Las empresas de servicios jurídicos de México

⎝⎜ ⎞

⎠⎟ = × =

( )

0 1 1 1600 1 3077 2 093 3

γ . , . kN -s/m

1.1.1.3  Characteristic Equation and Modal Model

To seek possible solutions of the homogeneous differential equation, assume that

x t Ae( ) = ^λt (1.16)

where A is a displacement amplitude. Then, substituting Equation 1.16 into Equation 1.7, it can be determined whether a possible solution described by Equation 1.16 exists. The idea of using the assumption such as described in Equation 1.16 is referred to as the semidefinite method. Here, λ is a complex number whose physical meaning is discussed later. With the help of Equation 1.16,

x t

( )

^=λAe^λt and x t

( )

^=λ2Ae^λt (1.17)

Substituting Equations 1.16 and 1.17, as well as Equations 1.8 and 1.9, into Equation 1.7 yields

(

λ²+2ξω λ ωⁿ + ²ⁿ

)

^Aeλt=0 ^(1.18)

Since Ae^λt does not equal zero, each side of Equation 1.18 is divided by this factor and results in

λ²+2ξω λ ω_n + ²_n =0 (1.19)

Equation 1.19 is referred to as the characteristic equation of the system. Solving Equation 1.19 for λ gives:

λ= −ξω_n± ξ²−1ω_n (1.20)

From Equation 1.20, it is seen that when the damping ratio ξ is smaller than 1, that is,

ξ <1, (1.21)

Equation 1.20 can then be rewritten as

λ1 2, = −ξωn±j 1−ξ ω2 n (1.22)

where

j = −1 (1.23)

In Equation 1.22, λ1 and λ2 are complex conjugates. Both λ1 and λ2 belong to two distinct vibra-tion modes, which implies the aforemenvibra-tioned energy exchange. In fact, the energy exchange occurs between the potential and the kinetic energies. Once again, a vibration system does not exist without the energy exchange. Furthermore, the energy exchange is described if and only if the inequality shown in Equation 1.21 holds. In this case, the system is underdamped. In engineering practice, underdamped systems are more common. In fact, ξ can be much smaller than 1. In that case, no matter what type of damping forces (e.g., viscous, viscoelastic, hysteretic, quadratic) actually exist, the response x will not be greatly influenced by the same “effective” damping ratio (see Section 1.3).

The assumption of viscous damping can provide fairly accurate results of responses. The approach of using linear equations should be, from a mathematical standpoint, the most convenient. This will be examined in more detail in Chapter 5, where specificities about the various damping devices are first presented.

From the above discussion, it is seen that the damping ratio and natural frequency are param-eters of the system itself, which will not be affected by external conditions. They are referred to as the basic dynamic parameters or eigen-parameters of the system; therefore, the term λ is called the eigenvalue. Figure 1.2 depicts the eigenvalues for the undamped case in the com-plex plane. Furthermore, the model described by Equation 1.19 is referred to as the modal model because it can be described by the complex conjugate modes determined by parameters ξ and ωn. Additionally, ξ and ωn, the damping ratio and the natural frequency, are referred to as the modal parameters.

Note that there are three basic parameters of the physical model for the SDOF system, whereas for the modal model there are only two.

When the dynamic behavior of linear multi-degree-of-freedom systems is studied, a complete set of modes or a modal model can be used to represent the entire system. This is discussed in Chapters 2 and 3. Here, for the SDOF system, there is only one mode for consideration in engineer-ing practice. Complex conjugates modes do not need to be distengineer-inguished in particular and are usu-ally referred to as a single identical mode.

On the other hand, when

ξ =1, (1.24)

Equation 1.20 can be rewritten as

λ1 2, = −ξωn (1.25)

Thus, the characteristic equation has two identical real-valued solutions, and the corresponding system is said to be critically damped. From Equation 1.13, it is known that when ξ = 1, c=2 km. To denote the critically damped case, this particular damping coefficient is shown with subscript

“cr” as

ccr=2 km (1.26)

With this notation, the damping ratio for a general system with damping coefficient c can be rewrit-ten as

ξ = c =

km c 2 ccr

(1.27)

Thus, the damping ratio is formally referred to as the critical damping ratio in many references.

In addition, from Equation 1.27, it is seen that the damping ratio denotes the ratio of the damping coefficient c and twice the geometric mean of the mass m and the stiffness k, namely, 2 km.

Lastly, when

ξ >1, (1.28)

Equation 1.20 can be rewritten as

λ1 2, = −ξωn± ξ2−1ωn (1.29)

Im

Re ξω_n

+j√1–ξ²ω_n

–j√1–ξ²ω_n

FIGURE 1.2  Eigenvalues of an underdamped system.

Thus, the characteristic equation has two distinct real-valued solutions, and the corresponding system is said to be overdamped.

Notice that in both the critically damped and the overdamped cases, complex-valued eigenvalues no longer exist. The systems are reduced to two first-order real-valued pseudo modes, and there will be no energy exchange between the two modes. In other words, both types of systems are no longer vibration systems. Although in these two cases, the mass can still be made to oscillate from external forces, these systems will no longer be able to oscillate in free vibration.

Example 1.3

Suppose a characteristic equation is given by

^{mλ2+c kλ+ =0}

Find the quantities of λ1λ2 and λ1 + λ2, where λ1 and λ2 are the solutions of the above equation.

From the general properties of the quadratic equation, it is known that

^{λ λ1 2= km}

and

^{λ1+λ2= − cm}

The above expressions are always true regardless of whether the system is underdamped or overdamped. However, if the system is underdamped, then

^{λ12, = −ξωn±j1−ξ ω2 n}

Therefore,

^{λ λ1 2 λ λ1 1 ω}

= ^*= 2n= k m

where superscript * stands for the operation of taking the complex conjugate. In this case, it is easily seen that

mk = ω_n²

Furthermore,

^{λ1+λ2= −ξωn+ −(ξωn)= −2ξωn} Thus,

−2ξωn= − c

m

or

mc =2ξωn

1.1.2  H^oMogEnEous s^olution, f^rEE-d^Ecay V^iBration, ^and tHE r^EsPonsE M^odEl

The solutions of Equation 1.1 represent the vibration responses, which can be classified as: (1) the transient response, including free-decay vibration and forced vibration excited by the transient forc-ing function; (2) periodic vibration, the steady-state response due to periodic excitations; and (3) random vibration, due to random excitations.

It is understandable that the system will not vibrate unless a certain external input is applied.

The input can be either the initial condition of velocity and/or displacement or the forcing func-tions. Note that a forcing function can also cause a forced initial condition. If the input is the initial conditions only, then free-decay vibrations will occur. That is, Equation 1.1 can have a free-decay solution, if the system, e.g., the cart in Figure 1.1a, is excited by an initial force, or has an initial velocity or displacement; and after the initial excitation, no external force is added to the system. In this case, Equation 1.7 will be used. An SDOF vibration system with m = 2, k = 100, and initial unit velocity is used as an example. Suppose there are two cases of damping magnitude, the first with a 5% damping ratio and the second with a 50% damping ratio. The responses of the two systems are plotted in Figure 1.3. It is seen that the vibration levels continuously decrease in both cases.

However, the vibration with a larger damping ratio decays much faster.

Using the above mentioned semi-definite method, the free-decay displacement under certain initial conditions is written as follows:

x t Ae( )= ^−ξωntsin(ωdt+ϕ) (1.30)

with the initial conditions

x d

x v

( ) ( ) 0 0

0 0

=

=

⎧⎨

⎩  (1.31)

0 1 2 3 4 5 6

–0.2 –0.15 –0.1 –0.05 0 0.05 0.1 0.15 0.2

Time (sec)

Displacement (m)

Comparison between free-decay responses 5% damping free decay curve 50% damping free decay curve 5% damping free decay envelop curve 50% damping free decay envelop curve

FIGURE 1.3  Free vibration with decay.

Here, d₀ and v₀ are, respectively, the initial displacement and velocity, and

ω_d= 1−ξ ω² _n (1.32)

is called the damped natural frequency, which is the imaginary part of the eigenvalue described in Equation 1.22. In Equation 1.30, A and φ are, respectively, the amplitude and phase angle constants and can be determined by the initial conditions given in Equation 1.31. That is,

A= _ω¹

(

v⁰+^ξω d⁰

)

²⁺

(

^ω d⁰

)

²

d n d (1.33)

and

ϕ ω

ξω ϕπ

= +

⎛

⎝⎜

⎞

⎠⎟+ tan−^{1 0}

0 0

d n

v d d h (1.34)

Note that the period of the tangent function is π, and the arctangent function has multiple val-ues. However, the period of the sine and cosine functions is 2π. Therefore, the Heaviside func-tion, h_φ, cannot be arbitrarily chosen. Based on the fact that most computational programs, such as MATLAB^•,^* calculate the arctangent by limiting the values from – π/2 to + π/2, hφ is defined as

h v d

v d

ϕ

ξω

= +ξω ≥

+ <

⎧⎨

⎩

0 0

1 ⁰0 ⁰0 0 ,

, ⁿn

(1.35)

As shown in Figure 1.4, the phase angle φ can have four cases, namely, the combination of ωdd0

and v0 + ξωnd0 to be positive and negative numbers. Despite the values of ωdd0, it is seen from Figure 1.4 and Equation 1.35 that the sign of v0 + ξωnd0 determines the choice of h_φ.

From Equation 1.30, it is seen that the vibration will have an envelope of Ae^−ξωnt. As time goes on, the level of the free vibration will decrease. The rate of the decay per cycle depends on the value

* The mathematical software package MATLAB is often referenced throughout this book. The reader is encouraged to become familiar with this useful tool, which is used in many of the exercises.

v₀ +ξω_nd₀ >0 v₀ +ξω_nd₀ <0

ω_dd₀ <0 ω_dd₀>0

Im

Re ϕ₁

ϕ₂

FIGURE 1.4  Phase angle φ.

of the damping ratio. In Figure 1.3, it is seen that the responses of the systems with a 5% damping ratio and with a 50% damping have corresponding envelopes.

In addition, when the damping ratio is larger, the first peak value is smaller. Furthermore, the peak value appears earlier and the corresponding damped frequency is smaller.

Taking the time derivative of the displacement in Equation 1.30, the velocity becomes

x t

( )

⁼A e^ωn ^−ξωntcos

(

^ωnt^{+ +ϕ θ}

)

(1.36)

where

θ ξ

= ξ

−

⎛

⎝⎜ ⎞

⎠⎟ tan−¹ ₂

1 (1.37)

Comparing Equation 1.30 and Equation 1.36, it is seen that the time variables are sine and cosine functions. The two functions, sin(ωdt + φ) and cos(ωdt + φ + θ), are trigonometric functions with a phase difference of 90° + θ.

In many civil engineering structures, the damping ratio is a rather small number, therefore,

θ≈tan ( )⁻¹ ξ ≈ξ (1.38)

Furthermore, the velocity and displacement will have a phase shift θ close to 90°. That is, since the damping ratio is a small number, in the case of free-decay vibration, the velocity and the dis-placement will have a nearly 90° phase difference, and the smaller the damping is, the closer to 90°

the phase difference will be. Note that the above discussion is only for linear systems. This conclu-sion can be visualized by using the examples of free vibration decay shown in Figure 1.5. In Figure 1.5a, the damping ratio is taken to be 0.01 and in Figure 1.5b, the damping ratio is 0.3. Under initial conditions of d0 = 0.1 (m) and v0 = 0 (m/s), the velocity (shown in broken lines) of the system with a lower damping ratio is close to 90° ahead of the corresponding displacement (shown in solid lines).

Meanwhile, for the more heavily damped system, velocity noticeably leads displacement by < 90°.

Example 1.4

Suppose the free-decay peak responses of an SDOF system measured at the second cycle and the tenth cycle are, respectively, 20 and 0.1 (mm), whereas the damped natural frequency is 1 (Hz).

Find the damping ratio.

Denote the peak response at the second cycle and the tenth cycle to be x₂ and x₁₀, respectively.

Also, denote the corresponding time points to be t₂ and t₁₀. From Equation 1.30,

^{x2=Ae−ξωnt2}sin

(

^ωd^t2+ϕ

)

⁼²⁰

(

m m

)

and

^{x10=Ae−ξωnt10}sin

(

^ωd^t10+ϕ

)

^{=0 1}.

(

m m

)

Thus, the ratio of x2 to x10 is further taken to be

_{x x2 10}=_e^{ξωn(t t}10 2^{− )}=200

Therefore

^ln

(

^{x x2 10}

)

^=ξωn

(

^t10−t2

)

^=ln

(

^{200 5 298}

)

^{= .}

Hence,

^{ξ= ω ω}

( )

(

−

)

⁼

(

⁻

)

lnx x .

t t t t

2 10

10 2 10 2

5 298

n n

Note that from t₂ to t₁₀, there are eight complete cycles, with each cycle occupying the duration of one period 2π/ωd. Therefore,

ω π

ξ

π

n t10 t2 2 ξ2

2 10 2 1

16

− 1

( )

⁼

(

⁻

)

− =

− The damping ratio is calculated to be 10.5%.

Free-decay response (damping ratio = 0.01)

0.1 0 20

(a)

(b) 15 10 5 0 –5

Displacement (dm) Velocity (m/sec) Displacement (dm) Velocity (m/sec)

–10 –15

–20 0.2 0.3 0.4 0.5

Time (sec)

0.6 0.7 0.8 0.9 1

0.1

0 0.2 0.3 0.4 0.5

Time (sec)0.6 0.7 0.8 0.9 1 Free-decay response (damping ratio = 0.30)

5

0

–5

–10

–15 10

FIGURE 1.5  Free-decay velocities and displacements (a) ξ = 0.01 and (b) ξ = 0.30.

Example 1.5

Consider two cases of an SDOF system with different initial conditions. The first case has a zero initial velocity (v₀ = 0), while in the second case, zero displacement (d0 = 0) is imposed. Find the relationship between the damping ratio and amplitude A as well as the peak values of the cor-responding displacements for these two cases.

The first case when the initial displacement is d₀ is examined. From Equation 1.33,

^{A v d d}

Therefore, the amplitude of displacement A is approximately proportional to the term 1 + ξ²/2.

In other words, the larger the damping ratio, the larger the resulting amplitude A is in this case, despite the common sense notion that larger damping will always result in decreased response amplitude.

Note, however, that a larger amplitude A does not mean that the response will have a larger value, since the free-decay response due to initial displacement only will never be larger than the initial displacement. This can be proven by taking the derivative of the displacement described in Equation 1.38 while assigning v₀ = 0 and solving for time t. In fact, the left side of Equation 1.36 can be zero to locate the extreme value; that is,

^{A eωn −ξωntcos}

(

^{ωnt+ +ϕ θ}

)

⁼⁰

Therefore, t must be zero in order to obtain the extreme value, which is understood to be the peak value of the displacement, denoted by x_max. Additionally, when t = 0, the displacement is nothing but the initial displacement d0. That is,

^{xm ax=d0}

In other words, with zero initial velocity, the peak value of the displacement is always the initial displacement, independent of the value of the damping ratio. However, the damping ratio affects the amplitude A, and larger damping yields larger amplitude.

The second case, when the initial velocity v₀ is nonzero, is examined next. From Equation 1.33,

^{A v d d}

and from Equations 1.34 and 1.35 with positive v₀,

^{ϕ =0}

Thus, from Equation 1.30, the displacement is given by x t v e

( )

⁼_ω^{0 −ξωt}

( )

^ω t

d ⁿsin d

To find the peak value of the displacement, the derivative of the above equation is taken and

This equation yields time t when the displacement reaches peak value, which is

^t=

and with a small damping ratio the corresponding peak value of the displacement, xmax, is

From the above equation, the peak displacement of the SDOF system with initial velocity only is approximately inversely proportional to its natural frequency ωn and directly proportional to the amplitude of the initial velocity and the term e^{– ξπ/2}. That is, the larger the damping of the system, the smaller the peak response.

1.1.3  forcEd ViBration witH HarMonic Excitation

When the external input to the system continues to be applied, the system will vibrate in a forced vibration mode. The case when the linear system defined by Equation 1.1 is excited with harmonic forcing functions is discussed in the following subsections.

In document UNIVERSIDAD AUTÓNOMA DE BAJA CALIFORNIA FACULTAD DE CONTADURÍA Y ADMINISTRACIÓN (página 62-72)