility of a structure exceeds the displacement ductility. In the case of earthquakes the structure needs to behave sufficiently ductile even for repeated cyclic loading. Here is a difference to inelastic states, that are due to static loads. For steel structures a high ductility can be obtained, due to the good deformability of the material. However, this requires a careful constructive design to avoid local instabilities (as buckling or tipping). For concrete structures the plastic deformability is limited, nevertheless for a suitable reinforcement design, limited ductilities can be achieved. As unreinforced masonry structures show almost no ductility, in earthquake regions, a limitiation to few floors is necessary. For high earthquake danger, the ductility can be increased through reinforcement. The ductility for timber constructions is mainly given by the plastic deformability of the joints, the material itself shows almost no plastic deformability.

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15.3.2 Reduction factors For the reduction of the stress resultatnts corresponding the to the effect of a plastification, different approaches can be found in the literature. The approximate treatment of the nonlinearity is proposed “statically” by a comparison of the energy contents. Taking as a basis ideally elastic-plastic behavior, a linear calculation leads to a fictitious stress resultant Felmax , that would have to be resisted under earthquake loading. F µ=

Fel,max

upl uel

Fpl

uel

upl

u

Actually, the force cannot occur because of the plastification, but instead just Fpl as an approximation of the stress resultant for elastic-plastic behavior. The requirement of equal deformation energy in both cases leads to very rough and mechanically hardly convincing estimation u for the internal forces under plastification. The decisive value is the ductility µ = upl . el 1 Fel,max (Fel, max + Fpl ) · uel − 1 ≈ Fpl (upl − uel ) 2 Fpl !

Fel,max → Fpl = √ 2µ − 1 It needs to be considered, that with this static “energy balance”, the dynamic behavior can only be predicted in a relative rough approximation. There is furthermore the possibility to work with inelastic design response spectra, which already incorporate the corresponding reductions. Apparently the aforementioned procedures lead to plausible, but calculatory not exactly verifiable values (there it needs to be considered, that an “exact calculation” can nevertheless only give an illusory precision, given the uncertainty with respect to the uncertainties in the loading).

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15.3.3 The capacity design Example 15.1: Chain under static loading

To justify the capacity design, a chain under static loading is considered exemplarily. ∆

gleiche Glieder F1 F1 >> F2

ein duktiles Glied F2 i

With similarly elastic chain links, the necessary elongation would only be achievable by a relatively large force F1 . Every link in the chain is loaded by the same force. Under the assumption of brittle chain elements, all chain elements must remain in the linear domain and their resistance had to be set high above this force F1 . With a ductile chain link i, the same given end displacement ∆ can be achieved with a much lower force F2 , compared to the case where all links, including i, behave elastically. All links then have a much lower stress level and can be designed with less resistance. In the case of dynamic loading, the corresponding correctly designed and constructed ductile element, results in an additional high damping, thanks to the hysteretic behavior.

The reduction of the loading is only then permissible, if the structure has the necessary ductility. This requires, to already define the locations, where plastification shall occur purposefully. At the same time, all other structural elements shall still behave elastically and only fail, when the plastic deformation capacity of the dissipative elements is reached. This means, that a dissipative structural element has to be designed such that it deforms sufficiently and thereby dissipates energy. For the desired ductility, limit values have to be fulfilled for certain construction forms. The plastification should be preferably distributed over various locations, as depicted in the system below.

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Example 15.2: Choice of mechanisms

ungeeignet

geeignet

The procedure of the planing for the capacity design, is as follows: 1. design of the building 2. derivation of the structural model and calculation of the internal forces, e.g. with the response spectrum method 3. dimensioning of the structural elements • choice of a mechanism (where should the structure plasticize and where not!) • constructive design of the plasticizing regions. The ductile structural element must be designed constructively, such that it has the required high plastic deformation capability under alternating loading. The reduction of bearing capacity has to be considered. For example, it has to be taken care, that the concrete compression zone cannot burst (Intention is to create inelastic alternating demands). • constructive design of the remaining elastic parts using the rules of elastic design and considering possible post-limit stiffnesses of the plasticized regions. The elastic, comparably “brittle” elements have to designed such hat their behavior stays elastic with sufficient certainty, even if the yield strength of the plastic elements exceed their planned values. Therefore, a “strengthening factor” (between 1,5 and 2,0) is introduced. This relatively high value results, because the material strength of the ductile elements can be higher than the nominal values. .

For the “capacity design” further questions concerning the constructive design are significant, comp.,e.g., Bachmann [2013]. The procedure with elastic response spectra and damping coefficients is based on the “natural” ductility, that is specifically used by the capacity design. For important buildings under extreme earthquake loads very complex non-linear investigation can thus become necessary. For the necessary detailed dynamic calculations with time step

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procedures, accelerograms are needed. If appropriate recorded accelerograms are not available, artificial accelerograms can be created numerically, and scaled to match a specific design response spectrum.

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16 Numerical time step procedures 16.1 General remarks 16.1.1 Overview Different time step procedures (step by step Methods), such as the Newmark-β method, the Wilson-θ method or the Houbold method, and the Generalized-α method and the central differences method can be applied. In the scope of these methods, the time domain is discretized into finite time intervals and the equations of motion are then only fulfilled in an approximative manner. The accuracy of the solution can in general be increased, by applying smaller time intervals. However, the computing time strongly increases with the number of time steps. For large calculation tasks, it is necessary, to find an optimum between the requirements with respect to the accuracy and the required computation time.

16.1.2 Explicit - Implicit The approaches can—in general—be classified as explicit or implicit. With an explicit method, the response values are calculated only on the basis of the known values of the immediately preceding time step. These procedures move from one time step to the next. With an implicit procedure, e.g. the Newmark-β method, the response values are obtained from both, the immediately preceding time step and the following time step.

16.1.3 Errors Errors can occur due to rounding or too large time intervals, that are not able to capture the time dependency of the response. These errors can introduce phase errors, changes in the frequencies and also artificial (positive or negative) damping.

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16.2 Newmark -β method 16.2.1 General approach The general implicit description reads: w˙ n+1 = w˙ n + (1 − γ) ∆tw¨n + γ∆tw¨n+1 wn+1 = wn + ∆tw˙ n +



(16.1)

1 − β ∆t2 w¨n + β ∆t2 w¨n+1 2 

(16.2)

16.2.2 Constant acceleration over the time interval In the scope of this method, two parameters β und γ have to be chosen. For a simplified variant, they are chosen to be β = 14 und γ = 21 , which correspond to a constant acceleration over the time interval. w(τ ¨ ) w¨n+1

Acceleration (constant)

w¨n

w(τ ˙ ) Velocity (linear)

w¨average = 21 (w¨n + w ¨n+1 )

¨n+1 ) w(τ ˙ ) = w˙ n + τ2 (w¨n + w w˙ n+1 = w˙ n + w˙ n

w(τ )

w(τ ) = wn + w˙ n τ +

τ2 (w¨n 4

∆t (w¨n 2

+w ¨n+1 )

+w ¨n+1 )

Displacement (quadratic)

wn+1 = wn + w˙ n ∆t +

∆t2 (w¨n 4

+w ¨n+1 )

wn t=n

∆t

t + ∆t = n + 1

τ

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With γ = 21 , the acceleration can be written as 1 w(τ ¨ ) = (w¨n + w ¨n+1 ). 2 This leads to the velocity 1 w(τ ˙ ) = w˙ n + τ (w¨n + w ¨n+1 ), 2 and the displacement 1 ¨n+1 ). w(τ ) = wn + w˙ n τ + τ 2 (w¨n + w 4 These relations can be put into a explicit from (comp. with the expression in the above figure):

w¨n+1 =

4 4 (w − w ) − w˙ n − w¨n n+1 n ∆t2 ∆t

(16.3)

After inserting Eq. (16.3) into Eq. (16.1), it follows for the velocity: w˙ n+1 =

2 (wn+1 − wn ) − w˙ n ∆t

(16.4)

Further inserting the so obtained expressions for acceleration and velocity at time tn+1 into the dynamic equilibrium, it reads mw¨n+1 + cw˙ n+1 + kwn+1 = fn+1

(16.5)

This equation can be transformed into a linear equation, where all elements, that belong the the preceeding time step tn , are brought to the right hand side: ¯ n+1 = f¯n+1 Kw with ¯ = k + 2c + 4m K ∆t ∆t2 and 2wn 4 4 + w˙ n + m wn + w˙ n + w ¨n . f¯n+1 = fn+1 + c 2 ∆t ∆t ∆t 

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¯ is the same for all time steps. Thus, K, The so obtained linear equation can be solved for each time step. In order to calculate the generalized force f¯n+1 , the velocity w˙ n and the acceleration w¨n in the preceding time step are determined according to Eqs. (16.3) and (16.4), respectively. The acceleration w¨n+1 is then calculated from the equation of motion (Eq. (16.5)). In the first time step, w0 and w˙ 0 are determined from the initial conditions and w¨0 is calculated from the equation of motion. w¨0 =

1 (f0 − cw˙0 − kw0 ) . m

16.2.3 Predictor/ corrector approach Additionally a so called predictor/corrector approach is applicable. In the scope of this approach we start with the initial values for the velocity and the displacement (Eqs. (16.1) and (16.2)), whereby the terms, that correspond to the following time step (underlined terms), are neglected at first. 1 − β ∆t2 w¨n + β ∆t2 w¨n+1 wen+1 = wn + ∆tw˙ n + 2 e˙ w ¨n + γ∆tw¨n+1 n+1 = w˙n + (1 − γ) ∆tw 



(16.6) (16.7)

With these predictor values, the spring and damping forces are determined: f¯i,n+1 = c w˜˙n+1 + k w˜n+1 With these values, the acceleration at time tn+1 can be calculated. With mw¨n+1 + cw˜˙n+1 + k w˜¨n+1 = fn+1 follows: w¨n+1 =

 1  fn+1 − f¯i,n+1 m

The predictor values are used for the corrector step. For this purpose, the underlined terms in Eqs. (16.1) and (16.2) are added. w˙ n+1 = w˜˙n+1 + γ∆tw¨n+1

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and

wn+1 = w˜n+1 + β∆t2 w¨n+1

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16.2.4 Stability considerations For the investigation of possibly occurring numerical instabilities, the homogeneous solution fo the undamped motion is considered: w¨n+1 + ωE2 wn+1 = 0

(16.8)

Again assuming a constant acceleration (corresponding to Eq. (16.3)), it follows w¨n+1 =

4wn+1 4wn 4w˙ n − w¨n − − ∆t2 ∆t2 ∆t

(16.9)

and with Eq. (16.4) w˙ n+1 =

2wn+1 2wn − − w˙n ∆t ∆t

(16.10)

and finally, inserting Eq. (16.9) into Eq. (16.8), we obtain 

4 4wn 4w˙ n + ωE2 wn+1 − − − w¨n = 0. 2 ∆t ∆t2 ∆t 

This can alternatively be written as wn+1 =  with ωE =

1

4 ∆t2

2π . TE

4w˙ n 4wn  w ¨n + + . 2 ∆t ∆t2 + ωE 



(16.11)

The expression in Eq. (16.11), has the form

wn+1 = b31 w¨n + b32 w˙ n + b33 wn . Inserting the above into Eqs. (16.10) and (16.9), it follows w˙ n+1 = b21 w¨n + b22 w˙ n + b23 wn w¨n+1 = b11 w¨n + b12 w˙ n + b13 wn . In matrix notation this relationship is Zn+1 = BZn

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with w¨n   Zn = w˙ n  wn 







b11 b12 b13  B = b21 b22 b23  . b31 b32 b33

After n time steps, the solution is Zn = [B]n Zo . The stability of the method depends on the behavior of Bn for n → ∞. Since the coefficients of Bn depend on ωE and ∆t or TE , respectively, limit values can be defined, for which infinite growth of [B]n is excluded. The Newmark-β method with β = 41 and γ = 21 is (absolute) stable, independently on the choice of the time step ∆t. Absolute stability implies, that the result is limited independently on the choice of the time step. i.e., the calculated values lie in the range of the static deformation of the system, thtat result from wstat = fmax . k For β = 61 and γ = 21 (linear acceleration) the Newmark-β method is only conditionally stable. The linear acceleration approach leads for the same choice of ∆t und under consideration of the stability criteria to better results. Nevertheless, these conclusions are only valid for linear systems. In general, the time step size has to be chosen such that the load characteristic und the characteristic of the free vibration can be represented accurately enough.

16.3 The central difference method 16.3.1 Derivation of w ¨ and w˙ out of w With the central difference method, the numerical approaches for the discrete computation of the first and second derivative are used: w˙ n =

1 (−wn−1 + wn+1 ) 2∆t

(16.12)

w¨n =

1 (wn−1 − 2wn + wn+1 ) ∆t2

(16.13)

For both expansions, the error is of order (∆t)2 .

16.3.2 General approach The displacement for the time step n + 1 is obtained from the equilibrium at time step n: mw¨n + cw˙ n + kwn = f .

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(16.14)

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Inserting Eqs. (16.12) and (16.13) into Eq. (16.14), it follows: 

1 2 1 1 1 c wn+1 = f − k − 2 m wn − c wn−1 . m+ m− 2 2 ∆t 2∆t ∆t ∆t 2∆t 









This equation can be solved for wn+1 . The solution wn+1 thus depends on the application of the equilibrium condition or equations of motion at the time step n. For this reason, the method is referred to as an explicit integration method.

16.3.3 Start procedure From the previous derivation it is apparent, that for the calculation of wn+1 always wn and wn−1 are needed. For the computation at time step n = 0, thus, a separate start procedure is necessary, because w0 as well as w˙0 und w¨0 are known, but w−1 not. w−1 can then be calculated as follows: w−1 = w0 − ∆tw˙ 0 +

∆t2 w¨0 . 2

It should be mentioned, that w¨0 can be calculated from w0 and w˙ 0 with the help of the equilibrium equation.

16.4 Non-linear problems In the following figures, a non-linear SDOF system is depicted. The free body cut shows the corresponding forces. Non-linear dependencies between the forces and the displacements or the velocities are depicted below. f (t)

F (t)

m

FI Fc

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Fk

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302

Fc

Fk

wn wn+1

w

w˙ n

w˙

w˙ n+1

The dynamic equilibrium for the time steps tn+1 and tn are fIn+1 + fcn+1 + fkn+1 = fn+1 ,

(16.15)

f In + f cn + f k n = f n ,

(16.16)

with fIn+1 − fIn = m∆w¨ fcn+1 − fcn ≈ cn =

dfc dw˙

dfc ∆w˙ = cn ∆w(t) ˙ dw˙

(16.18)

!

(16.19) n

fkn+1 − fkn ≈ kn =

(16.17)

dfk dw

dfk ∆w = kn ∆w(t) dw

(16.20)

!

(16.21) n

fn+1 − fn = ∆fn

(16.22)

Subtracting Eq. (16.15) from Eq. (16.16), an incremental representation is obtained: m∆w¨ + cn ∆w˙ + kn ∆w = ∆fn . On this basis, now the Newmark-β method can be applied. For β = wn+1 = wn + ∆w

(16.23) 1 4

and γ =

1 2

and with (16.24)

and ¯ n+1 = f¯n+1 kw

(16.25)

¯ n = f¯n kw

(16.26)

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it follows: ¯ k∆w = ∆f¯,

(16.27)

with 2cn 4m k¯ = kn + + 2 ∆t ∆t   4w˙n ∆f¯ = ∆f + m +w ¨n + cn (w˙n ) . ∆t

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