Vibration and Shock Handbook 05

Vibration and Shock Handbook 05 Every so often, a reference book appears that stands apart from all others, destined to become the definitive work in its field. The Vibration and Shock Handbook is just such a reference. From its ambitious scope to its impressive list of contributors, this handbook delivers all of the techniques, tools, instrumentation, and data needed to model, analyze, monitor, modify, and control vibration, shock, noise, and acoustics. Providing convenient, thorough, up-to-date, and authoritative coverage, the editor summarizes important and complex concepts and results into “snapshot” windows to make quick access to this critical information even easier. The Handbook’s nine sections encompass: fundamentals and analytical techniques; computer techniques, tools, and signal analysis; shock and vibration methodologies; instrumentation and testing; vibration suppression, damping, and control; monitoring and diagnosis; seismic vibration and related regulatory issues; system design, application, and control implementation; and acoustics and noise suppression. The book also features an extensive glossary and convenient cross-referencing, plus references at the end of each chapter. Brimming with illustrations, equations, examples, and case studies, the Vibration and Shock Handbook is the most extensive, practical, and comprehensive reference in the field. It is a must-have for anyone, beginner or expert, who is serious about investigating and controlling vibration and acoustics.

5 Random Vibration 5.1 5.2 Random Vibration Single Degree of Freedom: The Response to Random Loads Formulation Correlations † † Derivation of Equations † Response Response Spectral Density Rutgers University 5-2 5.3 5.4 Response to Two Random Loads 5-7 Multi-Degree-of-Freedom Vibration 5-12 5.5 Multi-Degree-of-Freedom: The Response to Random Loads 5-17 Deterministic Vibration † Solution by Frequency Response Function † Modal Analysis Response due to a Single Random Force † Response to Multiple Random Forces † Impulse-Response Approach Modal Analysis Approach Haym Benaroya 5-1 5.6 † Continuous System Random Vibration 5-29 Transverse Vibration of Beams Vibration † Random Transverse Summary This chapter summarizes the key ideas of linear random vibration This discipline focuses on determining the response statistics of an oscillator or structure to input forces that are definable only in terms of their statistics Typical problems include the following: (1) given the power spectrum of the force, find the power spectrum of the response; (2) given the mean value and variance of the force, find the mean value and variance of the response The methodology is built upon the linear theory of vibration for discrete single- and multi-degree-of-freedom (DoF) systems, and continuous systems The approaches are essentially the direct method and the modal analysis method The direct method may also be called a transfer matrix method (see Chapter 2) Modal analysis (see Chapters and 4) has the same benefit in random vibration as is done in deterministic vibration studies: it can be computationally more efficient A number of examples are given, as are a list of representative references 5.1 Random Vibration The discipline of random vibration of structures was borne of the need to understand how structures respond to dynamic loads that are too complex to model deterministically Examples include aerodynamic loading on aircraft and earthquake loading of structures Essentially, the question that must be answered is: given the statistics (read: uncertainties) of the loading, what are the statistics (read: most likely values with bounds) of the response? Generally, for engineering applications the statistics of greatest concern are the mean, or average value, and the variance, or scatter These concepts are discussed in detail subsequently Suppose that we are aircraft designers currently working on the analysis and design of a wing for a new airplane As engineers, we are very familiar with the mechanics of solids and can size the wing for static loads Also, we have vibration experience and can evaluate the response of the wing to a harmonic 5-1 © 2005 by Taylor & Francis Group, LLC 5-2 Vibration and Shock Handbook or impulsive forcing However, this wing provides Force lift to an airplane flying through a turbulent atmosphere Even though we are not fluid dynamicists, we know that turbulence is a very complicated physical process In fact, the fluid (air) motion is so complicated that probabilistic models are required to model the behavior Here, a plausibly deterministic but very complicated Time dynamic process is taken to be random for the purposes of modeling Wing design requires force FIGURE 5.1 Turbulent force history data resulting from the interaction between fluid and structure Such data can be shown as the time history in Figure 5.1 The challenge is to make sense of such intricate fluctuations The analyst and designer must run scale model tests A wing section is set up in the wind tunnel and representative aerodynamic forces are generated Data on wind forces and structural response are gathered and analyzed With additional data analysis, it is possible to estimate the force magnitudes Estimates of the mean values of these forces can be calculated, as well as of the range of possible forces With these estimates, it is possible to study the behavior of the wing under a variety of realistic loading scenarios using the tools of probability and statistics to model this complex physical problem This text introduces the use of probabilistic information in mechanical systems, primarily structural and dynamic systems These tools are applicable to all the sciences and engineering, even though this text focuses on the mechanical sciences and engineering 5.2 Single Degree of Freedom: The Response to Random Loads Consider the second-order differential equation1 governing the linear motion of an oscillator ỵ 2zvn Xtị _ ỵ v2n Xtị ẳ Ftị Xtị 5:1ị where the input force per unit mass is given by stationary random process FðtÞ and the output displacement by random process XðtÞ: Note that it is customary to use upper case letters XðtÞ to represent random processes, and lower case letters xðtÞ to represent the realizations of a random process The notation here is standard: 2zvn ¼ c=m where c ¼ viscous damping and m ¼ mass; and v2n ¼ k=m where k ¼ stiffness We assume that the reader understands the concepts of impulse response and convolution We would also like to present a priori the results which are derived in the subsequent section; the mean value of the response mX and the spectral density of the output SXX ðvÞ : mX ¼ Hð0ÞmF SXX ðvÞ ¼ lHðvÞl2 SFF ðvÞ " #2 1=v2n lHvịl ẳ p v2 =v2n ị2 þ ð2zv=vn Þ2 These equations tell us that the mean value of the response mx is proportional to the mean value of the force mF , and that the response spectral density SXX ðvÞ is proportional to the force spectral density SFF ðvÞ: For both results, the proportionality constants depend on the structural or system parameters In this instance, the system parameters z and are deterministic and, therefore, so is the governing equation of motion If z and are either random variables or processes then the governing equation is random The case of random parameters is much more complicated because the system itself is random rather than just the forcing We would have to solve the problem for the “many randomly prescribed systems” rather than for just one system with randomly prescribed forces © 2005 by Taylor & Francis Group, LLC Random Vibration 5-3 By previewing the end results, the reader will hopefully be able to better follow the mathematical manipulations which follow 5.2.1 Formulation Consider the linear system defined by Equation 5.1 and assume random process input FðtÞ to be stationary, with mean mF and power spectrum SFF ðvÞ: The stationarity assumption for the forcing means that transient dynamic behavior cannot be directly considered here.2 Thus, the initial loading transients of an earthquake, a wind gust, or an extreme ocean wave cannot be considered as stationary Assuming that the character of the loading does not change, steady-state behavior can be assumed to be statistically stationary 5.2.2 Derivation of Equations Begin with the convolution equation for the deterministic response of a linear oscillator Xtị ẳ gtịFt tÞdt 21 ð5:2Þ where gðtÞ is the impulse response given by gtị ẳ 2zvn t e sin vd t vd and stationary random load per unit mass FðtÞ is applied at t ¼ 21; that is, long before the present time This ensures that the impulse response is stationary Beginning with Equation 5.2, take the expected value of both sides and use the linear property of mathematical expectation to interchange it with the integral: ð1 ð1 gðtÞdt E{XðtÞ} ¼ gðtÞE{Fðt tÞ}dt ¼ E{FðtÞ} 21 ¼ mF ð1 21 21 gtịdt ẳ mF H0ị where the fact that FðtÞ is stationary was utilized in the second and third equations,3 and Hvị is the frequency response function; H0ị ẳ Hvịlvẳ0 : Therefore, we arrive at the rst important result mX ẳ H0ịmF ẳ m v2n F; which shows us that since mF is time independent, then so must be E{xðtÞ}: In order to derive the output spectral density, we must work through the intermediate results involving the correlation function 5.2.3 Response Correlations For a stationary process, the autocorrelation function is given by xtịxt ỵ tịfX xịdx RXX tị ẳ 21 where fX xị is the probability density function of the process We cannot take the Fourier transform of this equation to find the response spectral density because we not know the response density There are, however, clever ways by which stationary solutions can be utilized in nonstationary cases One possibility is to multiply the stationary process by a deterministic time function such that the product is an evolutionary, or nonstationary, process For example, use AðtÞFðtÞ as the forcing function where AðtÞ is a deterministic transient function and FðtÞ is stationary The force is stationary and has a constant mean value © 2005 by Taylor & Francis Group, LLC 5-4 Vibration and Shock Handbook function fX ðxÞ: There are two other equivalent ways to proceed: (i) utilize the ergodic definition of the autocorrelation,4 or (ii) use Equation 5.2, utilizing available information on FðtÞ; as follows First, derive the cross-correlation between FðtÞ and XðtÞ: Multiply both sides of Equation 5.2 by Fðt a1 Þ and take the expected values of both sides: ð1 E{XðtÞFðt a1 ị} ẳ gt1 ịE{Ft t1 ịFt a1 Þ}dt1 21 where E{Fðt t1 ÞFðt a1 Þ} ¼ RFF ðt1 a1 Þ and E{XðtÞFðt a1 Þ} ¼ RXF ða1 Þ; the cross-correlation between loading FðTÞ and response XðtÞ: Thus, ð1 gðt1 ÞRFF ðt1 a1 Þdt1 ð5:3Þ RXF ða1 Þ ¼ 21 and RFF is known from experimental data Next, multiply both sides of Equation 5.2 by Xt ỵ a2 ị and take expected values of both sides: E{Xt ỵ a2 ịXtị} ẳ gt2 ịE{Xt ỵ a2 ịFt t2 ị}dt2 21 RXX a2 ị ẳ 21 gt2 ịRXF t2 ỵ a2 ịdt2 ð5:4Þ Substitute Equation 5.3 into Equation 5.4 to find ð1 gaịgbịRFF t ỵ b aịda db RXX tị ¼ 21 ð5:5Þ 21 which is a double convolution To evaluate the variance, we use s2X ẳ E{Xtị2 } E2 {Xtị} ẳ RXX 0ị ẵH0ịE{Ftị} While this information on the correlation is of interest, a more important result is the response spectral density that we derive in the next section 5.2.4 Response Spectral Density Begin with the Fourier transform relation between power spectrum and correlation function 1 SXX vị ẳ R tịe2ivt dt; 2p 21 XX and substitute Equation 5.5 for RXX tị; with l ẳ t ỵ b a : 1 2ivt ð1 e gðaÞgðbÞRFF ðlÞda db dt 2p 21 21 21 1 1 ẳ gaịe2iva da R lịe2ivl dl gbịeỵivb db Ê 2p 21 FF 21 21 ẳ HvịH p vịSFF vị SXX vị ẳ by denition, where p denotes complex conjugate and therefore SXX vị ẳ lHvịl2 SFF ðvÞ ð5:6Þ This is the fundamental result for random vibration and linear systems theory that allows us to evaluate the output spectral density, given the input spectral density and the system frequency response Rxx tị ẳ limT!1 ỵT xtịxt ỵ tịdt: 2T 2T â 2005 by Taylor & Francis Group, LLC Random Vibration 5-5 It is emphasized here that the derivation of Equation 5.6 made use of the convolution equation that is valid for linear systems and structures Any generalization for nonlinear behavior requires problemspecific approaches Example 5.1 Oscillator Response to White Noise Consider a simple application of the above ideas to an oscillator What is the response of a damped oscillator to a force with white noise density? Solution The governing equation of motion is ỵ 2zvn Xtị _ ỵ v2n Xtị ẳ Ftị Xtị where FðtÞ is the external force per unit mass, the system transfer function is given by Hvị ẳ v2n ỵ i2zvn v ỵ ivị2 and the squared magnitude of Hvị is given by lHvịl2 ẳ v2n v ị þ ð2zvn vÞ2 2 Therefore, given any input spectral density SFF ðvÞ; the response spectral density is SXX ðvÞ ¼ lHðvÞl2 SFF ðvÞ ¼ ðv2n SFF ðvÞ v ị2 ỵ 2zv n vị Suppose, for mathematical simplicity, that the forcing is white noise, SFF vị ẳ S0 : Then, SXX vị ẳ S0 v2n v2 ị2 ỵ 2zvn vị2 5:7ị and the mean-square response is given by pm2 S0 pS0 E{Xtị2 } ẳ SXX vịdv ẳ ẳ kc 2v3n z 21 The mean-square response can also be written in terms of a one-sided spectrum pm2 S0 m2 W0 E{Xtị2 } ẳ ẳ kc 4kc where the one-sided density W0 is related to the two-sided density by S0 ¼ pW0 =4: This integral is not standard, but can be found in texts on random vibration.5 Even though infinite mean-square energy is input to the system,6 it responds with finite mean-square energy See Figure 5.2 for plots of the components of Equation 5.7 Only the positive frequency ranges are plotted as they are For example, the integral of this example problem is a specialized version of B0 ỵ ivB1 pA0 B21 ỵ A2 B20 ị dv ẳ A0 A1 A2 21 A0 þ ivA1 v A2 where A0 ¼ v2n ; A1 ¼ 2zvn ; A2 ¼ 2; B0 ¼ 1; B1 ¼ 0: The energy input equals to the area under the spectral density, which for white noise is S0 dv ẳ 1: 21 â 2005 by Taylor & Francis Group, LLC 5-6 Vibration and Shock Handbook symmetric about the abscissae White noise is useful and frequently used, even though it is nonphysical, because it leads to good approximate results SFF Input Spectrum So ω Example 5.2 Response to Colored Noise Suppose the same system as in the last example is subjected to more complex loading, where the spectral density of the forcing is not a constant, but a function of v: How would the above analysis change? |H(iω)|2 System function ω Solution ωn The output spectral density becomes a more complex function of frequency, for example, if SXX the loading density is similar to those found for Response spectrum wind loads Then, the mean-square response must be evaluated numerically The applied problems are always solved numerically, although hopefully after some significant ω analytical exposition There are numerical ωn methods that are specifically geared to handling uncertainties Of particular note is the group FIGURE 5.2 SFF ðvÞ; lHðivÞl2 ; SXX ðvÞ: known as Monte Carlo methods These methods utilize the massive computational power available today to account for uncertainties This is accomplished by generating random numbers that are used to represent random parameters For each of these generated random values, the program recalculates the problem After running enough cycles to ensure the convergence of the statistics, these numerical realizations are averaged to find the mean value and variance of the relevant parameters We summarize the key results for the random vibration of a single-DoF linear oscillator in Table 5.1 to Table 5.3 Figure 5.3 shows some of the most important correlation and spectral density pairs TABLE 5.1 Mean-Value Response mX ẳ H0ịmF lHvịl2 ẳ 1=v2n q 2 2 v =vn ị ỵ ð2zv=vn Þ2 32 mF is known from force data TABLE 5.2 RXX tị ẳ s2X Output Correlation Function/Variance é1 é1 21 21 gaịgbịRFF t ỵ b aịda db ẳ RXX 0ị ẵH0ịE{Ftị} RFF t ỵ b aị is known from force data â 2005 by Taylor & Francis Group, LLC Random Vibration 5-7 TABLE 5.3 Output Spectral Density SXX vị ẳ lHvịl2 SFF vị lHðvÞl2 from structural parameters SFF ðvÞ is the input force power spectrum FIGURE 5.3 5.3 Correlation functions and corresponding power spectral densities Response to Two Random Loads Previously, the system responses were due to a single randomly varying force In general, the situation is more complicated, because more than one load may act on a system and the resulting response depends not only on the properties of each force, but also on the correlation between the two forces © 2005 by Taylor & Francis Group, LLC 5-8 Vibration and Shock Handbook Consider the response of the system to two Q(t) P(t) random loadings, PðtÞ and QðtÞ; acting simultaneously but at different points on the system, as X(t) shown in Figure 5.4 We are interested in calculating the response statistics of the displacement XðtÞ at an arbitrary point on the system Assume that E{Ptị} ẳ and E{Qtị} ẳ 0: Also, by utilizing available data, we are able to estimate FIGURE 5.4 Response to two random loads RPP ðtÞ and RQQ ðtÞ: Our interest is in evaluating RXX ðtÞ and its Fourier transform SXX ðvÞ: Using linear superposition and the convolution integral, the response due to both forces is given by gXP t1 ịPt t1 ị ỵ gXQ t1 ÞQðt t1 Þ dt1 XðtÞ ¼ 21 Similarly, for Xt ỵ tị : Xt ỵ tị ẳ 21 gXP t2 ịPt ỵ t t2 ị ỵ gXQ t2 ịQt ỵ t t2 ị dt2 where gXP is the impulse response function at coordinate X due to force P and gXQ is the impulse response function at X due to Q: Then, ð1 gXP ðt1 ÞPðt t1 ị ỵ gXQ t1 ịQt t1 ị dt1 RXX tị ẳ E{XtịXt ỵ tị} ẳ E 21 21 gXP t2 ịPt ỵ t t2 ị ỵ gXQ t2 ịQt ỵ t t2 ị dt2 Now expand the product, and then move the expectation operator to the random processes, as follows: ð1 ð1 gXP ðt2 ÞE{Pðt t1 ịPt ỵ t t2 ị}dt2 dt1 gXP t1 ị RXX tị ẳ 21 21 ỵ ỵ ỵ ð1 gXP ðt1 Þ 21 ð1 gXQ ðt1 Þ 21 ð1 gXQ ðt1 Þ 21 ð1 gXQ ðt2 ÞE{Pðt t1 ịQt ỵ t t2 ị}dt2 dt1 21 gXP t2 ịE{Qt t1 ịPt ỵ t t2 Þ}dt2 dt1 21 ð1 gXQ ðt2 ÞE{Qðt t1 ÞQðt þ t t2 Þ}dt2 dt1 21 In this expression E{Pt t1 ịPt ỵ t t2 ị} ẳ RPP t ỵ t1 t2 ị E{Qt t1 ịQt ỵ t t2 ị} ẳ RQQ t ỵ t1 t2 Þ The expectations in the second and third terms are cross-correlations of the form RPQ tị ẳ E{PtịQt ỵ tị}: Therefore, the autocorrelation of the response becomes RXX tị ẳ gXP t1 ị 21 ỵ þ þ © 2005 by Taylor & Francis Group, LLC 21 ð1 21 ð1 21 ð1 gXP ðt1 Þ gXQ t1 ị gXQ t1 ị gXP t2 ịRPP t ỵ t1 t2 Þdt2 dt1 21 ð1 21 ð1 21 21 gXQ t2 ịRPQ t ỵ t1 t2 ịdt2 dt1 gXP t2 ịRQP t ỵ t1 t2 ịdt2 dt1 gXQ t2 ịRQQ t ỵ t1 t2 Þdt2 dt1 Random Vibration 5-9 The importance of this result is primarily in the observation that we cannot derive RXX ðtÞ unless the cross-correlations RPQ ðtÞ and RQP ðtÞ are also known Using the Fourier transform relation between RXX ðtÞ and SXX ðvÞ; that is ð1 R ðtÞe2ivt dt SXX vị ẳ 2p 21 XX we obtain p p vịHXP vịSPP vị ỵ HXP vịHXQ vịSPQ vị SXX vị ẳ HXP p p ỵ HXQ vịHXP vịSQP vị ỵ HXQ ðvÞHXQ ðvÞSQQ ðvÞ ð5:8Þ where p HXP ðvÞHXP ðvÞ ¼ lHXP ðvÞl2 p HXQ ðvÞHXQ ðvÞ ¼ lHXQ ðvÞl2 and SPQ vị ẳ 1 R tịe2ivt dt 2p 21 PQ As expected, the evaluation of the output spectral density requires knowledge about the cross-spectra SPQ ðvÞ and SQP ðvÞ: If there are more than two forces, then we will have additional cross-spectra between each pair of forces Examining Equation 5.8 closely, we find that we can write SXX ðvÞ as " #" # h i SPP ðvÞ SPQ vị HXP vị p p SXX vị ẳ HXP vị HXQ ðvÞ ð5:9Þ SQP ðvÞ SQQ ðvÞ HXQ ðvÞ Example 5.3 Conjugates of Cross Spectra It was briefly mentioned that RPQ tị ẳ RQP 2tị: How are SPQ vị and SQP ðvÞ related? Solution By the definition of spectral density SPQ vị ẳ Replacing RPQ tị with RQP 2tị SPQ vị ẳ 1 R tịexp2ivtịdt 2p 21 PQ 1 R 2tịexp2ivtịdt 2p 21 QP Letting 2t ẳ t SPQ vị ẳ 1 R tịexpivtịdt 2p 21 PQ Then, SPQ vị ẳ SQP 2vị ẳ SpQP vị Example 5.4 Response Spectrum due to Two Random Loads Consider a mass– spring–damper system in Figure 5.5 subject to two random forces PðtÞ and QðtÞ: Find the response spectrum SXX ðvÞ assuming that SPP vị ẳ SP ; â 2005 by Taylor & Francis Group, LLC SPQ vị ẳ 0; SQQ vị ẳ SQ 5-10 Vibration and Shock Handbook Solution The equation of motion for this system is given by mX€ ỵ cX_ ỵ kX ẳ Ptị ỵ Qtị X(t) k First, assume that Qtị ẳ in order to rst obtain HXP ðvÞ: Taking the Fourier transform, the equation of motion is given by Q(t) m c P(t) 2mv2 ỵ civ ỵ kịXvị ẳ Pvị Then, the frequency response function HXP vị is HXP vị ẳ 2mv ỵ civ ỵ kị FIGURE 5.5 Single-DoF system subjected to two random loads or m2v2 ỵ 2vn ziv ỵ v2n Þ HXP ðvÞ ¼ Similarly, HXQ ðvÞ is obtained by setting Ptị ẳ and is also given by HXQ vị ẳ m2v ỵ 2vn ziv ỵ v2n Þ Then, the spectral density of the response is given by SXX vị ẳ h m v2n 2 v 2vn ziv 21 mv2n v2 ỵ 2vn zivÞÞ21 m v2n 2 v 2vn ziv 21 i SPP ðvÞ SQP ðvÞ SPQ vị SQQ vị ẳ SPP vị ỵ SPQ vị ỵ SQP vị ỵ SQQ vị m2 v2n v2 ị2 ỵ 2vn zị2 mv2n v2 ỵ 2vn zivÞÞ21 In our case, the spectral density is reduced to SXX vị ẳ SP ỵ SQ m2 v2n 2 v2 ỵ 2vn z Spectral densities are rather complicated expressions to evaluate in general A number of special cases will help better understand the effects of the cross-terms: PðtÞ and QðtÞ arise from independent sources and are, therefore, uncorrelated.7 Then, RPQ tị ẳ 0; RQP tị ẳ and SPQ vị ẳ 0; SQP vị ẳ 0: PðtÞ and QðtÞ are directly correlated; that is, QðtÞ ¼ kPðtÞ where k is a constant PðtÞ and Qtị are exponentially correlated, E{PtịQt ỵ tị} ẳ kPQ exp{2 at} where kPQ is a constant PðtÞ and QðtÞ are correlated in a “simplified” exponential; that is, with a triangular correlation dened by E{PtịQt ỵ tị} ẳ kPQ t=t1 Þ; 2t1 # t # t1 : Independence implies that E{Pt1 ịQt2 ị} ẳ E{Pt1 ị}E{Qt2 ị} They are uncorrelated if CovPt1 ịQt2 ịị ẳ E{Pt1 ịQt2 Þ} E{Pðt1 Þ}E{Qðt2 Þ} ¼ Independent processes are always uncorrelated whereas uncorrelated processes may not be independent © 2005 by Taylor & Francis Group, LLC 5-20 Vibration and Shock Handbook F1(t) X1(t) c1 X2(t) c2 m2 m1 k1 FIGURE 5.9 k2 Two-DoF system excited by a single random force Solution The equation of motion for this system has been obtained as Equation 5.14 with F2 tị ẳ 0: We found previously that the frequency response function for this system is given by " # 2m2 v2 ỵ ivc2 ỵ k2 ivc2 ỵ k2 ẵHvị ẳ detẵZ ivc2 þ k2 2m1 v2 þ ivðc1 þ c2 Þ þ k1 ỵ k2 ị where detẵZ ẳ v4 m1 m2 iv3 m1 c2 ỵ c1 m2 ỵ c2 m2 ị ỵ ivc1 k2 ỵ k1 c2 ị v2 m1 k2 ỵ c1 c2 ỵ k1 m2 ỵ k2 m2 ị ỵ k1 k2 Since F2 ẳ 0; the frequency response functions due to F2 ; H12 ðvÞ and H22 ðvÞ; are zero The frequency response functions due to F1 are given by H11 vị ẳ 2m2 v2 ỵ ivc2 ỵ k2 detẵZ H21 vị ẳ ivc2 ỵ k2 det½Z Using Equation 5.27, the response spectra are given by p vịH11 vịSF1 F1 vị ẳ S0 SX1 X1 vị ẳ H11 k2 m2 v2 ị2 ỵ vc2 ị2 detẵZ ị2 p vịH22 vịSF1 F1 vị ẳ S0 SX2 X2 vị ẳ H22 k22 ỵ vc2 ị2 detẵZ ị2 p vịH21 vịSF1 F1 vị ẳ S0 SX1 X2 vị ẳ H11 2m2 v2 ivc2 ỵ k2 ịivc2 ỵ k2 ị detẵZ ị2 p vịH11 vịSF1 F1 vị ẳ S0 SX2 X1 vị ẳ H21 2m2 v2 ỵ ivc2 ỵ k2 ị2ivc2 ỵ k2 ị detẵZ ị2 Note that SX2 X1 vị ẳ SX1 X2 2vị ẳ SpX1 X2 ðvÞ: 5.5.2 Response to Multiple Random Forces Consider a multi-DoF system subjected to multiple random forces The goal is to obtain the relation equivalent to Equation 5.27 for the general response for N DoF by first considering the response of a Single-DoF to N forces, and then generalizing to N DoF This will be done in two ways First, by extending the impulse-response method leading to the convolution integral, and then by utilizing the modal © 2005 by Taylor & Francis Group, LLC Random Vibration 5-21 analysis approach where the coupled physical equations of motion are decoupled in a new modal coordinate system 5.5.3 Impulse-Response Approach Let us start by expressing the response Xk ðtÞ using the impulse-response functions gki ðtÞ: Recall that gki ðtÞ is the impulse response of mass mk to force Fi ðtÞ: By linear superposition, the total response Xk ðtÞ of mass mk is equal to the sum of the individual responses to each of the N forces Xk tị ẳ N X iẳ1 Xki tị ¼ N ð1 X i¼1 21 gki ðtÞFi ðt tÞdt Assuming that the forces are stationary with respective means mFi and cross-correlations RFi Fj ðtÞ; the mean and correlation of the response can be found as follows: N ð1 X E{Xk tị} ẳ 21 iẳ1 N X mXk ẳ iẳ1 mFi gki tịE{Fi t tị}dt 21 gki tịdt ẳ N X iẳ1 mFi Hki 0ị In matrix form mX ẳ HXX 0ịmF This is a static response and, therefore, can be ignored here and added on at the end of the computations We next evaluate the expressions for the response correlations and spectral densities, from which we can evaluate the mean-square response By definition, the correlations are given by ( N ) N X X Xkm ðtÞ Xjn t ỵ tị RXk Xj tị ẳ E{Xk tịXj t ỵ tị} ẳ E ( ẳE mẳ1 N X mẳ1 ẳ gkm z ịFm t z ịdz N X N ð1 ð1 X m¼1 n¼1 ¼ 21 21 21 N X N ð1 ð1 X m¼1 n¼1 21 21 n¼1 N ð1 X n¼1 21 ) gjn j ịFn t ỵ t j ịdj gkm z ịgjn j ịE{Fm t z ịFn t ỵ t j Þ}dz dj gkm ðz Þgjn ðj ÞRFm Fn t j ỵ z ịdz dj The response spectral density is by denition SXk Xj vị ẳ 1 R ðtÞexpð2ivtÞdt 2p 21 Xk Xj Before substituting the correlation function into this equation, multiply it by expð2iv½z j ị Ê expivẵz j ị: Also, dene n ẳ t j ỵ z with dn ẳ dt: All of these manipulations allow us to put the spectral density in the following form: SXk Xj vị ẳ ẳ N X N ð1 X m¼1 n¼1 N X N X mẳ1 nẳ1 21 gkm zịexpivzịdz p vịHjn vịSFm Fn vị Hkm © 2005 by Taylor & Francis Group, LLC ð1 21 gjn ðjÞexpð2ivjÞdj £ ð1 ðnÞexpð2ivnÞdn R 2p 21 Fm Fn 5-22 Vibration and Shock Handbook where the star denotes complex conjugate In matrix form SXk Xj vị ẳ Hpk ðvÞSFF ðvÞHTj ðvÞ ð5:28Þ In this notation, Hpk ðvÞ is a row vector of dimension £ N p Hpk vị ẳ Hk1 vị ã ã ã HjT vị is a column vector of dimension N £ 6 HTj vị ẳ 6 and SFF ðvÞ is a matrix of dimension N £ N 6 SFF vị ẳ 6 p vị HkN Hj1 ðvÞ HjN ðvÞ 7 7 SF1 F1 ðvÞ ··· SF1 FN ðvÞ SFN F1 ðvÞ · · · SFN FN ðvÞ 7 7 Equation 5.28 can now be generalized for any Xk and Xj SXX vị ẳ Hp vịSFF vịHT vị where 6 Hvị ẳ and 6 SXX vị ẳ 6 H11 vị ããã H1N vị HN1 ðvÞ ··· ð5:29Þ 7 HNN ðvÞ SX1 X1 ðvÞ ··· SX1 XN ðvÞ SXN X1 ðvÞ · · · SXN XN ðvÞ 7 7 Let us recall Equation 5.6, Equation 5.9, and Equation 5.27 Equation 5.6 is the response spectral density of a single-DoF system to a single random force, Equation 5.9 is the response spectral density of a singleDoF system to two random forces, and Equation 5.27 is the response spectral densities of a two-DoF system due to a single force Comparing them with Equation 5.29, we realize that they are all special cases of this equation For a zero mean, the mean-square response of the jth coordinate is given by the relation SXj Xj vịdv s2Xj ẳ 5:30ị 21 Example 5.6 Spacecraft Design Problem (based on Wirsching et al., p 228) A rocket mounting is to be designed The rocket of mass m1 ¼ 181 kg is mounted on a supporting structure of mass m2 ¼ 362 kg using structural brackets that can be modeled as springs of constant k ¼ 1:40 MN/m There is also damping with coefficient c ¼ 3:50 kNsec/m When the engine fires, the thrust can be interpreted as being composed of a mean value thrust and a random fluctuating thrust © 2005 by Taylor & Francis Group, LLC Random Vibration 5-23 The random variations are due to factors such as fluctuations in fuel mix and eccentricities in thrust direction The mean thrust is given by mT ¼ 16 kN and the random component is given by T, which is a random function of time T is characterized as white noise of intensity 1980 N2/Hz over the frequency range –50 Hz The mean thrust can be viewed as a static load while the random thrust as the dynamic load The design must consider that components in the supporting structure will be damaged if the acceleration is too large Also, the structural brackets will fail if the relative displacement between the rocket and the supporting structure is too large Both these design issues are difficult because the input is random Therefore, there is a trade-off between the cost of making a component stronger and making it more reliable The probabilistic framework allows the analyst to quantitatively understand the trade-off Solution The goal of the analysis is to determine the acceleration of the supporting mass X€ and the relative displacement Z ¼ X2 X1 : First we solve the mean value problem The mean value acceleration is a¼ F 16;000 29:46 ¼ ¼ 29:46 m=sec2 ẳ ẳ 3:00g m1 ỵ m2 181 ỵ 362 9:81 The mean force in the bracket is found by applying the equation of motion to m2 mFb ¼ m2 a2 ¼ 362 £ 29:46 ¼ 10665 N ¼ 10:66 kN The mean bracket deflection is mZ ¼ mFb 10:66 £ 103 ¼ ¼ 7:61 £ 1023 m ¼ 7:61 mm k 1:40 £ 106 While the mean value is the single most important descriptor of randomness, the variance is important because it is a measure of the scatter of possible results In order to calculate the variance, we need to derive the response spectral density and the frequency response function (see Equation 5.30) This we compute next The equations of motion are given by m1 X€ kðX2 X1 Þ cðX_ 2 X_ ị ẳ Ftị m2 X ỵ kX2 X1 ị ỵ cX_ 2 X_ ị ẳ Given the problem statement, the variables of interest are ðZ; X2 ị: Transform variables using Z ẳ X2 X1 : The equations of motion become m1 X€ 2 m1 Z kZ cZ_ ẳ Ftị m2 X ỵ kZ ỵ cZ_ ẳ To derive the frequency response function, assume harmonic excitation " # F0 i vt Ftị ẳ e and harmonic response of the form " Ztị X # " ẳ Z0 X # eivt Substitute these equations into the matrix equation of motion to nd m1 X ỵ m1 v2 Z0 kZ0 icvZ0 ¼ F0 m2 X€ ỵ kZ0 ỵ icvZ0 ẳ â 2005 by Taylor & Francis Group, LLC 5-24 Vibration and Shock Handbook The frequency response function is then found to be " # m2 2m1 Hvị ẳ Dvị 2k icv m1 v2 k icv Dvị ẳ m1 m2 v2 km1 ỵ m2 ị icvm1 ỵ m2 ị There is only one applied force, and the spectral density function matrix is " # S0 SF vị ẳ 0 Apply Equation 5.29 to find " SXX ðvÞ ¼ SZZ SZ X€ SXZ € SX€ X€ # S0 ẳ lDvịl2 " m22 2km2 ỵ icm2 v 2km2 ỵ icm2 v k2 ỵ cvị2 # To determine the root mean square of relative displacement ZðtÞ; note that the spectral density function of ZðtÞ is the top term on the diagonal of SXX vị: SZZ vị ẳ S0 m22 lDvịl2 The mean-square value of Ztị is then s2Z ẳ 21 SZZ vị ẳ pS0 m22 ckm1 ỵ m2 ị2 The above are for a two-sided spectrum The given spectrum is one-sided, therefore, S0 ¼ W0 =4p: Then, with appropriate substitutions, we find sZ 0:212 mm: In summary, the relative displacement has mean and root mean-square mZ ¼ 7:61 mm; sZ ¼ 0:212 mm The force in the spring has mean and root mean-square mFb ¼ kmZ ¼ 1:40 £ 106 £ 7:61 ¼ 10:65 kN sFb ¼ ksZ ¼ 1:40 £ 106 £ 0:212 ¼ 0:297 kN Finally, the spectral density of the acceleration is the second diagonal term of SX vị: SX X ẳ S0 s2X ẳ k2 ỵ cvị2 lDvịl2 21 SX X vịdv and s2X ẳ pS0 k c ỵ cm1 ỵ m2 ị2 m1 ỵ m2 ịm1 m2 Upon making the appropriate substitutions, we find sX€ ¼ 0:852 m=sec2 ¼ 0:087g In summary, the mean and standard deviation of X€ are mX€ ¼ 3g; sX€ ¼ 0:087g See Example 5.1 for a discussion on the above nonstandard integrals A designer would now use these mean and standard deviation numbers to decide how reliable to make the components For example, if we assume that Fb and X€ are normal density, we can specify a design based © 2005 by Taylor & Francis Group, LLC Random Vibration 5-25 fX (x) on m ^ 3s, which tells us the probability of exceeding this 6s range is less than 0.1% If this design is too expensive, then the designer will try a design based on m ^ 3s; one that will have a higher probability of failure In this way, a trade-off study is performed weighing cost against probability of failure Figure 5.10 demonstrates a design based on 5% failure probability 0.95 mX −1.96σX mX FIGURE 5.10 probability 5.5.4 mX + 1.96σX Normal density with 5% failure Modal Analysis Approach The response of a multi-DoF discrete system loaded by random forces is generally a very complicated problem to formulate and solve Here, to demonstrate one possible approach, we study the modal analysis of such a structure and work through some of the intricacies We will start the analysis at the point where the modal equations of motion have been formulated and the assumption of proportional damping has been made From Equation 5.22, for a random vibration, the modal equations in indicial notation are Z i ỵ 2zi vi Z_ i ỵ v2i Zi ¼ Qi ðtÞ; i ¼ 1; 2; …; N ð5:31Þ where the parameters are familiar We further assume that the modal forces Qi ðtÞ are ergodic random excitations The transformation between physical and modal spaces is Xj tị ẳ N X iẳ1 uji Zi tị and the transformation between physical and modal forces is Qj tị ẳ N X iẳ1 where ui is the modal vector for the ith DoF and uij Fi tị ẳ uTj F u1i 5:32ị 7 ui ¼ uNi In matrix form, we can write X ¼ PZ Q ¼ PT F where P is the modal matrix given by P ẳ ẵ u1 ããã uN For a two-DoF system, we can be specific Xj ðtÞ ẳ Qj tị ẳ â 2005 by Taylor & Francis Group, LLC X i¼1 X i¼1 uji Zi tị ẳ uj1 Z1 tị ỵ uj2 Z2 tị; j ¼ 1; ð5:33Þ uij Fi ðtÞ ¼ u1j F1 tị ỵ u2j F2 tị; j ẳ 1; 5-26 Vibration and Shock Handbook for each DoF j of a two-DoF structure Our goal in this analysis is to evaluate the statistics of the response, that is, to find autocorrelations and cross-correlations, RXk Xj ðtÞ; and its Fourier transform, the power spectrum, SXk Xj ðvÞ: Begin with the definition of the autocorrelation and substitute Equation 5.33 for Xj ðtÞ: (n n ) X X ukl ujm Zl tịZm t ỵ tị 5:34ị RXk Xj tị ẳ E{Xj tịXj t ỵ tị} ẳ E lẳ1 mẳ1 where Zi tị is the solution to Equation 5.31: Zi tị ẳ gi tị ẳ ðt Qi ðtÞgi ðt tÞdt ð5:35Þ expð2zi vi tịsin vdi t vdi 5:36ị vdi ẳ vi z2i Þ1=2 ð5:37Þ Note that, since the impulse response function gðtÞ is zero for t , 0; the lower limit on the integral defining ZðtÞ can be made 21 without changing the value of the integral Substituting Equation 5.35 to Equation 5.37 into Equation 5.34 and letting the expectation operate only on the stochastic terms, we have RXk Xj tị ẳ n X n tỵt t X 21 lẳ1 mẳ1 21 ukl ujm E{Ql u1 ịQm u2 ị}gl t u1 ịgm t ỵ t u2 ịdu1 du2 ð5:38Þ where u1 and u2 are dummy time variables, and RQl Qm u2 u1 ị ẳ E{Ql u1 ịQm ðu2 Þ} owing to the assumed ergodicity (and thus stationarity) of the forcing If the system is lightly damped and has well-separated modal frequencies, as is the case in many engineering structures, the response due to Ql ðtÞ is almost statistically independent of the response due to Qm ðtÞ: The cross-correlation terms that arise in Equation 5.38 are then almost zero, with the only nonzero terms arising for m ¼ l : RQl Ql u2 u1 ị ẳ E{Ql ðu1 ÞQl ðu2 Þ} Now that we have the correlation function for the response in terms of the correlation function for the random forcing, we can proceed to evaluate the response spectral density, from which we can evaluate probabilities of occurrence To this, the following transformation of variables is first necessary: u1 ; t u1 ; u2 ; t ỵ t u ; du1 ẳ 2du1 ; du2 ¼ 2du2 resulting in the response correlation RXk Xj tị ẳ n 1 n X X lẳ1 m¼1 21 21 ukl ujm RQl Qm ðu1 u2 þ tÞgl ðu1 Þgm ðu2 Þdu1 du2 The power spectral density for response XðtÞ is equal to the Fourier transform of the correlation function SXk Xj 21vị ẳ ẳ ð1 R ðtÞe2ivt dt 2p 21 Xk Xj (n n X ð1 ð1 ð1 X 2p 21 l¼1 m¼1 © 2005 by Taylor & Francis Group, LLC 0 ) ukl ujm RQl Qm u1 u2 ỵ tịgl ðu1 Þgm ðu2 Þdu1 du2 e2ivt dt Random Vibration 5-27 or, recalling and utilizing the assumption that the processes are ergodic, and, therefore, by averaging in time SXk Xj ðvÞ ¼ n X n X l¼1 m¼1 ukl ujm lim T!1 ðT ðT gl ðu1 Þdu1 lim gm ðu2 Þdu2 T!1 2p 2T 2p 2T ðT RQl Qm u1 u2 ỵ tịe2ivt dt lim T!1 2p 2T where the lower limits on the integrals were set to 2T since gðtÞ is zero for t , 0; and thus the change in lower limit does not affect the values of the integrals Using the change of variables g ; u1 u2 ỵ t; dg ẳ dt we obtain SXk Xj vị ẳ n n X X l¼1 m¼1 lim T!1 ukl ujm lim T!1 ðT g ðu Þexpðivu1 Þdu1 2p 2T l 1 ðT T2u2 ỵu1 RQl Qm gịe2ivg dg gm u2 ịexp2ivu2 ịdu2 lim T!1 2p 2T2u2 ỵu1 2p 2T In the last integral, we make the physical argument that RQj Qm ðgÞ ! as lgl increases and, therefore, the limits can be replaced by 2T and T; respectively.10 Then, Hl 2vị ẳ lim T!1 Hm vị ẳ lim T!1 SQl Qm vị ẳ lim T!1 T g u ịexpivu1 Þdu1 2p 2T l 1 ðT g ðu Þexpð2ivu2 ịdu2 2p 2T m T2u2 ỵu1 gịe2ivg dg R 2p 2T2u2 ỵu1 Ql Qm with the resulting response spectral density SXk Xj vị ẳ n X n X lẳ1 mẳ1 ukl ujm Hl 2vịHm vịSQl Qm vị For a two-DoF system, we would have " p #" #" #" # uj1 SQ1 Q1 ðvÞ SQ1 Q2 ðvÞ H1 vị 0 H1 vị SXk Xj vị ẳ uk1 uk2 p uj2 SQ2 Q1 ðvÞ SQ2 Q2 ðvÞ H2 vị H2 vị ẳ uk1 uj1 H1p vịH1 vịSQ1 Q1 vị ỵ uk1 uj2 H1p vịH2 vịSQ1 Q2 vị ỵ uk2 uj1 H2p vịH1 vịSQ2 Q1 vị ỵ uk2 uj2 H2p ðvÞH2 ðvÞSQ2 Q2 ðvÞ Suppose that we wish to find SXX for a two-DoF system Then " SXX ẳ u11 u12 u21 u22 #" H1p vị 0 H2p ðvÞ 10 #" SQ1 Q1 ðvÞ SQ1 Q2 ðvÞ SQ2 Q1 ðvÞ SQ2 Q2 ðvÞ #" H1 ðvÞ 0 H2 ðvÞ #" u11 u21 u12 u22 # Physically, we are stating that, as time difference (trial mode) increases, there will be an exponentially decaying correlation This is borne out by experiments on physical systems © 2005 by Taylor & Francis Group, LLC 5-28 Vibration and Shock Handbook In general, we can write SXX vị ẳ PHp vịSQQ vịHvịPT where 6 Hvị ẳ H1 vị 0 0 Hi vị ẳ ð5:39Þ 7 HN ðvÞ v2 ỵ i2zi vi v ỵ v2i The spectral densities of the modal forces SQl Qm ðvÞ can be obtained from SFF ðvÞ using Equation 5.32 The autocorrelation RQl Qm ðtÞ is defined as N N N X N