Απόκριση σε Αρμονικές και Περιοδικές Διεγέρσεις. Απόστολος Σ. Παπαγεωργίου

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1 Απόκριση σε Αρμονικές και Περιοδικές Διεγέρσεις

2 VISCOUSLY DAMPED SDOF SYSTEM: RESPONSE TO HARMONIC EXCITATION Μονοβάθμιο Σύστημα με Ιξώδη Απόσβεση: Απόκριση σε Αρμονική Διέγερση HARMONIC VIBRATION OF UNDAMPED SYSTEMS: Equation of Motion: sin Ω Initial Conditions: 0 & 0 Solution: For a particular solution, we try: sin Ω Substitution in the Equation of Motion: Ω sin Ω sin Ω Ω Ω Therefore: cos sin Ω Λύση ομογενούς εξισώσεως NOTE: The complementary solution represents the transient part of the solution, and for the slightest amount of damping resistance it dies out with time. This will become evident when we consider damped harmonic response. The unknown constants & are determined so that satisfies the initial conditions &. The reader is warned against simply assuming &. sin Ω

3 Introducing the initial conditions, we obtain: where: frequency ratio Λόγος συχνοτήτων Transient Vibration: It depends on & It exists even if 0 cos Παροδική ταλάντωση Steady-State Vibration: It may be expressed as follows, where: Therefore: where: sin sin Ω Μόνιμη ταλάντωση sin Ω static deformation sin Ω sin Ω sin Ω deformation (or displacement) response factor Στατική παραμόρφωση Παράγων απόκρισης παραμορφώσεων 0 Ω ω ω Ω phase angle Γωνία φάσης

4 By following a procedure identical to that outlined above, it can be shown that when the excitation is the steady-state response is given by and the total response is given by cos Ω cos Ω cos sin Παροδική ταλάντωση cos Ω Μόνιμη ταλάντωση

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6 RESONANT RESPONSE OF AN UNDAMPED SYSTEM: We assume that the system starts from rest, i.e., 0 0 0and that: 0 sin Ω 0 sin Then: 0 2 sin sin Resonance : lim lim 0 cos sin sin cos The response is periodic with period. The amplitude of the response continues to grow indefinitely. A measure of the rate of growth may be obtained by taking the difference of amplitude at two successive peaks: sin 0,,2, Peaks occur at cos 0

7 RESONANT RESPONSE OF AN UNDAMPED SYSTEM (continued): For general (i.e., not necessarily zero) initial conditions, the response to 0 sin is given by 0 cos 0 sin 0 2 sin cos Then: For large : and 0 sin 0 cos 0 2 sin 2 0 cos sin 2 Notice that is in phase with 0 sin, power input is always positive: sin2

8 RESONANT RESPONSE OF AN UNDAMPED SYSTEM (continued): Alternative derivation of the response: Excitation: 0 sin Ω Initial Conditions: Response: 0 Ω 2 cos Ω Ω sin When the forcing frequency is equal to or very close to the natural frequency of the system, then we proceed as follows: Let: Ω2( = small quantity) We may re-write the response in the following form: 0 Ω 2 Ω 2 sin Ω sin Ω 2 2 sin Ω sin 0 Ω Ω Ω 2 Ω 2 Ω cos sin Ω sin cos sin 2 cos cos sin Evaluating the above equation in the limit, as 0, we have: Ω 2 lim 0 2 In phase-angle form, this expression becomes: where 2 0 cos sin 0 cos 2 & tan Therefore, in the limiting case when, the amplitude of vibration increases indefinitely with time (see FIGURE).

9 The solid curve is a dimensionless plot of equation: lim cos sin whereas the dashed curve is a similar plot of the first term only. It can be seen that after a short time the first term represents a good approximation to total response, as follows: lim cos The curve in the FIGURE shows that the system theoretically attains an infinite amplitude of forced vibration at resonance in the absence of damping, but this amplitude requires infinite time to build up.

10 When the frequency,, of the forcing function, is close to (but not exactly equal to) that of the vibrating system, a phenomenon known as beating may be observed. Equation, 0 2 sin cos cos sin represents this situation, and we may obtain a good approximation to the response by considering a simplified form of the first term, as follows: 0 sin cos 2 Because the quantity is small, the function varies slowly; and its period, equal to, is large. Therefore, the above expression may be recognized as representing vibrations with period and a variable amplitude equal to. This kind of vibration builds up and diminishes in a regular pattern of beats, as indicated in the FIGURE. The period of beating, equal to, increases as (i.e., as ). At resonance, the period of beating becomes infinite.

11 HARMONIC VIBRATION WITH VISCOUS DAMPING: Equation of Motion: 0 sin Ω Initial Conditions: 0 0 & 0 0 Solution: As a particular integral, we try a solution of the form: Ειδική (ή μερική) λύση cos Ω 2 sin Ω Introducing the above trial solution in the Equation of Motion, we obtain: Ειδική (ή μερική) λύση Particular Solution: Ω 2 2 Ω cos Ω 2 Ω 2 Ω 2 sin Ω 0 sin Ω Ω 2 2 Ω Ω 2 Ω sin Ω 2 cos Ω

12 Complete Solution: Πλήρης λύση cos sin sin Ω 2 cos Ω Introducing the initial conditions () &, we obtain:

13 Πλήρης λύση Complete Solution (cont d): Therefore, the total response is expressed as follows: 0 cos 0 0 sin cos sin sin Ω

14 Particular Solution/Integral: Ειδική (ή μερική) λύση sin Ω 2 cos Ω 2 2 The phase-angle form of the particular solution: where sin Ω tan 2 2 Displacement (or Deformation) Response Factor: Παράγων απόκρισης παραμορφώσεων

15 , max,, [NOTE: Observe that for 2 no peak occurs for.]..,, 2 Notice that,, i.e., the maximum steady-state response is achieved at amplitude resonance and not at phase resonance. It must be recognized, however, that the largest deformation peak may occur before the system has reached steady-state.

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18 We examine the rate at which steady-state is attained in the case of phase resonance, i.e.,. Consider the viscously damped SDOF system starting from rest and subjected to 0 sin Ω. The complete response is: 0 2 cos 2 sin cos For lightly damped systems (i.e., ) the term 2 sin is small and ; thus: cos Περιβάλουσα συνάρτηση where: = peak amplitude after cycles of vibration.

19 COMPLEX FREQUENCY RESPONSE FUNCTION Μιγαδική συνάρτηση συχνοτικής απόκρισης SDOF System with Viscous Damping: Equation of Motion: Steady-State Response: Ω Ω The real part of represents the response to the real part of. The imaginary part of represents the response to the imaginary part of. Introducing in the Equation of Motion, we obtain: Ω Ω Ω 2 Ω Ω Ω 2 2 = Complex Frequency-Response Function (the subscript denotes that the function describes the amplitude of the (harmonic) displacement).

20 COMPLEX FREQUENCY RESPONSE FUNCTION (cont d) Μιγαδική συνάρτηση συχνοτικής απόκρισης (συνέχεια) NOTE: Complex frequency-response functions can be similarly derived for other response quantities, quantities, e.g., velocity, acceleration, elastic restoring force, etc. Notice that, being a complex number, may be expressed as follows: where: Ω tan Ω Ω Ω Ω

21 System Linearity: Γραμμικότητα του συστήματος Ω Ω Ω Therefore, the steady-state response to may be expressed as follows: 0 Ω Ω 0 Ω Ω 0 Ω Ω

22 VECTOR AND COMPLEX NUMBER REPRESENTATION OF HARMONIC MOTIONS It is convenient to represent a harmonic motion by means of a rotating vector of constant magnitude at a constant angular velocity. A mathematically convenient way to represent such a rotating vector is using a complex number representation in the complex (Argand) plane as follows: cos sin The displacement of from on the axis is: The displacement of from on the axis is: cos sin

23 If a harmonic displacement is, then: Displacement: cos Velocity: 2 cos 2 Acceleration: cos The representation of displacement, velocity and acceleration by rotating vectors is illustrated in the FIGURE above. Since the given displacement is a cosine function, or along the real axis, the velocity and acceleration must be along the real axis. Hence the real parts of the respective vectors give the physical quantities at a given time. If the given harmonic displacement is a sine function (i.e., ), then the physical quantities at a given time are the imaginary parts of the respective vectors (i.e., projection of vectors on the -axis).

24 Harmonic motions can be added algebraically (and graphically) by means of vector addition. As an example, let us consider two harmonic motions having the same frequency : cos 2 2 cos The complex number/vector representation is: 2 2 The sum of the above two harmonic motions/functions may be expressed as follows: cos 2 sin where: 2 cos 2 2 sin 2 tan 2 sin 2 cos Since the given harmonic motions, &, are along the real axis, their sum is: cos

25 FORCE BALANCE DIAGRAM OF STEADY-STATE RESPONSE 0 Ω Ω where: tan Ω Spring Force: Damping Force: Ω Ω 2 Inertia Force: Ω 2 Ω

26 To compare the magnitudes of,, for the various ranges of vectors by :, divide the magnitudes of the i.e.,, 2, 2, 0 and recall that: (dynamic equilibrium)

27 RESPONSE TO PERIODIC LOADINGS Any periodic loading of period 0 [i.e., 0 ] can be written as a sum of exponentials: where: For real periodic functions : (the * indicates complex conjugate) Συζυγής μειγαδικός Ω 0 Ω Ω 0 0

28 RESPONSE TO PERIODIC LOADINGS (cont d) We know that: Ω Ω Ω where: Ω 2 2 Therefore, the steady-state response to a periodic loading is given by superposition: υπέρθεση Ω 0 Ω Ω 0