Harmonic and Periodic Excitations

periodic excitation fourier series and periodic excitation response and periodic excitation vibrations and periodic table excitation energy periodic excitation force definition
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Dr.NaveenBansal,India,Teacher
Published Date:25-10-2017
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5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations 5.1 INTRODUCTION 5.2 RESPONSE TO HARMONIC EXCITATION 5.2.1 Excitation Applied from t  0 5.2.2 Excitation Present for All Time 5.2.3 Response of Undamped System and Resonance 5.2.4 Magnitude and Phase Information 5.3 FREQUENCY-RESPONSE FUNCTION 5.3.1 Introduction 5.3.2 Curve Fitting and Parameter Estimation 5.3.3 Sensitivity to System Parameters and Filter Characteristics 5.3.4 Relationship of the Frequency-Response Function to the Transfer Function 5.3.5 Alternative Forms of the Frequency-Response Function 5.4 SYSTEMS WITH ROTATING UNBALANCED MASS 5.5 SYSTEMS WITH BASE EXCITATION 5.6 ACCELERATION MEASUREMENT: ACCELEROMETER 5.7 VIBRATION ISOLATION 5.8 ENERGY DISSIPATION AND EQUIVALENT DAMPING 5.9 RESPONSE TO EXCITATION WITH HARMONIC COMPONENTS 5.10 INFLUENCE OF NONLINEAR STIFFNESS ON FORCED RESPONSE 5.11 SUMMARY EXERCISES 5.1 INTRODUCTION In Chapter 3, the governing equation of a single degree-of-freedom system was derived. From this equation, the solution for the system subjected to 181 181182 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations forcing and initial conditions was determined in Appendix D. In the absence of forcing, the responses of a vibratory system subjected to different types of initial conditions were studied. In this chapter, we address those situations in which a physical system is subjected to an external force f(t) and the initial displacement and the initial velocity are zero. In particular, we consider the responses to harmonic and other periodic excitations. The notion of frequency-response function is introduced and related to the notion of trans- fer function, which is used for system design. In Figure 5.1, a vibratory system subjected to a forcing f(t) is illustrated along with a conceptual illustration of the system’s input-output relationship. In this chapter, we determine this relationship in the time domain and in the transformed domain for periodic excitations. In Chapter 6, we determine this relationship in the time domain and in transformed domains for arbitrary excitations. Responses of vibratory systems with a rotating unbalanced mass are also studied in this chapter. Physical systems subjected to base motions are ana- lyzed, and in this context, the device called an accelerometer, which is used for acceleration measurements, is introduced. Vibration isolation is examined at length. Energy dissipation for different damping models is discussed and the notion of equivalent viscous damping is introduced. The influence of stiff- ness nonlinearities on the forced response is also treated. Referring to Figure 5.1, we consider the case where f(t) varies periodi- cally. The governing equation of motion is of the form recall Eq. (3.22) 2 f1t2 d x dx 2  2zv v x  (5.1a) n n 2 m dt dt The initial conditions are taken to be zero; that is, x102  0 and x102  0 (5.1b) Noting that the initial conditions are zero for this system, the general solution is given by Eq. (D.17) for 0 z 1; that is, t 1 zvh n x1t2  e sin1vh2f1t h2dh d  mv d 0 t 1 zv1th2 n  e sin1v3t h42f1h2dh (5.2) d  mv d 0 x(t) k x(t) f(t) f(t) m m, k, c c Input System Output FIGURE 5.1 Vibratory system subjected to forcing and conceptual illustration of system.5.2 Response to Harmonic Excitation 183 1 whereh is the variable of integration and 2 v v21 z d n Next, the response of a linear vibratory system subjected to a harmonic excitation is considered. First, the response is studied when the excitation is initiated at time t  0, and then the response is studied when the excitation is present for all time. In this chapter, we shall show how to: • Analyze the responses of single degree-of-freedom systems to harmonic excitations of various durations. • Determine the frequency response and phase response of a single degree- of-freedom system. • Interpret the response of a single degree-of-freedom system for an exci- tation frequency less than, equal to, and greater than the system’s natural frequency. • Determine the system parameters from a measured frequency response. • Analyze the responses of single degree-of-freedom systems with rotating unbalance and with base excitation. • Use accelerometers to measure the responses of single degree-of-freedom systems. • Isolate vibrations of single degree-of-freedom systems. • Analyze the responses of single degree-of-freedom systems to excitations with multiple harmonic frequency components. • Determine energy dissipation and define equivalent damping. 5.2 RESPONSE TO HARMONIC EXCITATION 5.2.1 Excitation Applied from t  0 In this section, responses to sine harmonic and cosine harmonic excitation are considered. It will be shown that although the initial conditions are zero, the fact that the excitation is suddenly applied at t  0 results in a response that consists of a transient portion and a steady-state portion. These transients are typical of situations where a motor is started or where an excitation is inter- mittently turned on and off. In the absence of damping, the response of a vi- bratory system cannot be characterized as having a transient portion and a steady-state portion. Case 1: Sine Harmonic Excitations We first consider the periodic forcing function f1t2  F sin1vt2u1t2 (5.3) o 1 In writing Eqs. (5.2), the following property of convolution integrals was used: t t h1h2f1t h2dh  h1t h2f1h2dh   0 0184 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations 2 where u(t) is the unit step function u1t2  0 t  0 u1t2  1 t  0 (5.4) When Eq. (5.3) is substituted into Eq. (5.2), the result is t zv t n F e o zvh n x1t2  e sin1v3t h42 sin1vh2dh (5.5) d  mv d 0 Further, we introduce the dimensionless time t v t and rewrite Eq. (5.5) as n t zt F e o zj 2 x1t2  e sin121 z3t j42 sin1j2dj (5.6)  2 k21 z 0 where the nondimensional excitation frequency v/v , the nondimen- n sional time variable of integration j vh, and we have used the definition n ofv given by Eq. (4.5). Note that when the excitation frequency is at the d natural frequency—that is,v v or 1. n 3 Solution for forced response After performing the integration, Eq. (5.6) leads to x1t2 3x 1t2  x 1t24u1t2 (5.7) strans sss where the steady-state portion of the response is given by F o x 1t2  H12sin1t u122 sss k 2 2 2 D11 2 12z2 1 1 H12   2 2 2 2D12 211 2 12z2 2z 1 u12  tan (5.8a) 2 1  and the transient portion of the response is given by 2 The unit step function is used as a compact form that simplifies the way that we can express a function like Eq. (5.3), which is non-zero only in a specific interval. Without using the unit step function, Eq. (5.3) would have been written as F1t2  sin1vt2 t  0  0 t  0 3 The MATLAB function int from the Symbolic Toolbox was used.5.2 Response to Harmonic Excitation 185 zt F H12e o 2 x 1t2  sin1t21 z u122 strans t 2 k 21 z 2 2z21 z 1 u12  tan (5.8b) t 2 2 2z 11 2 After a long period of time, which here means after many cycles of forc- ing, the response reduces to F o lim x1t2  x 1t2  x 1t2  H12 sin1t u122 (5.9) ss sss tq k In Eqs. (5.8a), the quantity H() is called the amplitude response and the quantityu() is called the phase response, which provides the phase relative to the forcing f(t). We see that the steady-state portion varies periodically at the nondimensional frequency , the frequency of the applied force f(t), and with an amplitude F H()/k. In addition, the displacement response is de- o layed by an amount u() with respect to the input. The amplitude response H() and phase response u() are plotted in Figure 5.2 for several values ofz. We shall discuss the significance of these quantities in Section 5.3. The transient response x (t) varies periodically with a frequency strans v /v and its amplitude decreases exponentially with time as a function of the d n damping ratio z. In addition, the response is shifted by an amount u () with t respect to the input. Duration of transient response For practical purposes, we define the time durationt beyond which the system can be considered as having reached d steady state; that is, x(t)  x (t) for t t ss d 6 180 160 5 ζ  0.02 140 ζ  0.02 4 120 ζ  0.1 100 Ω 3 ζ  0.3 ζ  0.7 80 Ω ζ  0.5 ζ  0.5 2 60 ζ  0.7 ζ  0.3 40 1 ζ  0.1 20 ζ  0.02 0 0 0 0.5 1 1.5 2 2.5 0 0.5 1 1.5 2 2.5 Ω Ω (b) (a) FIGURE 5.2 Harmonic excitation applied directly to the mass of the system: (a) amplitude response and (b) phase response. H( ) θ ( ) (deg)186 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations To obtain an estimate of this duration, let the envelope of the transient decay to a value d at a nondimensional time t , which is given by the relation d ztd e  d (5.10a) 2 21 z If 0 d0  1, that is, when the amplitude of the transient portion of the dis- placement is much less than the amplitude of the steady-state portion of the displacement, the transient is said to have “died out” and only the steady-state portion of the response remains. Solving for the nondimensional time t ,we d obtain 2 1 d21 z t  lnc d (5.10b) d z  We can also express the nondimensional time t , which corresponds to d the dimensional time t , in terms of the number of periods N at the excitation d d frequency v that it takes for the normalized displacement x(t)/(F H()/k) o to decay to d. Since the period of the excitation t  2p/v, then the period v t  N t  2pN /v, and d d v d v 2pN 2pN n d d t v t   (5.10c) d n d v  Therefore, Eqs. (5.10) lead to 2  d21 z N  lnc d z  1, 0 d0  1 (5.11) d 2pz  Some typical values of N obtained from Eq. (5.11) are provided in Table 5.1. d It is interesting to note from this table that for a given damping factor z, the number of cycles that the transient lasts increases with the excitation fre- quency. As expected, for a given excitation frequency, the duration of the tran- sient portion of the response decreases for an increase in the damping factor. Representative system responses for sine harmonic excitation The three sets of graphs shown in Figure 5.3 give the normalized displacement response x(t)/(F /k) defined by Eq. (5.7) for three values of , and at each value of , o the response is obtained for three different values of z. The first set shown in Figure 5.3a corresponds to an excitation frequency that is less than the natural frequency:  1. The second set, which is shown in Figure 5.3b, corresponds to an excitation frequency that is equal to the natural frequency:  1. The third set, which is shown in Figure 5.3c, corresponds to an TABLE 5.1 ↓Z/→ 0.01 1 2 Some Values of N for d 0.05 0.51 12.5 29.3 d  0.02 0.3 0.09 02.1 04.9 0.7 0.04 00.97 02.32 → x (τ) 2% H(0.2)  1.04 ss 0 N  1.47,  0.05 d 2 2 x (t) 2% → H(0.2)  1.04 ss 0 N  0.37,  0.2 d 2 2 → x (t) 2% H(0.2)  1 ss 0 N  0.12,  0.7 d 2 0 40 80 120 160 200 τ (a) 20 (t) 2% →x H(1)  10 ss 0 N  12.5,  0.05 d 20 →x (t) 2% 5 ss H(1)  2.5 0 N  3.13,  0.2 d 5 →x (t) 2% H(1)  0.714 1 ss 0 1 N  0.966,  0.7 d 0 20 40 60 80 100 t (b) FIGURE 5.3 Normalized response of a system to a suddenly applied sine wave forcing function when the transient envelope parameter d  0.02 and for different values of z: (a)  0.2 and (b)  1.0. x(τ)/(F /k) x(τ)/(F /k) x(τ)/(F /k) x(t)/(F /k) x(t)/(F /k) x(t)/(F /k) o o o o o o188 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations 0.5 H(3)  0.125 0 N  47.9,  0.05 d 0.5 0.5 → x (t) 2% ss H(3)  0.124 0 N  12,  0.2 d 0.5 0.5 → (t) 2% x ss H(3)  0.111 0 N  3.65,  0.7 d 0.5 0 12 24 36 48 60 t (c) FIGURE 5.3 (continued) (c) 3.0. excitation frequency that is greater than the natural frequency: 1. For each of these nine combinations of values, the values of H() and N are given. d As t increases, the transient portion dies out, and the amplitude of the displacement response approaches the magnitude H() (the steady-state value). The response magnitude is within 2% (d  0.02) or less of this steady- state value after N periods or, equivalently, when t t . It is noted that d d when 1 or  1, the system response decays to within d of its steady- state value. When  1, the displacement response increases until it reaches within d of its steady-state value. Furthermore, during the portion of the response where the transient portion is pronounced, the response is not periodic. However, when t has elapsed, each of the responses approaches d periodicity with the period determined by the excitation frequency; that is, t  2p/v. v Case 2: Cosine Harmonic Excitations For completeness, consider the periodic forcing function f(t)  F cos(vt)u(t) (5.12) o After substituting Eq. (5.12) into Eq. (5.2), the result is x(t)/(F /k) x(t)/(F /k) x(t)/(F /k) o o o5.2 Response to Harmonic Excitation 189 t zt F e o zj 2 x1t2  e sin121 z 3t j42 cos1j2dj (5.13)  2 k21 z 0 where the nondimensional frequency  and the variable of integration j are as defined for Case 1. 4 Solution for forced response Performing the integration in Eq. (5.13), we arrive at the displacement response x1t2 3x 1t2  x 1t24u1t2 (5.14) css ctrans where the steady-state portion of the response is given by F o x 1t2  H12cos1t u122 (5.15a) css k and the transient portion of the response is given by zt F H12e o 2 x 1t2  cos1t21 z u 122 ctrans ct 2 k 21 z 2 z11 2 1 u 12  tan (5.15b) ct 2 2 1  1221 z In Eqs. (5.14) and (5.15), the amplitude response H() and the phase u() are given by Eqs. (5.8a), and it is noted that the proper quadrant must be consid- ered for determining u (). Again, as in the case with the sine harmonic ex- ct citation, after a long time (many cycles of forcing) the response settles down to the steady-state form; that is, F o lim x1t2  x 1t2  x 1t2  H12 cos1t u122 ss css tq k The displacement response given by Eq. (5.14) is plotted in Figure 5.4 for three values of , and at each value of , the response is determined for three different values of z. For each of these nine combinations of values, the val- ues of H() and N are given. As in the case of the sine harmonic excitation, d the transient is initially aperiodic in each of the nine time histories, and then the response becomes periodic after the system settles down. Although the shapes of the transient responses are different than those obtained for the sine wave forcing function, the times it takes for the transients to die out are the same as in the corresponding cases. 4 The MATLAB function int from the Symbolic Toolbox was used.2 →x (t) 2% H(0.2)  1.04 ss 0 N  1.47,  0.05 d −2 2 →x (t) 2% H(0.2)  1.04 ss 0 N  0.37,  0.2 d −2 2 → x (t) 2% H(0.2)  1 ss 0 N  0.12,  0.7 d −2 0 40 80 120 160 200 t (a) 20 →x (t) 2% H(1)  10 ss 0 N  12.5,  0.05 d −20 5 →x (t) 2% H(1)  2.5 ss 0 N  3.13,  0.2 d −5 1 H(1)  0.714 → x (t) 2% ss 0 N  0.966,  0.7 d −1 0 20 40 60 80 100 t (b) FIGURE 5.4 Response of a system to a suddenly applied cosine wave forcing function when the transient envelope parameter d  0.02 and for different values of z: (a)  0.2 and (b)  1.0. x(t)/(F /k) x(t)/(F /k) x(t)/(F /k) x(t)/(F /k) x(t)/(F /k) x(t)/(F /k) o o o o o o5.2 Response to Harmonic Excitation 191 H(3)  0.125 0.2 0 0.2 N  47.9,  0.05 d →x (t) 2% ss 0.2 H(3)  0.124 0 0.2 N  12,  0.2 d →x (t) 2% 0.2 H(3)  0.111 ss 0 N  3.65,  0.7 0.2 d 0 12 24 36 48 60 t (c) FIGURE 5.4 (continued) (c) 3.0. EXAMPLE 5.1 Estimation of system damping ratio to tailor transient response A single degree-of-freedom system with a natural frequency of 66.4 rad/s is intermittently cycled on and off. When it is on, it vibrates at 5.8 Hz. What should the damping ratio be in order for the system to decay to within 5% of its steady-state amplitude in 150 ms each time that the forcing is applied? Assuming that the system settles down to the rest state in between the forcing cycles, from Eq. (5.10b), which is applicable when the forcing is turned on, we have that 2 d21 z 1 t v t  lnc d d n d z  2 1 0.0521 z 166.4210.152  lnc d z 12p  5.82/66.4 1 2 9.96  ln30.091121 z4 z 5 Solving numerically , we obtain z  0.244. 5 The MATLAB function fzero was used. x(t)/(F /k) x(t)/(F /k) x(t)/(F /k) o o o192 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations EXAMPLE 5.2 Start up response of a flexibly supported rotating machine When a rotating machine starts from rest, the rotation speed usually increases  e linearly until it reaches its operating speed v at t  t . Then, from s o Figure 5.5, we see that the excitation frequency of the machine can be ex-  s pressed as v1t2 v1t/t23u1t2  u1t  t24 v u1t  t2 (a) e s o o s o Then, for the system shown in Figure 5.6, the forcing on the inertia element t t o of the system is FIGURE 5.5 f1t2  F sin1v1t2t2 o e Excitation frequency ramped up to 2  F sin1v1t /t23u1t2  u1t  t24 v tu1t  t22 (b) o s o o s o its operating frequency v at time t . s o or 2 f1t2  F sin1t /t3u1t2  u1t t24 tu1t t22 (c) o s o o s o F sin( (t)t) o e where  v /v is the ratio of the final rotational speed of the machine to s s n the natural frequency of the system,t v t, and t v t  2pt /T is n o n o o n m proportional to the ratio of the time it takes to reach the operating speed to the period of undamped free oscillation of the system. Then, the system shown in Figure 5.6 is governed by Eq. (3.23), which is c k F o 2 x  2zx  x  sin1t /t3u1t2  u1t t24 tu1t t22 (d) s o o s o k FIGURE 5.6 where the over dot denotes the derivative with respect to the nondimensional Single degree-of-freedom system timet. Because of the form of the argument of the sine function, this equa- subjected to an excitation whose 6 tion has to be solved numerically for x(t)/(F /k). The results are shown in o frequencyv (t) ramps up from zero e Figure 5.7 for z  0.1 and for all combinations of   0.25, 1.0, and 2, and to the operating frequency v . s s t /2p  0.25, 1.0, 2.0. At each value of  , the corresponding steady-state re- o s sponse is given by Eqs. (5.17) and (5.18); that is, H( ) and u( ). s s The results shown in Figure 5.7 have transient characteristics during an initial phase that is followed by a steady-state phase, as seen in Figure 5.3. When the final value of the excitation frequency is lower than the natural fre- quency, the steady-state amplitude is not much different from the maximum amplitude of the transient motions. However, when the final value of the ex- citation frequency is equal to the natural frequency, a build up from the tran- sient motions to the steady-state motions is noticeable. When the final value of the excitation frequency is higher than the excitation frequency, it is seen that the transients decay to the final steady-state motions. 5.2.2 Excitation Present for All Time In the previous section, it was shown that for a harmonic periodic excitation initiated at time t  0, the response of the vibratory system consists of a 6 The MATLAB function ode45 was used.⎫ ⎬ ⎫ ⎬ 5.2 Response to Harmonic Excitation 193   /8,   0.25, H( )  1.07   /8,  1, H( )  5   /8,   2, H( )  0.33 o s s o s s o s s 5 5 5 0 0 0 5 5 5 0 50 100 0 50 100 0 50 100      ,   0.25, H( )  1.07   ,  1, H( )  5   ,   2, H( )  0.33 o s s o s s o s s 5 5 5 0 0 0 5 5 5 0 50 100 0 50 100 0 50 100      8,   0.25, H( )  1.07   8,  1, H( )  5   8,   2, H( )  0.33 o s s o s s o s s 5 5 5 0 0 0 5 5 5 0 50 100 0 50 100 0 50 100    FIGURE 5.7 Response of a single degree-of-freedom system to an excitation whose frequency ramps up from zero to a final nondimensional frequency  . s transient part and a steady-state part. After a nondimensional time t , only the d steady-state part of the response remains. This observation is taken advantage of to characterize linear systems in terms of frequency-response functions and transfer functions. Once a frequency-response function is determined for a linear vibratory system from a harmonic forcing, this frequency-response function can be used to determine the response of a linear vibratory system for any combination of harmonic inputs. In order to proceed in this direction, first, the previously determined results for the steady-state portion of the re- sponse found in Section 5.2.1 are revisited. When the periodic forcing is given by f(t)  F sin(vt) (5.16a) o or equivalently, in terms of the nondimensional time variable t,as f(t)  F sin(t) (5.16b) o that is, the harmonic excitation is present for all time, the associated steady- state portion of the response is given by Eqs. (5.8) and (5.9). Thus, F o x 1t2  H12 sin1t u122 (5.17) ss k ⎭ ⎭ Steady-state Steady-state amplitude phase x()/(F /k) x()/(F /k) x()/(F /k) o o o x()/(F /k) x()/(F /k) x()/(F /k) o o o x()/(F /k) x()/(F /k) x()/(F /k) o o o194 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations where 1 1 H12   2 2 2 2D12 211 2 12z2 2z 1 u12  tan (5.18) 2 1  The steady-state velocity and steady-state acceleration are, respectively, given by dx1t2 Fv o n v 1t2  a b H12 cos1t u122 ss dt k Fv o n a b H12 sin1t u12 p/22 k 2 2 d x1t2 Fv o n 2 2 2 a 1t2  a b  H12 sin1t u122 v x1t2 (5.19) ss n 2 k dt We see that for harmonic oscillations, the magnitude of the acceleration is equal to the square of the excitation frequency times the displacement mag- nitude and the acceleration response is 180° out of phase with the displace- ment response. The magnitude of the velocity is equal to the excitation fre- quency times the magnitude of the displacement and the velocity response is 90° out of phase with the displacement response. EXAMPLE 5.3 Forced response of a damped system Consider the electric motor shown in Figure 5.8a. The output of the motor is connected to two shafts whose opposite ends are fixed. The motor provides a harmonic drive torque directed along the direction of the unit vector k. This torque has a magnitude M  100 N  m and the driving frequency o 2 v is 475 rad/s. The rotary inertia of the electric motor J is 0.020 kg  m , o the torsional stiffness of the shafts are k  2500 N  m/rad and k  t1 t2 3000 N  m/rad, and the overall damping experienced by the rotor can be . .. C J t o J o k k t1 t2 k k t1 t2 kk M cos (t) o (a) (b) FIGURE 5.8 (a) Electric motor driven restrained by two shafts and (b) free-body diagram.5.2 Response to Harmonic Excitation 195 quantified in terms of a torsional damper with the damping coefficient c  t 1.25 N  m  s/rad. We shall determine the form of the steady-state response and the amplitude and the phase of the steady-state response. The governing equation of the motor is derived based on the principle of angular momentum balance. Consider the free-body diagram shown in Figure 5.8b, which includes the inertial moment Jfk. The principle of an- o gular momentum applied to the center of the motor leads to the following governing equation Jf  cf 1k  k 2f  M cos1vt2 (a) o t t1 t2 o Dividing Eq. (a) by the rotary inertia J , we obtain the following equation o whose form is similar to Eq. (5.1) M o 2 f  2zvf vf  cos1vt2 (b) n n J o where the system natural frequency and damping factor are given by, respectively, k  k t1 t2 v  n B J o c t z  (c) 2Jv o n Based on Eq. (5.17), the solution of Eq. (b) for steady-state motion is given by M o f 1t2  H12 cos1t u122 (d) ss k  k t1 t2 and the non- where the nondimensional excitation frequency  v/v n dimensional time t v t. For the given parameter values, the calculations n lead to the following: k  k  5500 N m/rad t1 t2 5500 N m/rad v   524.40 rad/s n 2 B 0.020 kg m 1.25 N m s/rad z   0.06 2 2  0.020 kg m  524.40 rad/s 475 rad/s v    0.91 v 524.40 rad/s n 1 H12   4.91 2 2 2 211  0.912 12  0.06  0.912 2  0.06  0.91 1 u12  tan a b  0.57 rad (e) 2 1  0.91196 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations Hence, from the values provided in Eqs. (e), the steady-state response of the electric motor given by Eq. (d) is written as 100 N m f 1t2   4.91  cos1475t  0.572 ss 5500 N m/rad  0.09 cos1475t  0.572 rad (f) Thus, the amplitude of the harmonic steady-state motion is 0.09 rad at a fre- quency of 475 rad/s, and the phase lag relative to the excitation is 0.57 rad. 5.2.3 Response of Undamped System and Resonance When a linear vibratory system is undamped,z  0, the governing equation given by Eq. (5.1), reduces to 2 f1t2 d x 2 v x  (5.20) n 2 m dt Response When v v ( 1) n For the excitation given by Eq. (5.3), and for zero initial conditions, the re- sponse is determined from Eqs. (5.7) and (5.8) with z  0. Thus, we obtain F o x1t2  5 sin1t2  sin1t26 (5.21a) 2 k11 2 or, equivalently, F o x1t2  5 sin1vt2  sin1v t26 (5.21b) n 2 k11 2 which is valid when the excitation frequency is different from the natural frequency; that is, when  1 (or v v ). It is clear from the form of n Eqs. (5.21) that for an undamped system excited by a sinusoidal forcing at a frequency that is not equal to the natural frequency, the response consists of a frequency component at the excitation frequency v and a frequency compo- nent at the natural frequency v . Unless the ratio of v/v is a rational number, n n the displacement response is not periodic. Response When v v ( 1) n The response of an undamped system when  1, is obtained from Eq. (5.6) withz  0 and  1. Thus, we arrive at t F o x1t2  sin1t j2 sin1j2dj (5.22)  k 0 Evaluating the integral in Eq. (5.22) results in F o x1t2  5 sin1t2 tcos1t26 (5.23) 2k5.2 Response to Harmonic Excitation 197 The displacement response is not periodic, because the amplitude of the sec- ond term in Eq. (5.23) increases as time increases. Resonance and Stability of Response For the case where  1, the response of the undamped system given by Eq. (5.21) always has a finite magnitude, since sin(t) and sin(t) have finite values for all time. Thus, it follows that 0 x1t20 A t 0 (5.24) where A is a positive finite number. Hence, for  1, an undamped system excited by a finite harmonic excitation is stable in the sense of boundedness introduced in Section 4.3. However, this is not true when  1. When  1, the term tcos(t) in Eq. (5.23) grows linearly in amplitude with time and, hence, it becomes unbounded after a long time. This special ratio  1 (v v ) is called a resonance relation; that is, the linear system given by n Eq. (5.20) is said to be in resonance when the excitation frequency is equal to the natural frequency. From a practical standpoint, the question of boundedness is important. Since there is always some amount of damping in a system, it is clear from Eqs. (5.7) to (5.9) that the response remains bounded when excited at the nat- ural frequency; that is, for  1 F o lim x1t2  sin1t p/22 (5.25) tq 2zk The response given by Eq. (5.25) satisfies the boundedness condition given by Eq. (5.24), even though as the damping decreases in magnitude the re- sponse increases in magnitude. EXAMPLE 5.4 Forced response of an undamped system Consider translational motions of a vibratory system with a mass of 100 kg and a stiffness of 100 N/m. When a harmonic forcing of the form F sin(vt) o acts on the mass of the system, where F  1.0 N, we shall determine the re- o sponses of the system and plot them for the following cases: i) v  0.2 rad/s, ii)v  1.0 rad/s, and iii) v  2.0 rad/s. Let the variable x be used to describe the translation of the mass. Then, from Eq. (5.20), F o 2 x v x  sin1vt2 (a) n m The natural frequency of the system is k 100 N/m v    1 rad/s (b) n B m B 100 kg198 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations Hence, for Cases i and iii, the excitation frequency is different from the natu- ral frequency, and for Case ii, the excitation frequency is equal to the natural frequency of the system. In Cases i and iii, the solution for the displacement response is given by Eq. (5.21); that is, F o x1t2  5sin1vt2  sin1v t26 (c) n 2 k11 2 On the other hand, in Case ii, the response is given by Eq. (5.23); that is, F o x1t2  5sin1v t2 v t cos1v t26 (d) n n n 2k For the given values, the responses are as follows: Case i 1 N x1t2  5sin10.2t2 10.2/1.02sin11  t26 2 100 N/m 11 10.2/1.02 2  0.015 sin10.2t2  0.2 sin t6 m (e) Case ii 1 N x1t2  5sin t  t cos t6 2  100 N/m  0.0055sin t  t cos t6 m (f) Case iii 1 N x1t2  5 sin12t2 10.2/1.02 sin1t26 2 100 N/m 11 10.2/1.02 2 0.0035 sin12t2  2sin t6 m (g) The graphs of Eqs. (e), (f), and (g) are provided in the Figure 5.9. From the graphs, it is clear that the response of the undamped system remains bounded when the excitation frequency is away from the natural frequency, as in Cases i and iii. However, in Case ii, the response becomes unbounded after a long period of time, since the amplitude increases with time. 5.2.4 Magnitude and Phase Information In the case of an undamped linear vibratory system, there is a phase shift of 180° in the response as one goes from an excitation frequency that is less than the natural frequency to an excitation frequency that is greater than the natu- ral frequency. This can be discerned from Eq. (5.21) by noting that the change 2 in sign is brought about by the term 1  . For a linear damped vibratory system excited at the resonance frequency  1, the response lags the excitation by 90° as shown by Eq. (5.25) and in5.2 Response to Harmonic Excitation 199 0.02 v  0.2 rad/s 0 0.02 0 50 100 150 200 0.5 v  1.0 rad/s 0 0.5 0 20 40 60 80 100 0.01 v  2.0 rad/s 0 0.01 0 10 20 30 40 50 t (s) FIGURE 5.9 Displacement response of an undamped system subjected to harmonic forcing at three frequencies: i) v  0.2 rad/s; ii) v  1.0 rad/s; and iii) v  2.0 rad/s. Figure 5.2b. This observation is used in experiments to determine if the exci- tation frequency is equal to the undamped natural frequency of the system; that is, v  v . The magnitude of the response is also large at v  v n n ( 1) when z is small. Response Characteristics in Different Excitation Frequency Ranges Additional characteristics of the response of an underdamped, linear vibra- tory system can be determined by examining the steady-state response x (t) ss in the frequency ranges  1 and  1 and at the frequency location  1; that is, in a region considerably below the natural frequency, in a re- gion well above the natural frequency, and at the natural frequency, respec- tively. The examination is performed by studying the values of H() and u() in these three ranges.   1 In this region, Eqs. (5.18) lead to H() → 1 u() → 0 (5.26) and, therefore, from Eq. (5.17), we obtain F 1 o x 1t2  sin1t2  f1t2 (5.27) ss k k x(t) x(t) x(t)200 CHAPTER 5 Single Degree-of-Freedom Systems Subjected to Periodic Excitations Since the only system parameter that determines the displacement response is the stiffness k, we call this region the stiffness-dominated region. In this re- gion, the displacement is in phase with the force. The velocity response is determined from Eq. (5.27) to be v F v F n o n o v 1t2  cos1t2  sin1t p/22 (5.28) ss k k Therefore, as expected, the velocity response leads the displacement response byp/2; that is, the maximum value of the velocity occurs before the maxi- mum value of the displacement. These results are illustrated in Figure 5.10.  1 At this value of , Eqs. (5.18) simplify to 1 H112  2z p u112  (5.29) 2 Therefore, the displacement response given by Eq. (5.17) takes the form F F o o x 1t2  sin1t p/22  sin1t p/22 (5.30) ss cv 2kz n Thus, for a given excitation amplitude F and natural frequency v , the am- o n plitude of the displacement is determined by the damping coefficient c. This 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 0 50 100 150 Ω t FIGURE 5.10 Phase relationships among displacement, velocity, and force in the stiffness-dominated region: 1. — f(t)/F ; x(t)/(F /k); v(t)/(v F /k). o o n o Amplitude