dynamic analysis foundations

7/27/2019 Dynamic Analysis Foundations

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Dynamic Analysis of Structures for the Finite ElementMethod

Francisco J. Montns, Ivan Muoz

May 7, 2013


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Contents

Contents

I Fundamentals of dynamics of structures 1

1 Single-Degree-of-Freedom Systems 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Free undamped vibrations . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.1 Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Harmonic motion: more formal approach solution . . . . . . . 7

1.3 Free damped vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.1 Underdamped motion . . . . . . . . . . . . . . . . . . . . . . 12

1.3.2 Overdamped motion . . . . . . . . . . . . . . . . . . . . . . . 151.3.3 Critically damped motion . . . . . . . . . . . . . . . . . . . . 15

1.4 Response to harmonic excitation . . . . . . . . . . . . . . . . . . . . 161.4.1 Undamped system . . . . . . . . . . . . . . . . . . . . . . . . 171.4.2 Damped system . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4.3 Dynamic response factor . . . . . . . . . . . . . . . . . . . . . 251.4.4 Frequency response function method . . . . . . . . . . . . . . 261.4.5 Laplace transform based analysis . . . . . . . . . . . . . . . . 28

1.5 General forced response . . . . . . . . . . . . . . . . . . . . . . . . . 281.5.1 Impulse response function . . . . . . . . . . . . . . . . . . . . 291.5.2 Response to an arbitrary excitation . . . . . . . . . . . . . . 30

2 Multi-Degree-of-Freedom Systems 332.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.1.1 Two-Degree-of-Freedom system . . . . . . . . . . . . . . . . . 342.1.2 Mathematical modeling of damping . . . . . . . . . . . . . . 362.1.3 Solutions of the equation of motion . . . . . . . . . . . . . . . 37

2.2 Vibration absorber: an application of Two-Degree-of-Freedom systems 382.2.1 The undamped vibration absorber . . . . . . . . . . . . . . . 412.2.2 The damped vibration absorber . . . . . . . . . . . . . . . . . 43

2.3 Free undamped vibrations of MDOF systems . . . . . . . . . . . . . 472.3.1 Natural frequencies and mode shapes . . . . . . . . . . . . . . 48

2.3.2 Orthogonality of mode shapes . . . . . . . . . . . . . . . . . . 492.3.3 Modal matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 512.3.4 Normalization of modes . . . . . . . . . . . . . . . . . . . . . 522.3.5 Response of undamped MDOF systems . . . . . . . . . . . . 53

2.4 Free damped vibrations of MDOF systems . . . . . . . . . . . . . . . 562.5 Response of MDOF systems under arbitrary loads . . . . . . . . . . 592.6 Response of MDOF systems under harmonic loads . . . . . . . . . . 602.7 Systems with distributed mass and stiffness . . . . . . . . . . . . . . 64

2.7.1 Vibration of beams . . . . . . . . . . . . . . . . . . . . . . . . 652.7.2 Vibration of plates . . . . . . . . . . . . . . . . . . . . . . . . 72

2.8 Component mode synthesis . . . . . . . . . . . . . . . . . . . . . . . 76

2.8.1 The fixed interface method . . . . . . . . . . . . . . . . . . . 76

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2.8.2 The free interface method . . . . . . . . . . . . . . . . . . . . 78

3 Introduction to signal analysis 813.1 Introduction to signal types . . . . . . . . . . . . . . . . . . . . . . . 81

3.1.1 Deterministic signals . . . . . . . . . . . . . . . . . . . . . . . 813.1.2 Random signals . . . . . . . . . . . . . . . . . . . . . . . . . . 82

3.2 Fourier Analysis of signal . . . . . . . . . . . . . . . . . . . . . . . . 823.2.1 The Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . 823.2.2 The Fourier integral transform . . . . . . . . . . . . . . . . . 833.2.3 Digital signals . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

3.3 State-space analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

II Finite element procedures for the dynamic analysis of struc-tures. 89

4 Finite element discretization of continuous systems. Stiffness andMass matrices 914.1 Stiffness matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4.1.1 Beam elements . . . . . . . . . . . . . . . . . . . . . . . . . . 924.1.2 Continuum elements . . . . . . . . . . . . . . . . . . . . . . . 95

4.2 Mass matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994.2.1 Consistent mass matrix . . . . . . . . . . . . . . . . . . . . . 99

4.2.2 Lumped mass matrix . . . . . . . . . . . . . . . . . . . . . . . 101

5 Computational procedures for eigenvalue and eigenvector analysis.1055.1 The modal decomposition revisited. Mode superposition analysis . . 1055.2 Other eigenvalue and eigenvector problems . . . . . . . . . . . . . . 1075.3 Computation of modes and frequencies . . . . . . . . . . . . . . . . . 1105.4 Reduction of the general eigenvalue problem to the standard eigen-

value problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1135.5 Static condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1195.6 Model order reduction techniques: the Guyan reduction . . . . . . . 1215.7 Inclusion of damping matrices . . . . . . . . . . . . . . . . . . . . . . 1255.8 Complex eigenvalue problem: complex modes . . . . . . . . . . . . . 128

5.8.1 Formulation in nonsymmetric standard form . . . . . . . . . 1305.8.2 Formulation in general symmetric form . . . . . . . . . . . . 133

6 Computational algorithms for eigenvalue and eigenvector extrac-tion 1356.1 Some previous concepts . . . . . . . . . . . . . . . . . . . . . . . . . 136

6.1.1 Matrix deflation . . . . . . . . . . . . . . . . . . . . . . . . . 1366.1.2 Rayleigh quotient . . . . . . . . . . . . . . . . . . . . . . . . . 1376.1.3 Courant minimax characterization of eigenvalues and Sturm

sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1396.1.4 Shifting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1426.1.5 Krylov subspaces and the Power method . . . . . . . . . . . . 143

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6.2 Determinant search method . . . . . . . . . . . . . . . . . . . . . . . 151

6.3 Inverse iteration method . . . . . . . . . . . . . . . . . . . . . . . . . 1526.4 Forward iteration method . . . . . . . . . . . . . . . . . . . . . . . . 154

6.5 Jacobi method for the standard eigenvalue problem . . . . . . . . . . 156

6.6 The QR decomposition and algorithm . . . . . . . . . . . . . . . . . 161

6.7 Jacobi method for the generalized eigenvalue problem . . . . . . . . 164

6.8 Bathes subspace iteration method and Ritz bases. . . . . . . . . . . 170

6.9 Lanczos method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

7 Transient analyses in linear elastodynamics 185

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

7.2 Structural dynamics and wave propagation analyses. The Courantcondition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

7.3 Linear multistep methods. Explicit and implicit algorithms. Dahlquisttheorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

7.4 Explicit algorithms: central difference method . . . . . . . . . . . . . 192

7.5 Implicit algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

7.5.1 Houbolt method . . . . . . . . . . . . . . . . . . . . . . . . . 202

7.5.2 Newmark- method . . . . . . . . . . . . . . . . . . . . . . . 204

7.5.3 Collocation Wilson- methods . . . . . . . . . . . . . . . . . . 218

7.5.4 Hilbert-Hughes-Taylor (HHT) method . . . . . . . . . . . 2257.5.5 Bathe-Baig composite (substep) method . . . . . . . . . . . . 227

7.6 Stability and accuracy analysis . . . . . . . . . . . . . . . . . . . . . 2367.7 Consistent initialization of algorithms . . . . . . . . . . . . . . . . . 250

8 Transient analysis in nonlinear dynamics 251

8.1 The nonlinear dynamics equation . . . . . . . . . . . . . . . . . . . . 251

8.2 Time discretization of the nonlinear dynamics equation . . . . . . . 252

8.3 Example: The nonlinear Newmark-algorithm in predictor-multicorrectord-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

8.4 Example: The HHT method in predictor-multicorrector a-form . . . 257

8.5 Example: The Bathe-Baig algorithm in predictor-multicorrector d-form.259

9 Harmonic analyses 2719.1 Discrete Fourier Transform revisited . . . . . . . . . . . . . . . . . . 271

9.2 Harmonic analysis using the full space. . . . . . . . . . . . . . . . . . 272

9.3 Harmonic analysis using mode superposition . . . . . . . . . . . . . . 277

10 Spectral and seismic analyses 281

10.1 Accelerograms and ground excitation . . . . . . . . . . . . . . . . . . 281

10.2 The equation of motion for ground excitation. Accelerometers andvibrometers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

10.3 Elastic Response Spectra: SD, SV, SA, PSV, PSA. . . . . . . . . . . 284

10.4 Modal superposition methods for spectral analysis. Modal mass . . . 288

10.5 Static correction or mode acceleration method . . . . . . . . . . . . . 289

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Contents

11 Bibliography 297

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Part I

Fundamentals of dynamics ofstructures

Ivan M. DazAssociate Professor. Universidad Politcnica de Madrid, Spain

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1 Single-Degree-of-Freedom Systems

1.1 Introduction

The principal distinctive feature of dynamic analysis (compared with static analysis)is the consideration of inertial forces. That is, forces acting on the structure willcause accelerations such that inertial forces cannot be neglected in the analysis. Ifloading is such that the accelerations caused can be neglected, a static analysis canbe performed.

Generally, real systems are continuous and their parameters are distributed.However, in many cases it is possible to simplify the analysis by representing asystem with distributed parameters by a discrete one. Hence, mathematical modelscan be divided into two types:

1. continuous systems, or distributed-parameter systems, and

2. discrete systems, or lumped models.

The number of unknown displacements (or velocities or accelerations) whichare of interest is called the number of degrees of freedom (DOFs) of the system.As a general rule, a DOF must be associated with each point on the structurehaving significant inertial forces. Thus, a continuous model represents an infiniteDOF system because there is an infinite number of points (each having a differentposition coordinate) having mass and stiffness properties associated with them. Thismeans that the unknown displacements at each point are represented by a continuousfunction ( ) of the position and time

On the other hand, in a discrete or lumped model, the whole system is assumedto be represented by a number of point masses at particular pre-determined positionson the structure The finite element modelling technique is currently the most widelyused method for developing discrete parameter models. Mathematically, the behav-ior of discrete parameter systems is described by ordinary differential equations,whereas that of distributed-parameter systems is generally governed by partial dif-ferential equations. In the case of ordinary differential equations, unknown functions(typically displacement) depend on time as the only variable. In the case of partialdifferential equations, time and at least one position coordinate are the variablesof the unknown functions. As an example, Figure 1 shows a distributed-parametermodel and a discrete-parameter model of a cantilever beam

Of the discrete models, the simplest model is the one described by a first or secondorder ordinary differential equation with constant coefficients. This system is usuallyreferred to as a single-degree-of-freedom (SDOF) system. Such a model is often usedas an approximation for a generally more complex system. However, its importanceis coming from the fact that in cases in which a technique known as Modal Analysiscan be employed, the mathematical formulation associated with many linear multi-degree-of-freedom (MDOF) discrete systems and continuous systems can be reducedto sets of independent second-order differential equations, each having exactly thesame form and means for solution as the equation of a SDOF system. Hence, athorough study of SDOF linear systems is clearly justified and will be carried outhere.

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Figure 1: (a) Distribuited-parameter model. (b) Discrete-parameter model for acantilever beam

The elements of a SDOF systems are (see Figure 2): spring of stiffness , whichrelates forces to displacements, dashpot with a viscous damping coefficient , whichrelates forces to velocities, and mass of value , which relates forces to accelerations.The physical properties of a SDOF system are constants, , and playing therole of modeling parameters. It should be noted that springs and dampers possessno mass and masses are assumed to behave as rigid bodies, i.e., they do not deform.

As an example of modeling a real system by a SDOF system, consider the ele-vated water tank of Figure 3. It is of interest to obtain the lateral vibration underground motion or under lateral wind excitation. This structure can be initiallymodelled as a lumped mass supported by a massless structure with stiffness . Anenergy dissipating mechanism has been included into the structural idealization tofeature the decaying motion observed during free vibration after an excitation oc-curred. The most commonly used damping is the viscous damping since it is thesimplest to deal with mathematically

The main objective of theoretical dynamic analysis is to study the behavior ofstructures subjected to given excitations. The behavior of the structure is char-acterized by the motion caused by these excitations and is commonly referred toas system response. The motion is generally described by displacements, velocitiesor accelerations. When displacements are known at each instant of time, velocitiesand accelerations (i.e., the first and second derivatives of the time-varying functiondescribing displacements) and other types of structural responses to excitation, suchas stresses and strains, can be calculated.

The excitations can be in the form of initial displacements and velocities, or inthe form of externally applied forces. The response of systems to initial conditionsis known as free response or free vibration, whereas the response to continuous ex-ternally applied forces is known as forced response or forced vibration. Moreover,although all real structures have some form of damping which will produce decayingfree vibrations, mathematical treatment of vibrating systems can be either with nodamping (undamped vibrations) or with damping (damped vibrations), dependingon the importance of the corresponding damping force.

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1.1 Introduction

Figure 2: Elements of SDOF system

Figure 3: (a) Water elevated tank. (b) Idealized model. (c) Spring-mass systemwith damping

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Figure 4: Undamped SDOF oscillator and its free body diagram

1.2 Free undamped vibrations

1.2.1 Equation of motion

Consider Figure 4 as a schematic form of a SDOF spring-mass oscillator. Assumethat = 0, i.e. there is no damping. According to Newtons second law of motion,the inertial force is proportional to the acceleration () through the mass .The restoring force due to the spring is proportional to the spring stiffness

() + () = 0 () +

() = 0 (1)

Equation (1) represents simple harmonic motion and is analogous to

() + 2 () = 0 with = r

(2)

In order to predict the response, Equation (2) must be solved. The solution ofthis equation is of the following form

() = sin( + ) (3)

where is the amplitude, is the angular natural frequency [rad/s], which deter-mines the interval in time during which the function repeats itself and , called thephase, determined the initial (at = 0) value of the sine function. From successivedifferentiation of the displacement Equation (3), the velocity and accelerationfunctions are obtained

() = cos( + ) () = 2 sin( + ) (4)

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Then, by substituting Equation (4) into (2), it can be seen that this equation is

satisfied. This corresponds to a second-order linear differential equation. Thus,there are two constants of integration to evaluate. These are and and aredetermined by the initial state of motion (initial conditions): (0) = 0 and (0) =0. Substitution of these initial conditions on the solution (3) yields

(0) = sin(0 + ) = sin and (0) = 0 = cos(0 + ) = cos (5)

The solution of these two equations for the unknown constants and leads to

=

p2

20 +

20

= tan1

0

0

(6)

Thus, the solution (3) of the equation of motion for the spring-mass system is givenby

() =

p2

20 +

20

sin

+ tan

1

00

(7)

An example of a solution () is plotted in Figure 5. This solution is known asfree response of the system since no external force is applied. The motion of thespring-mass system is called simple harmonic motion.

Considering the simple harmonic motion given by Equations (3) and (4), thevelocity is 90 (or 2 radians) out of phase with the displacement, while the ac-celeration is 180

(or radians) out of phase with the displacement and 90

out of

phase with the velocity (see Figure 5) Additionally, the velocity response amplitudeis greater (or smaller, depending if is greater or smaller than 1) than that of thedisplacement response by a multiple of , and the acceleration response is greaterby a multiple of 2. The angular frequency, , describes the repetitiveness of theoscillation and the time the cycle takes to repeat itself is the period , which isrelated to the natural frequency by

=2

[s] (8)

Quite often the frequency is measured and discussed in terms of cycles per second,which is called Hertz. The frequency in Hertz [Hz], denoted by , is related to thefrequency in radians per second by

=1

=

2

[Hz] (9)

1.2.2 Harmonic motion: more formal approach solution

The solution given by (7) was obtained assuming that the response was harmonic.However, the form of the response can also be derived by following the theory ofelementary linear differential equations. This approach is reviewed here.

Assume that the solution of () is as follows

() = (10)

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0 1 2 3 4 5 6 7 8 9 102

1

0

1

2

Time (s)

u(t)

Free response of an undamped SDOF system

0 1 2 3 4 5 6 7 8 9 104

2

0

2

4

Time (s)

v(t)

0 1 2 3 4 5 6 7 8 9 105

0

5

Time (s)

a(t)

Figure 5: Free response of an undamped SDOF.

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where and are constants to be determined. The velocity and acceleration are

() = and () = 2. Substitution of the assumed exponential form inthe equation of motion (1) yields

2 +

= 0 2 + = 0 (11)whose solutions are

12 =

r

=

r

= (12)

where =1 is the imaginary number. The substitution of the two solutions of

(12) into (10) gives two solutions for ()

() = 1, and () = 2

(13)

Then, since the system is linear, the sum of two solutions is also a solution

() = 1 + 2

(14)

where 1 and 2 must be complex constants in order to make () real after mul-tiplying by and . Taken into account the Euler relations =cos sin , Equation (14) can be written as

() = 1 cos + 2 sin (15)

where 1 and 2 are real constants, or

() = sin( + ) (16)

and being real constants. Then Equations (14), (15) and (16) are three equiva-

lent ways of the solution of (1) subjected to non-zero initial conditions.The relationships between the three sets of constants are shown now. It is

recommended for the reader to prove these relationships.

For sets ( ) and (1

2

):

=q

21 + 22 and = tan

1

12

For sets (1 2) and (1 2):

1 = 1 + 2 and 2 = (1 2)

For sets (1 2) and (1 2):

1 =

1 +22 and 2 =

1

22

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Finally, we mention some quantities which are usually employed in structural dy-

namic analysis. The peak value defined as the maximum displacement, or magnitude of Equation (6). Other quantities useful are the mean value and the mean-squarevalue

= lim

1

Z0

() and 2 = lim

1

Z0

2 ()

The square root of the mean-square value, called the root mean square (RMS)value, is commonly used in dynamic analysis

=

2

Because the peak value of the velocity and acceleration are multiples of thenatural frequency and its square, respectively, times the displacement amplitudesee Equation (3) and (4), these three magnitudes usually differ by an orderof magnitude or more. Therefore, logarithmic scales are popular when presentinggraphs. A common unit of measurement for amplitude and RMS values is the decibel(dB). The decibel is defined as the base 10 logarithm of the ratio of the square ofthe amplitudes of two signals, thus

10log10

12

2= 20 log10

12

In many cases it is useful to employed dB scale to improve graphical representation.

1.3 Free damped vibrations

The undamped SDOF system is an ideal system which does not have a mechanismfor dissipating energy in such a way that the total energy initially supplied to thesystem remains constant. However, damped systems have mechanisms for reducingthe total energy supplied to the system. A clear distinction should be made betweenmechanisms of damping in real systems (such as heating, radiation, friction, etc.)and mathematical models for representing damping mechanisms (viscous, hysteretic,proportional of Rayleigh type, non-proportional, etc.). Viscous damping is the most

commonly used one in mathematical models and will be treated here.Consider Figure 6, in which a viscous damper has been added to Figure 4. Theforce generated by the damper is proportional to the velocity and in an oppositedirection of motion: = (), where is a constant parameter called dampingcoefficient, whose units are N s m. It should be noted that, in real systems, usually represents an equivalent damping effects and it is considered proportionalto velocity due to mathematical convenience. From the free body diagram of Figure6 and using the Newtons 2nd law of motion, the equation of motion of a freelyvibrating damped SDOF system is derived

() + () + () = 0 (17)

with the given initial conditions (0) = 0 and (0) = 0

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Figure 6: Damped SDOF oscillator and its free body diagram.

To solve Equation (17), the same approach used for solving Equation (1) is usednow. The substitution of Equation (10) into (17) yields

2 + +

= 0 2 + + = 0 (18)since 6= 0. This equation is known as characteristic equation and has twosolutions

12 = 2

1

2

p2 4 (19)

By examining these solutions, it can be seen that the roots will be real or complexdepending on the value of 2 4. At this point the critical damping coefficientand damping ratio are defined as follows

= 2 = 2

=

=

2=

2

(20)

respectively. Then, the solutions of the characteristic equation (19) can be rewritten

using relations (20) as

12 = p

2 1 (21)and the equation of motion (17) as

() + 2 () + 2 () = 0 (22)

This equation is sometimes known as the standard form of the equation of motion.It is now clear that the damping ratio determines if the roots are complex orreal. For positive mass, damping and stiffness coefficients (as they are for realproblems), there are three cases that are analyzed here: underdamped, overdampedand critically damped motion.

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1.3.1 Underdamped motion

In this case the damping ratio is less than 1 (0 1) and 2 1 is negative in(21). Then, the two solutions are complex conjugate pair of roots (see (21))

12 = p

1 2 (23)Following the same argument as before, a linear combination of solutions is also asolution

() = 11 + 2

2 =

1

12 + 2

12

(24)

Using the Euler relations, the solution can be rewritten as

() = h

1 sin

p

1 2

+ 2 cos

p

1 2i

= sin( + )

(25)in which and are real constants to be determined and is called the dampednatural frequency defined as

= p

1 2, with =r

(26)

The period of the damped vibration is

=2

= p1 2

(27)

being the natural period given in Eq. (8). Setting = 0 in Equation (25)

(0) = 0 = sin (28)

and differentiating Equation (25) and setting = 0

() = sin( + ) + cos( + ) (29)i.e.

() = 0 = 0 + 0 cot (30)Solving the last expression for , it is obtained

tan =0

0 + 0(31)

and the sine of is

sin =0q

(0 + 0)2 + (0)

2(32)

Finally, taken into account that from Equation (28) = 0 sin , the value of both and are given

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Time (s)

u

(t)

u1

u2u3

u4

Figure 7: Free response of an underdamped SDOF system considering 0 = 1,0 = 0, = 2 1 and = 01

=

q(0 + 0)

2 + (0)2

= tan1

00 + 0

(33)

where 0 and 0 are the initial displacement and velocity. A plot of () versustime for this underdamped case is shown in Figure 7. The motion is oscillatory withdecaying amplitude governed by the exponential term exp(). The dampingratio sets the rate of decay. As a check, if one sets = 0 in expressions (33) , werecover the undamped casesee expressions (6).

Next, the logarithmic decrement is defined. The ratio between successive peaksof the underdamped free response is related to the damping ratio. The ratio between

the displacement at time to its value a period later is independent of time andit is as follows using (25)

()

( + )= [(+)] = (34)

This expression is also valid if () and ( + ) are successive peaks: and +1.Then, the logarithmic decrement is defined as the natural logarithm of the ratiobetween these to successive peaks

= ln

+1

= =

2

p1 2

= 2

p1 2

(35)

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' 2 (36)

Then, if the damping ratio is small ( 02, which is the case of most civil engi-neering structures), it can be estimated from

=

2=

1

2ln

+1

(37)

If the decay of motion is slow, it is recommended to use the ratio between twoamplitudes separated several cycles. For instance, if it assumed amplitudes separated cycles

1

+1 =

1

2

2

3

+1 =

Hence

=1

ln

1

+1

' 2 (38)

and the damping ratio for lightly damped systems can be determined as

=1

2ln

1

+1

or =

1

2ln

1

+1

(39)

The second is a similar equation in terms of the acceleration, which is valid for

lightly damped systems.

Example 1 Consider a single-span footbridge. Assume that a mass of 5000 kg issuspended from mid-span of the bridge producing a static sag of 0016m. Then, thismass is suddenly released causing a free decay response. The response is basicallygiven by one vibration mode, then, SDOF vibration response can be assumed. Themaximum peak is0078m s2 at 03 s and18 cycles later the amplitude is0031m s2

at 201 s. From these data compute the following: (1) damping ratio; (2) naturalperiod; (3) stiffness; (4) mass; (5) damping coefficient ; and (6) number of cyclesrequired for the acceleration to decrease to 0005m s2.Solution: The damping ratio can be obtained as follows:

= 1218

ln00780031

= 0008 = 08%The assumption of small damping is valid since it is smaller than 20%The natural period can be obtained from the time between peaks and the number ofcycles, and assuming the natural and the damped periods are approximately the same

=201 03

18= 11 s; ' = 11 s

The stiffness can be obtained from the static equation = , in which is thestatic sag, then

= 50000016

= 312500N m

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The angular natural frequency and the natural frequency are given by

=2

=

2

11= 57120 rad s or =

2

=57120

2= 0909Hz

The mass can be obtained using the expression for the natural frequency

=

2= 95779 kg

The damping coefficient can be obtained as

=

2

= 8753 N s m

The number of cycles to get the signal reduced to 0005m s2 can be obtained fromthe approximated expression of the damping ratio

=1

2ln

1

+1

= 1

2ln

1

+1

=

1

20008ln

0078

0005

= 546 ' 55 cycles

1.3.2 Overdamped motion

In this case, the damping ratio is greater than 1 ( 1) and 2 1 is positive.Then, the two solutions (21) are real roots

12 =

p2 1 (40)The solution of (17) is as follows

() = 11 + 2

2 =

121 + 2

21

(41)

which is a non-oscillatory motion and 1 and 2 are real constants determined, again,by the initial conditions. It is quite straight-forward to obtain the value of 1 and2 (the reader is encouraged to try to derive these values) in terms if the initialdisplacement 0 and velocity 0

1 = 0 + 0 +p2 12

p2 1

and 2 = 0 + 0 +p2 12

p2 1

A plot of () versus time for this overdamped case with three different initialconditions is shown in Figure 8. The motion does not involve oscillation and thesystem returns to its rest position exponentially.

1.3.3 Critically damped motion

In this last case, the damping ratio is exactly 1 ( = 1) and 2 1 is zero. Then,the two solutions are equal and real roots

12 = (42)

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Figure 8: Free response of an overdamped system considering = 2 03 and = 11 . Initial conditions are 0 = 1 0 = 0 for the continuous curve, 0 = 0 0 = 1for the dashed curve and 0 = 1 0 = 0 for the dash-dotted curve.

and the solution is as follows (it is recommended to the reader to revise second-orderdifferential equations with identical roots of the characteristic equation)

() = 11 + 2

2 = (1 + 2) (43)

where, again, the constants 1 and 2 are determined by the initial conditions,

1 = 0, and 2 = 0 + 0

It can be noted that critically damped systems represent systems with the small-est damping ratio that produces non-oscillating response and provides the fastestreturn to zero without oscillation.

1.4 Response to harmonic excitation

This section considers the response to harmonic excitation of a SDOF system. Har-monic excitations are sinusoidal external forces of a single frequency applied to thesystem. Note that periodic excitations can be represented as the sum of harmonicfunctions due to the Fourier series decomposition theorem and the principle of su-perposition. The harmonic analysis is an important issue in structural dynamicsbecause:

1. the response can be calculated in a quite straight forward manner,

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Figure 9: Damped SDOF system acted by an external force () and its free diagram

2. harmonic excitations are simple to produce, and then, they are often used in

experimental dynamic analyses and

3. harmonic inputs are very useful in studying damping and stiffness propertiesof systems, that is, to characterize the dynamic properties of systems.

Consider Figure 6 with an external excitation () Figure 9 applied in thedirection of () being a sine or cosine of a single frequency

() = 0 cos (44)

where 0 is the amplitude and is the frequency of the applied force. The latteris also known as input frequency or driving frequency, depending on the particularapplication.

1.4.1 Undamped system

Recall Figure 2, in which (), defined by Equation (44), is applied in the directionof (). The equation of motion for the undamped case is as follows

() + () = 0 cos or () + 2 () = 0 cos with 0 =

0

(45)

In order to solve this equation, which is a second-order linear non-homogeneousdifferential equation, the solution is written as the sum of the homogeneous solution

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() (i.e., the solution for 0 = 0, which has been studied in the before subsection)

and a particular solution ()

() = () + () (46)

The particular solution is assumed to be in the same form as the excitation

() = cos (47)

Substituting (47) into (45), it is obtained

2cos + 2cos = 0 cos =0

2

2

provided 6= , then

() =0

2 2cos

The solution of (45) is

() = 1 sin + 2 cos | {z }homogeneous

+0

2 2cos | {z }

particular

(48)

in which the form of the homogeneous solution given by (15) has been considered.

The initial conditions have to be considered to obtain 1 and 2. The following twoequations are then obtained

(0) = 0 = 2 +0

2 2, and () = 0 = 1

Solving these two equations, 1and 2 are obtained and the solution (48) is rewritten

() =0

sin +

0 0

2 2

cos +0

2 2cos (49)

Figure 10 shows an example of the total response of an undamped system to har-

monic excitation and given initial conditions.It is now considered the case of = . The particular solution given by (47) is

no longer valid since this function is also a solution of the homogeneous part of (46).The particular solution is now as follows (it is recommended to see a book whichdeals with ordinary differential equations)

() = sin (50)

Substituting (50) into (45), is obtained that = 02 and then

() =02

sin (51)

and the total solution for = is

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Figure 10: Response of an undamped system to harmonic excitation considering = 2 1rad s, an excitation at = 2 3rad s, 0 = 0, 0 = 0005m s and0 = 1 N kg

() = 1 sin + 2 cos +02

sin (52)

Again, imposing the initial conditions, the solution is written as

() =0

sin + 0 cos +02

sin (53)

A plot of () is shown in Figure 11 for zero initial conditions. It can be observedthat () grows unboundly. It is here defined the resonance phenomenon, i.e.,the response becomes unbounded for excitation frequency equal to the naturalfrequency . Obviously, this is an academic result and should be interpreted assuch. At some point in time, the system would fail

1.4.2 Damped system

Take Figure 9, in which () is as defined by Equation (44). The equation of motionconsidering viscous damping is as follows

() + () + () = 0 cos or () + 2 () + 2 () = 0 cos (54)

with =

p, = (2), and 0 = 0. The same procedure as before is

used now to solve this differential equation. In this case, the particular solution isas in the undamped case excepts for a phase shift compared with (47), then

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0

2n

ft

w

Figure 11: Response of an undamped system excited harmonically at its naturalfrequency considering = = 2 1rad s, 0 = 0 = 0, and 0 = 1 N kg

() = cos( ) (55)or in this alternative form

() = 1 sin + 2 cos (56)

The derivatives of () are

() = 1 cos 2 sin , and () = 12 sin 22 cos (57)

Then, ,, and are substituted into the equation of motion (54):

22 + 12 + 22 0 cos + 12 22 + 21 sin = 0This equation must hold at every time. It is particularized for = (2) and = 0,yielding these two equations

(2) 1 +

2 2

2 = 0

2 2

1 (2) 2 = 0

Solving these two linear equations, it is achieved

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1 = 20(2 2)2 + (2)2

, and 2 = 2 2 0(2 2)2 + (2)2

(58)

From Equations (55) and (56), it can be seen that =p

21 + 22 and = tan

1 12

,then, the particular solution can be written as

=0q

(2 2)2 + (2)2cos

tan1 2

(2 2)

(59)

The total solution is again the sum of the particular solution and the homogeneousone. For the underdamped case (0 1) see Equation (25) this results in

() = sin( + )| {z }Transient

+ cos( )| {z }steady state

(60)

in which and are determined by the initial conditions following the same pro-cedure as the undamped case. The result is

= tan1 (0 cos )

0 + (0 cos ) sin , and =

0 cos sin

(61)

Figure 12 illustrates a typical response of a lightly damped SDOF system subjectedto a harmonic excitation. Note that for large values of , the first term of Equation

(60), which is the homogeneous solution, goes to zero and the total solution ap-proaches the particular solution. Thus () is known as the steady-state responseand the first term is called the transient response. The difference between both solu-tions comes from the exponential term, which make the transient solution negligibleafter a while. If the system has a relatively large damping, the transient part will goto zero quickly. Then, the steady-state response will be more important. It shouldbe recognized that the largest amplitude peak may occurred before the system hasreached its steady state.

It is now examined the case in which = for a damped SDOF system. If itis considered = and a lightly damped system, that is ' , the response ofthe system can be approximated as

() =0

22

1

| {z }

envelope function

cos

in which we have used Equation (60) and the corresponding assumptions (The readeris encouraged to demonstrate this expression). The response varies with time as acosine function, with its amplitude increasing according to the envelope function.Note that the envelope function is strongly affected by the damping ratio, and then,the steady state response of the system will be strongly affected by the dampingratio, as well. Figure 13 illustrates this case.

Bearing this in mind, it is of interest to consider the magnitude and the phase as a function of the exciting frequency (see Equation (59))

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0 1 2 3 4 51.5

1

0.5

0

0.5

1

1.5

Time (s)

u(t)

Response of a damped SDOF system to harmonic excitation

Total response

Steadystateresponse

Figure 12: Response of damped system to harmonic force considering = 012and = 004

0 1 2 3 4 50.2

0.15

0.1

0.05

0

0.05

0.1

0.15

Time (s)

u(t)

Response of a damped SDOF system to harmonic excitation atn

envelope cuve

steadystateamplitude

Figure 13: Response of a damped system with = 01 to a sinusoidal force offrequency = and zero initial conditions

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= 0q(2 2)2 + (2)2

, and = tan1 2(2 2) (62)

Equations (62) can be rewritten as

= =

1q(1 2)2 + (2)2

,and = tan12

(1 2) (63)

with = and = 02 = 0 is the static displacement. Thus, = is known as the normalized magnitude to the static displacement. Figure14, shows the normalized magnitude and phase against the excitation frequency(normalized to the natural frequency) for several values of the damping ratio. Theplot of the amplitude of a response quantity against the excitation is called

frequency-response curve. Damping reduces and then, the amplitude of theresponse at all frequencies. The magnitude of this reduction depends importantlyon the excitation frequency. Note that as the exciting frequency approaches ,the magnitude reaches its maximum value when the damping ratio is very small( 01). Observe also that the phase shift crosses through 90 and that the phaselies between 0 and . This defines resonance.

From the normalized magnitude in Figure (14), it can be seen that as goesto zero (implying that the force is "slowly varying"), the is slightly larger than 1and is almost independent of damping. Thus the amplitude goes to 0

2, which

is the static displacement,

'02

=

0

= 0

= static displacement

This result implies that the amplitude of the dynamic response is almost the sameas the static response and is essentially controlled by the stiffness property.

On the other hand, as becomes large (implying that the force is "rapidlyvarying"), tends to zero and is essentially unaffected by damping. The amplitude can be approximated as follows

'0

2

12

2

=0

2

Observe that the response is now governed by the mass value.When ' , is highly affected by the damping ratio. For smaller damping

values, can be several times larger than 1, implying that the dynamic responsecan be much larger than the static one. As the damping becomes smaller, the peakof becomes sharper, and finally, if damping is zero, the peak is infinite. This isin accordance with the unbounded undamped response at resonance see Figure10. As the damping increases, decreases and finally, it disappears. Then, in thiscase, the system response is governed by the damping.

For the undamped case, resonance occurs when = . However, the dampedcase, resonance does not occur exactly at the natural frequency. It can be shownthat the maximum value of takes place at = p1 22 if0 12 = 0707

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0 0.5 1 1.5 20

2

4

6

8

10

Frequency ratio /n

Normalizedma

gnitude

Normalized magnitude of a damped SDOF system

0 0.5 1 1.5 20

0.5

1

1.5

2

2.5

3

Frequency ratio /n

Phase

Phase of a damped SDOF system

=0.05

=0.1

=0.2

=0.4

=0.7

=0.05

=0.1

=0.2

=0.4

=0.7

Figure 14: Plot of the normalized magnitude20

and phase of the steady-state

reponse versus frequency ratio for several damping ratios.

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and at = 0 if 1

2. Then, the value of the exciting frequency for which

reaches its maximum value is the peak frequency

= p

1 22 for 0 0707Note that as damping tends to zero the peak frequency tends to the natural fre-quency. As damping increases from zero, the peaks occur farther left from thenatural frequency or = 1. Finally, when damping ratio is 0707, the peak occursat = 0.

It is interesting to examine how the peak magnitude changes with respect to thedamping ratio. Then, substituting =

p1 22 into , it is obtained

max =1

2p1 2 = 02p1 2and when = 1

( = 1) =1

2 = 0

1

2

1.4.3 Dynamic response factor

The displacement, velocity and acceleration response factors are introduced here.These are dimensionless and define the amplitude of these three quantities. The

steady-state displacement can be written as follows considering Equations (59) and(63)

()

0= cos( ) (64)

where the displacement response factor is the ratio between the amplitude ofthe dynamic response to the static displacement .

Differentiating Equation (64), it is obtained the velocity response

()

0

= sin( ) (65)

where the velocity response factor is related to as follows

=

=

Differentiating Equation (65), it is obtained the acceleration response

()

0= cos( ) (66)

where the acceleration response factor is related to as follows

= 2

= 2

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The dynamic response factors , and are plotted as a function of

in Figure 15. The displacement response factor is unity at = 0, the peaks arereached by 1, and approaches to zero as . The velocity responsefactor is zero at = 0, the peaks are reached by = 1, and approaches tozero as . The acceleration response factor is zero at = 0, the peaksare reached by 1, and approaches to unity as .

1.4.4 Frequency response function method

Up to now, we used the method of undetermined coefficients to solve the differentialequations. An alternative method is to treat the solution of Equation (17) as acomplex function. Then, a frequency domain analysis can be carried out. This

is useful for problems involving many degrees of freedom. Consider the complexrelation

= cos + sin

in which the real and complex parts are considered separately when solving complexequations. Thus, the equation of motion for a harmonic excitation can be rewrittenas a complex equation

() + () + () = 0 (67)

Here, the real part of the complex solution is the physical solution of (). Thisrepresentation is very useful in solving MDOF systems.

The particular solution of (67) is

() = (68)

Substitution of this equation into (67) yields

2 + + = 0 = 1( 2) + 0 = () 0 (69)

where

() =1

( 2) +=

1

(2 2) +2(70)

is known as the frequency response function (FRF). Multiplying by the complexconjugate over itself and taking the modulus of the result, it is obtained

=0q

( 2)2 + ()2 with = tan1

( 2) (71)

Substituting the value of into the particular solution (68) yields the solution

() =0

q( 2)2 + ()2

() (72)

The real part of this expression corresponds to the solution given by (62).

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0 0.5 1 1.5 20

1

2

3

4

5

6

Frequency ratio /n

Rd

=0.05

=0.1

=0.2

=0.4

=0.7

0 0.5 1 1.5 20

1

2

3

4

5

6

Frequency ratio /n

Rv

0 0.5 1 1.5 20

1

2

3

4

5

6

Frequency ratio /n

Ra

Figure 15: Displacement, velocity, and acceleration response factors for a dampedsystem excited by harmonic forces

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1.4.5 Laplace transform based analysis

Another method is to consider the Laplace transform to solve the particular solutionof the forced harmonic equation of motion (54). The Laplace transform is (it isrecommended to the reader to consult a specialized book)

2 + +

() =

0

2 + 2(73)

where is the complex variable and () is the Laplace transform of (). Then

() =0

(2 + + ) (2 + 2)(74)

The solution () is obtained by the inverse Laplace transform of (), equivalentto the solutions (62) and (72).Consider the equation of motion with any excitation and its Laplace transform,

() + () + () = () L 2 + + () = () (75)then, it is obtained

() =()

()=

1

2 + + (76)

() is the transfer function between the output (the system response) and theinput (the excitation). Note that if the value of is restricted to lie along theimaginary axis ( = ), the transfer function becomes FRF (70)

() =1

2 + (77)

Hence, the FRF of the system is the transfer function of the system evaluated along = .

1.5 General forced response

Up to now, the dynamic response of a SDOF system has been studied for given ini-tial conditions (initial disturbances) and/or forced harmonic excitations. Harmonicexcitation refers to an applied force that is sinusoidal with a single frequency. Inthis section the response of a SDOF system to different forces is considered, as wellas a general formulation for calculating the forced response to any applied load.Since the SDOF system considered here is linear, the principle of superposition canbe used to calculate the response to combinations of forces based on the individualresponse to a specific force.

Periodic forces are those that repeat in time. An example is an applied forceconsisting of the sum of two harmonic forces at different frequencies where onefrequency is an integer multiple of the other. A non-periodic force is one thatdoes not repeat itself in time. A step function is an example of a force that is anon-periodic excitation. A transient force is one that reduces to zero after a finite,

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usually short, period of time. An impulse or a shock loading are examples of transient

excitations. All of the aforementioned classes of excitation are deterministic (i.e.,they are known precisely as a function of time). On the other hand, a randomexcitation is one that is unpredictable in time and must be described in terms ofprobability and statistics. This section introduces a sample of these various classesof force excitations and outlines strategies for the calculation and analysis of theresulting motion when applied to a SDOF spring-mass-damper system.

Then, the objective is to find the solution of the differential equation of motion

() + () + () = () (78)

subjected to initial conditions

(0) = 0 () = 0 (79)

in which () is a force varying arbitrarily with time.

1.5.1 Impulse response function

A common excitation is the application of a short duration forced known as impulse.An impulse excitation is a force applied for a very short length of time. If() = 1,with a duration of starting at = (Figure ??). As 0 the magnitude of theimpulse tends to infinitive, but the area remains equal to unity, (1) = 1. Thisforce is known as unit impulse.

The application of the Newtons second law of motion to a force acting on abody mass and the integration of both side yields

( ()) = () ( (2) (1)) =

Z21

() (80)

In the case of a short unit impulse, which was not moving before the impulse wasapplied ( (0) = 0 and () = 0) at 1 = 0 and 2 = , the velocity at 2 = can becalculated as

() =1

Z2

1

() =1

(81)

After the application of the impulse, the force is removed and the response of theSDOF system is a free vibration response disturbed by ()0 = (0

+) = 1.Due to the short duration of the impulse, it can be assumed (0+) = 0. UsingEquations (26) and (33) of a free damped vibrations, it can be obtained

=1

and = 0

Then, using (25), the response of the system at time to a unit impulse applied at ( ) is as follows

() = ( ) = 1 () sin ( ) (82)

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Figure 16: Time history of an impulse applied at time and the correspondingresponse

which is illustrated in Figure 16. The response () is usually referred as ( ),which is called impulse response function.

1.5.2 Response to an arbitrary excitation

The response to a SDOF system to an arbitrary time-varying excitation can becalculated as a sum of the responses to a series of short impulses, as shown Figure17. Thus, the response at can be calculated considering the contributions of allimpulses applied at times before time Then

() =

X=1

() ( ) (83)

As 0, the latter expression can be written as

() =

Z0

() ( ) (84)

which is known as convolution or Duhamel integral.For underdamped SDOF system, the Duhamel integral (84), considering (82), is

as follows

() =1

Z

0 h() () sin ( )

i (85)

assuming zero initial conditions.

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t

dt

F(t)

t

Response to impulse 1

t

. . .

t

Response to impulse 2

u(t)

t

Total response

.

.

.

Figure 17: Schematic explanation of convolution integral

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2 Multi-Degree-of-Freedom Systems

Figure 18: a) Simple undamped model of the horizontal vibration of a three-story

building. b) Restoring forces.

matrix equation is that a linear time-invariant dynamic system is being analyzed.This leads to mass M, damping C and stiffness K matrices being time independent.The superposition principle and Maxwells reciprocity theorem, therefore, apply.

Similar as for a SDOF system, matrix equation (87) can be understood as astatement that equilibrium of inertial F (), damping F (), elastic (stiffness, orrestoring) forces F () and external forces F (), acting in the direction of eachDOF, is satisfied at each instant of time

F () +F () +F () = F ()

where

F () = Mu () F () = Cu () F () = Ku ()

For -DOFs, the matrix equation of motion is formed by second-order ordinarylinear differential equations. Generally, these equations are coupled which meansthat the unknown displacement () can be in more than one differential equation.

2.1.1 Two-Degree-of-Freedom system

The most simple MDOF system is now analyzed. It corresponds to a lumped modelof a two-story building frame subjected to external forces 1 () and 2 (). The

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2.1 Introduction

Figure 19: Two-story shear frame and forces acting on the masses

forces acting in each floor mass are shown in Figure 19. These include twoexternal forces 1 () and 2 (), the elastic stiffness forces 1 () and 2 (), andthe damping forces 1 () and 2 (). The external forces are supposed to actalong the positive direction. The elastic and damping forces are shown acting in theopposite direction since they are internal forces that resist the motions.

The application of the Newtons second law of motion gives the following equationfor each mass:

() () () = () , or () + () + () = () ; = 1 2

or, for both degrees of freedom in a matrix form

1 0

2

1 ()2 ()

+

1 ()2 ()

+

1 ()2 ()

=

1 ()2 ()

(88)

Assuming linear behavior, the elastic resisting forces are related to the relative dis-placement between the storeys and the storey stiffness. The storey stiffness is thesum of the lateral stiffnesses of all columns in the storey. For a storey of height ,and a column with modulus and second moment of area , the lateral stiffness of

a column with clamped ends, implied by the shear building idealization, is 123

.Thus, the storey stiffness is

=X

no of columns

12

3

which relates the elastic stiffness forces with the displacement. The force 1 at thefirst floor is made of two contributions: 1 from the story above, and

1 from the

story bellow. Thus

1 = 1 +

1 with

1 = 2 (2 1) and 1 = 11

When developing these equations it is assumed that force acts in the direction asshown in Figure 19. Therefore, the sign "" in 1 is because positive inter-storey

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drift (2

1) generate a force which acts on the floor in the left-to-right direction,

which is a negative direction, as defined in Figure 19. Therefore,

1 = 11 + [2 (2 1)] or 1 = (1 + 2) 1 22 (89)

Finally, similarly as for 1, the force 2 is

2 = 2 (2 1) (90)Equations (89) and (90) can be written in matrix form as

12

=

1 + 2 2

2 +21

2

or F = Ku (91)

The damping forces 1 () and 2 () are related to the storey horizontal ve-locities 1 () and 2 (), and their difference (2 () 1 ()). Following the sameprocedure as for elastic stiffness forces, it can be shown that

12

=

1 + 2 22 +2

12

or F = Cu (92)

When Equations (91) and (92) are substituted into Equation (88), the followingmatrix equation can be obtained

1 0

2

1 ()2 ()

+

1 + 2 22 +2

12

+

1 + 2 2

2 +2

12

=

1 ()2 ()

(93)

This matrix differential equation can be presented as a set of two second orderdifferential equations

11 () + (1 + 2) 1 () + (1 + 2) 1 () (22 () + 22) = 1 ()22 () + 22 () + 22 () (21 () + 21) = 2 ()

The analysis of these equations indicates that: (i) the two equations of motiondescribing the motion of two DOFs 1 and 2 are coupled because the unknownfunction 2 () depends on the dynamic equilibrium conditions of DOF 1 and viceversa, and (ii) the coupling is physically provided by stiffness 2 and damping coef-ficient 2.

2.1.2 Mathematical modeling of damping

Whereas the formulation of mass and stiffness matrices M and K is based on thesummation of physical properties of the individual discretized elements, the damp-ing matrix C cannot be formulated in the same way. Nevertheless, it is usuallyconvenient to assume that the damping is viscous meaning that the damping force

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u () =X=1

() e (95)

where () is known a generalized coordinate which is a function of time. Therefore,at each instant of time the displacement vector u () can be presented as thefollowing linear combination

u () =X=1

() e

where () can be seen as a constant which will change at the next instant of time.

The set of vectors e is known as basis of an -dimensional vector space definedby the MDOF vibration problem. The key is then how to find sets of () ande in order to obtain the solution given in Equation (95). To do this, and followinga procedure similar as for SDOF systems, a special case of an initially disturbedundamped system will be studied initially.

2.2 Vibration absorber: an application of Two-Degree-of-Freedomsystems

The goal of this section is to provide the theoretical background for the design ofpassive vibration absorbers. Damping devices can be categorized as: passive devices,active devices and semi-active devices. Within passive devices, one can find a largenumber of different devices: viscous damper, viscoelastic damper, Coulomb frictiondamper, hysteretic damper, tuned mass damper (TMD) or tuned liquid damper.

A passive TMD or tuned vibration absorber is basically an energy dissipationdevice that in its simplest form consists of a mass (secondary mass) that is attachedto a structure (primary system) with spring and damper elements. Figure 20 showsa CAD model of a TMD and Figure 21 shows the installation of a TMD underthe deck of a footbridge and a detail view of the TMD. Due to the damper, energydissipation happens when the secondary mass of the TMD oscillates. This is achievedby transferring as much energy as possible from the primary system to the TMDby a careful tuning of the natural frequency and damping ratio of the TMD. Sincethe mass of the TMD is significantly smaller than that of the primary system,transferring energy from the primary system to the TMD generates a great relativeoscillation of the mass of the TMD. A TMD operates efficiently only in a narrowfrequency band, That is, high energy transfer occurs when the natural frequency ofthe TMD is tuned to the natural frequency of the primary structure. Therefore, ifthe TMD is attached to a continuous structure, the TMD mitigates only one specificvibration mode.

The concept of a TMD without an integrated damper was firstly developed byFrahm in 1909 to reduce the rolling motion of ships (US Patent No. 989958). Inthe third edition of Den Hartogs book "Mechanical Vibrations" (1947), an analysisfor the optimal design of the TMD parameters considering the TMD damper waspresented. Thus, the model of a primary mass-spring-damper system, modeling the

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2.2 Vibration absorber: an application of Two-Degree-of-Freedom

systems

Figure 20: CAD model of TMD designed to cancel vertical vibrations.

Figure 21: Typical implementation of a TMD for mitigation of vertical vibrationson bridges.

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Figure 22: Two-degree-of-freedom model of a vibration absorber (TMD) attachedto a primary system considering an excitation force () acting on primary massand an excitation through base acceleration ().

primary structure, with an attached secondary mass-spring-damper system, model-ing the TMD, is represented by the two-degree-of freedom system shown in Figure22.

The equation of motion of the two-degree-of-freedom model of Figure 22 is (thereader is encouraged to derive these equation of motion)

11 () + (1 + 2) 1 () + (1 + 2) 1 () 22 () 22 () = () 1 ()(96)

22 () + 22 () + 22 () 21 () 21 () = 2 () (97)

These two equation can be written as a matrix equation as was done before (93)

1 0

2

1 ()2 ()

+

1 + 2 2

2 +2

12

+

1 + 2 2

2 +2

12

= (98)

=

()

0

12

() (99)

where 1 is the displacement of the primary system and 2 is the displacement ofthe TMD mass. Parameters 1 1 1 are the mass, damping and stiffness of the

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systems

primary mass and 2 2 2 are those of the TMD. () is the force acting on theprimary system and is the base acceleration.

2.2.1 The undamped vibration absorber

Let us consider model 22 with 1 = 2 = 0 and subjected only to a harmonicexcitation on the primary mass () = 0 sin(). The equation of motion are nowas follows

11 () + (1 + 2) 1 () 22 () = 0 sin()22 () + 22 () 21 () = 0

Let 1 = 1 sin() and 2 = 2 sin() (considering only the steady state re-

sponse, that is, the particular solution for an undamped system), and thus 1 =21 sin(), 2 = 22 sin(). Substituting these values into the equation ofmotion,

(1 + 2 12)1 22 = 021 +

2 2

2 = 0

Solving,

1 = 0 2 22

(1 + 2 12) (2 22) 22In order to cut down the amplitude of the vibration of the primary mass 1, 1 = 0,

2 22

must be equal to zero. Hence 2 = 22 and 2 = 22.

Then, the absorber must be designed such that its natural frequency is equalto the frequency of the applied force. When this happens, the amplitude of thevibration of the primary mass is practically zero. In general, a TMD is used onlywhen the natural frequency of the original system is close to the excitation frequency.Hence, 11 = 22

Example 3 A small reciprocating machine weighs25 and runs at a constant speedof 6000 rpm.(See Figure 23) After it was installed, it was found that that excitation

frequency was to close to the natural frequency of the system. What vibration ab-sorber should be added if the nearest natural frequency of the system should be at least20% from the excitation frequency?

Solution. After the absorber is added to the machine, the whole system becomesa two-degree-of-freedom system, which is simplified and represented by Figure (22.The amplitudes of the steady state vibration of the two masses are given by

=0

2 2

(1 + 2 2) (2 2) 22 = 02

(1 + 2 12) (2 22) 22

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Figure 23: Illustration of a reciprocating machine.

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systems

in which 1 = and 2 = . The natural frequencies of the whole system areobtained from

1 + 2 2

2 2 22 = 0

and these frequencies should be at least 20% from the excitation natural frequency inorder to avoid resonant behavior. Dividing the last expression by 12 yields

1 + 21 21

1 2 2 21 = 0

On the other hand, 1 = 2 = 2 must be fulfilled. If it is defined

2 = 22, the former equation can be rewritten as

4 (2 + 21) 2 + 1 = 0

and since = 08, then 21 = 021. Since 1 = 2, then 21 = =021. Therefore, = 021 = 525kg. Thus the obtained absorber should weigh525kg and have a spring of stiffness 0211.

2.2.2 The damped vibration absorber

Consider now the two-degree-of-freedom system represented in Figure 22 includingdamping in both, primary and secondary mass.

Example 4 Determine the general motion of a damped two-degree-of-freedom sys-tem subjected to a harmonic excitation on the primary mass.Solution: From Equations (96), the motion is described by the following equations

11 () + (1 + 2) 1 () + (1 + 2) 1 () 22 () 22 () = 0 sin()22 () + 22 () + 22 () 21 () 21 () = 0

Using the frequency response method (studied in Chapter 1), we substitute0 sin()by 0 and 1 = U1 and 2 = U2 . Rearranging and dividing by , theequation of motion can be rewritten as

(1 + 2) 12 + (1 + 2)

U1 (2 +2) U2 = 0 (2 +2) U1 +

2 22 +2

U2 = 0

Using the Cramers rule,

U1 =

0 (2 +2)0

2 22 +2

[(1 + 2) 12 + (1 + 2) ] [2 22 +2] (2 +2)2

which is of the form ( +) (+) or ( +), which modulus is 1 =q2+2

2+2 , giving

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1 =

s20 (2 22)2 + 222

(124 122 212 122 + 12)2

+ (12 + 21 123 213 + 223)2

2 =

s20

22 222

(124 122 212 122 + 12)2

+ (12 + 21 123 213 + 223)2

From the complex variable theory, we can write

U1 = 0 ( +) = 11 = 1 (cos 1 + sin 1)

U2 = 0 (+) = 22 = 2 (cos 2 + sin 2)

where 1 = tan1

and 2 = tan

1

. The excitation force is 0 sin() =

Im

0

, therefore

1 () = Im

U1

= Im

1(+1)

= 1 sin( + 1)

2 () = Im U2 = Im 2(+2) = 2 sin( + 2)If the excitation force were 0 cos() = Re

0

, therefore

1 () = Re

U1

= Re

1(+1)

= 1 cos( + 1)

2 () = Re

U2

= Re

2(+2)

= 2 cos( + 2)

Example 5 From Example 1, it was obtained the stiffness, mass and damping co-efficient of single-vibration-mode-response of a footbridge. Consider a harmonic ex-citation of unity amplitude and a frequency 5% higher from the natural frequency ofthe structure. Obtain the steady state response. Imaging now, a secondary mass is

added of value5% of structure (primary) mass. Choose the stiffness of the secondarymass in such a way that the natural frequency of the secondary mass is the same asthe primary mass. Obtain the response of the resulting two-degree-of-freedom system

for different values of the damping coeficient of the secondary mass. Plot then theamplitude of the displacement of the primary mass versus the damping coefficient ofthe secondary mass. Which is the effect of the secondary mass on the primary massdisplacement? Which is the effect of the damping coefficient of the secondary masson the primary mass displacement?

Up to now, many optimization criteria has been proposed for the design ofdamped vibration absorbers for damped structure. One of the most common cri-

teria is the one based on the norm optimization criteria. The norm of anFRF () represents the maximum amplitude of the absolute value | ()|. The

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systems

norm of the FRF between the primary mass displacement and the harmonicexcitation on the primary mass is minimized as follows

min22

| ()| = min22

max

| ()|

It is usually defined the frequency ratio =q

2211

, the mass ratio = 21 and

damping coefficients 1 and 2. The optimization problem is then stated as

min2

| ()| = min2

max

| ()|

The minimum value of the | ()| is achieved if and only if there exist two localmaxima and both have exactly the same amplitude (see Figure 24). If one assumethat 1 = 0, that is, the damping of the primary mass is vanished , there existat least two frequencies where | ()| is invariant with respect to 2 ( is fixed).This observation was used by Den Hartog to develop the well-known fixed pointsmethod to design absorbers for structures with vanishing damping. Thus, simpleclosed form approximations can be derived. These approximations are sufficientlyaccurate for primary structures with damping ratios less than 10% (1 010). Ifthe primary mass damping is high, the fixed points method does not work, andthe optimization problem should be solved numerically. Thus,the optimal absorbingcapacity is obtained if both peaks have the same height and for a structure with

vanishing damping the optimal parameters and 2 have been determined by DenHartog as

=1

1 +

2 =

s3

8 (1 + )3

Example 6 An example of a TMD design for the cancelation of a single-degree-of-freedom vibration of structure. Figure 25 shows the dynamic response factor betweenthe primary mass displacement and the excitation on the primary mass.

%Vibration absorber design

close all; clear all; clc

%Structure modal properties

f1=3.51; %natural frequency of the structure

w1=3.51*2*pi; %in rad/s

m1=18500; %modal mass associated the mode

k1=m1*((w1)^2);

c1=m1*2*0.007*(w1);

%Vibration absorberrmass=0.02; %mass ratio

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Figure 24: Illustration of the Den Hartogs fixed point method for primary structurewith vanishing damping 1 = 0.

m2=(m1*rmass); %secondary mass value

dampratio=sqrt(3*rmass/(8*(1+rmass)^3));freqratio=1/(1+rmass);

f2=f1*freqratio

w2=w1*freqratio

c2=2*dampratio*w2*m2

k2=(w2^2)*m2

%FRFs

s=tf(s);

%FRF without absorber

G=(1/m1)/(s*s+s*(c1/m1) + (k1/m1));

%FRF with vibration aboserber

A=s*s*m1 + s*(c1+c2) + (k1 + k2);B=((s*c2+k2)*(s*c2+k2))/(s*s*m2 + s*c2 +k2);

Gtotal=minreal(1/(A-B));

%BODE diagram

W=2*pi*linspace(2.5,4.5,1000);

bodemag(G,k-,Gtotal,b,W)

xlabel(Frequency,FontName,Times New Roman,FontSize,12);

ylabel(|G(jw)|,FontName,Times New Roman,FontSize,12);

title(TMD design,FontName,Times New Roman,FontSize,12);

legend(Without TMD,Tuned,"FontName,Times New Roman,FontSize,12)

The reader can easily analyze the effects of poor frequency tuning and poordamping tuning. For instance, it can be seen that the effect of the TMD is very

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2.3 Free undamped vibrations of MDOF systems

TMD design

Frequency (Hz)2.5 3 3.5 4 4.50

1

2

3

4

5

6

7

8

9x 106

|G(jw

)|(abs

)

Without TMD

With TMD

Figure 25: Example of TMD design.

sensitive to an inaccurate frequency tuning, while the tuning of the damping ratiois much less critical. Plot the dynamic amplification factor considering an error of

5% in the frequency, 2 = 0951 and 2 = 1051. Plot the dynamic amplificationfactor considering an error of 50% in the damping.


If the system is undamped, C = 0 and it vibrates freely after an initial disturbance,the set of equations that describe the dynamic of this system is

Mu () +Ku () = 0 (100)

with the corresponding initial conditions at = 0

u ( = 0) = u0 (101)

u ( = 0) = u0 (102)

It is assumed that after initial disturbance the motion of the MDOF system isharmonic (as in the SDOF case, which is nothing else but a limiting MDOF casewhen = 1, where all points vibrate with the same frequency, but have differentamplitude in the case of all DOFs. Therefore, similarly as for SDOF systems

u () = a sin ( + ) , u () = a2sin ( + ) with a R1

After replacing the displacement and acceleration vectors in Equation (100), thefollowing matrix equation is obtained

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2M + Ka sin ( + ) = 0since sin( + ) is not zero for every , the only option for this equation to besatisfied is that the following set of linear homogenous equations is satisfied2M + Ka = 0 or 2M a = Ka (103)where the unknowns are the vibration amplitudes in a. Equation (103) is known asan undamped eigenvalue problem. Possible solutions are:

1. A trivial solution a = 0, which implies no motion, so it is not appropriate, or

2. To find which would guarantee the existence of a non-trivial solution of thesystem. In this case, it must be satisfied

det2M + K = 0 (104)

When this determinant is expanded, a polynomial of order in 2 is obtained.Equation (104) is known as the characteristic equation. This equation has realand positive roots 2 ( = 1) for

2. This is because structural mass andstiffness matrices are symmetric and positive definite.

2.3.1 Natural frequencies and mode shapes

Each solution 2 of (104) leads to a non-trivial solution of (103)

a = = 1

so that 2M + K = 0, = 1 (105)Therefore, considering that every vector , which is a constant multiplying ,is also a solution, there are time domain solutions of Equation (100), each one

being as follows

u () = sin( + ) (106)

where and are two constants to be determined from the initial conditions.Equation (106) can also be written as

u () = ( sin() + cos()) (107)

where and are the new constant to be determined.Equation (107) shows that the solution of an undamped problem is in fact a col-

lection of sinusoidal functions each one having frequency , which is called a natural

frequency. Such equation also indicates that during this oscillation the structuretakes a shape defined by vector which is known as a mode shape (or natural mode

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of vibration or normal mode or characteristic vector). The pair and define an mode of vibration.

The process of extracting the pair and is usually known as eigenvalueextraction where =

2 is called eigenvalue and is eigenvector of the following

eigenproblem M1K

= , = 1 (108)

Remind that an eigenproblem is defined as the problem of finding vectors b andconstants , for a given matrix A, such that Ab= b.

As an example, it is considered the three-story shear model building of Figure18 with the following system properties are assumed

1 = 2 = 3 = 5000 kg

1 = 2 = 3 = 6000 N m

The natural frequencies and modes shapes calculated by solving the eigenproblemare shown in Figure 26. The natural frequencies calculated are

1 = 0487 rad s 1 = 0077Hz2 = 1366 rad s

2 = 0217Hz

3 = 1974 rad s 3 = 0314Hz

and the modal shapes (in which the largest value is scaled to unity)

1 =

04450802

1

2 =

10445

0802

3 =

08021

0445

2.3.2 Orthogonality of mode shapes

It can be shown that mode shapes are orthogonal with regard to mass and stiffness

matrices, that is, they are independent and can be used to express the solution of(100) as

u () =X=1

() (109)

where () is termed generalized or modal coordinate. This expression can be re-written in a matrix form

u () = q()

The proof of this key property is as follows. If and , and and are

two different modes of vibration with different frequencies 6= , then Equation(105) for mode is

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Figure 26: Plot of the three modes shapes of the system of Figure18

2M = K (110)

By multiplying by the transpose of from the left

2 M =

K

and transposing both sides of the equation, it is obtained

2M =

K (111)

where we considere that M and K are symmetric matrices. Equation (105) is nowre-written for mode and multiplied by the transpose of

2M =

K (112)

Right-hand sides in Equations (111) and (112) are the same, which means, afterdeducting them, that

2 2M = 0

As 6= , the only option to fulfil the latter equation is

M = 0 with 6= (113)

Similarly, it can be proven that

K = 0 with 6= (114)

Then, the orthogonality of the modes with different frequencies is proven. The or-thogonality of natural modes implies that the following square matrices are diagonal

K = K, M = M (115)

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where the diagonal elements are positive since M and K are positive definite

K= 0 and M = 0 (116)

As we see below, these diagonal terms are related by

= 2 (117)

The conclusion is that natural modes are orthogonal with respect to matrices Mand K and linearly independent, so they can be used in Equation (109) to describethe solution vector u (). This procedure is called modal expansion.

2.3.3 Modal matrices

As previously mentioned, considering Equation (105), each of the natural fre-quencies and mode shapes can be re-written in the form

K = M2 (118)

If the modal matrix is as an matrix, in which each column is a modalshape

= 1 and the spectral matrix is defined as an diagonal matrix containing squarednatural frequencies on its diagonal

2 =

21

2

then all modal equations (118) can be written in a compact form as the followingsingle matrix equation

K = M2

K = M with 2 (119)

Due to the orthogonality of modes of vibration, the following square matrices arediagonal

K =

1

= K, M =

1

= M

Then, multiplying Eq. (119) by from left side, it is obtained

1

= 1

21

2

or = 2

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We now pay attention to the de-coupling of the equations of motion. Fromequation of motion (100), and after differentiating the assumed solution (109), thenew equation of motion in the generalized (modal) coordinates is

X=1

(M () + K ()) = 0

If both sides are multiplied from the left by , this equation becomes

X=1 M () +

K () ()

=

0

Due to orthogonality conditions

() + () = 0 (120)

This operation can be performed for every = 1, which means that itis possible to establish equations (120) which can be arranged in the followingmatrix form

1

q() +

1

q() = 0 or Mq() + Kq() = 0 (121)

This means that the system of coupled equations of motion in physical coordinatesu () has been transformed into a system of independent (de-coupled) equationsof motion in the generalized coordinates q().

2.3.4 Normalization of modes

Being solutions of a homogenous system of equations, mode shapes do not have anabsolute value and, after eigenvalue extraction, they can be scaled (i.e. multipliedby a constant) in a number of ways. Two scaling or normalization methods areusually used:

1. A unity-scaled mode shape which is scaled in such a way its maximum valuehas an amplitude of 10, or

2. A mass-normalized mode shape , which is scaled using any mode shape obtained via eigenvalue extraction, as follows

=

with = M (122)

Scaled in this way, a mass-normalized mode shape leads to the following diagonal-ization of the mass matrix

1

M1

= 1

2 M =

1

= 1kg

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Therefore, considering Equation (117), in the case of mass normalized mode shapes

= 21 =

2

Hence option 2 is the most used one.

2.3.5 Response of undamped MDOF systems

As has been seen previously, the time-domain solution of the undamped system anafter initial disturbance can be written as (see Equation (107))

u () = ( sin() + cos()) (123)

Since each u () is a solution of Equation (100), a linear superposition of theseindividual solutions is also a solution, so the final solution is

u () =X=1

u () =X=1

( sin() + cos()) (124)

with initial conditions at = 0

u (0) =X=1

(125)

and

u (0) =X=1

(126)

With the vectors of initial displacements and velocities, and natural frequenciesand mode shapes known, each of equations (125) and (126) is a set of linearalgebraic equations in the unknown constants and .

From the modal expansion of the vectors u () and u () Equation (109) itfollows that:

u ()=

X=1

() (127)

u ()=X=1

() (128)

and then

u (0)=X=1

(0) (129)

u (0)=

X=1

(0) (130)

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By comparing Equations (129) and (130) with Equations (125) and (126), it is clearthat (0) = and (0) = , where is the known natural frequency.

By multiplying both sides of Equation (129) by M, and changing the sum-mation index from to , yields

M u (0) =X=1

M (0) = (0) (131)

Mu (0) =X=1

M (0) = (0)

Therefore

(0) = M u (0)

(132)

(0) = Mu (0)

Thus, Equations (132) convert initial conditions given in physical coordinates intoinitial conditions of a generalized coordinate for each of the modes. Finally,considering that

= (0)

= (0)

and based on Equation (124), the final solution can be expressed as

u () =X=1

(0)

sin() + (0)cos()

(133)

Also, by replacing (0) = sin and (0) = cos , the following relation-ship can be obtained (the same as SDOF system)

u () =X=1

sin( + ) (134)

where

=

p2

20) +

2 (0)

(135)

= tan1

(0)

(0)

The initial (modal) conditions of the generalized coordinate () are calculated

from Equation (132).The following observations regarding the above set of equations can be made:

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0 20 40 60 80 100 120 140 160 180 2000.1

0.05

0

0.05

0.1

Time (s)

u1

(m)

First story displacement

0 20 40 60 80 100 120 140 160 180 2000.1

0.05

0

0.05

0.1

Time (s)

u

2(m)

Second story displacement

0 20 40 60 80 100 120 140 160 180 2000.1

0.05

0

0.05

0.1

Time (s)

u3

(m)

Third story displacement

Figure 27: Undamped response of each of the three floors for the building modelgiven in Figure 18 to a initial displacement 3 (0) = 01

1. Equations (132) and (133) indicate that any scaling of a mode shape does notaffect the final amplitude of the response because the same scaling is usedwhen calculating modal mass, so the selected scaling factors are canceled.

2. A total response o

dynamic analysis foundations

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