Fundamentos de Vibraciones Mecánicas

Transcription

1 Fundamentos de Vibraciones Mecánicas Vibracion forzada general. Dr. Diego Ledezma Universidad Autónoma de Nuevo León Facultad de Ingeniería Mecánica y Eléctrica

2 Introducción

3 Diferentes tipos de excitación Se aplican los fundamentos de la respuesta a una excitación sinusoidal a distintos tipos de excitación. Periódica F(t) No Periódica m x(t) Sinusoidal k c Aleatoria.40. of Compleja Impacto

4 Principio de superposición y respuesta al impulso unitario

5 Principio de superposición Para sistemas lineales. La respuesta a un conjunto de entradas cuando se aplican a la vez, es igual a la suma de las respuestas de las entradas aplicadas individualmente. X1 Y1 Entradas X Y Salidas X1+X Y1+Y Sistema Lineal Además, si el sistema es lineal, la naturaleza de la respuesta (salida) será igual que la excitación (entrada).

6 Respuesta al impulso unitario Aproximación a un impulso aplicado instantáneamente, es decir un cambio de velocidad instantánea. Modelado usando la función Delta de Dirac. F(t)!(t-a) F(t) 1 " " a t a t f(t) = : t = a 0 : t = a Aplicando el sistema de superposición la respuesta al impulso unitario es útil para encontrar la respuesta de excitaciones arbitrarias.

7 Respuesta al impulso unitario En mecánica el impulso se define como el producto de masa por velocidad, por lo tanto un cambio de velocidad instantánea se define como: o displacem I = m v Entonces las condiciones iniciales son: x(0) = 0 ẋ(0) = 1/m Por lo tanto la respuesta al impulso de un sistema MK es: h(t) = 1 mω n sinω n t

8 Respuesta al impulso unitario x a t h(t) = 1 mω n sinω n t!! Para un sistema amortiguado: h(t) = 1 mω d e ζω nt sinω d t

9 Integral de convolución Método que permite encontrar la respuesta a una excitación arbitraria aplicando la respuesta al impulso y el principio de superposición en sistemas lineales. Amplitude Excitación arbitraria Time Amplitude Time Respuesta a un impulso en el tiempo Suma de la respuestas a pulsos individuales con amplitud escalada definida por la excitación x(t) = t 0 f(τ)h(t τ)dτ Integral de Convolución o Duhamel

10 Excitación compleja y análisis espectral

11 Excitación compleja En sistemas lineales, uso de la descomposición de Fourier y aplicación del principio de superposición Periodic Force Excitation Fourier Series decomposition Force excitation frequency components Solution Periodic Response Time Domain Fourier Series sumation Response frequency components Frequency Domain

12 Análisis Espectral Aplicación de la Transformada de Fourier para la obtención de la respuesta a excitaciones complejas. f(t) = x(t) = Transformada de Fourier de la excitación F (ω)e iωt dω Relación entre entrada, salida y función de respuesta en frecuencia. X(ω) =H(ω)F (ω) Transformada inversa de Fourier de la respuesta H(ω)F (ω)e iωt dω Input f(t) Convolution Response h(t) Output x(t) Time Domain Fourier Transform Inverse Fourier Transform Frequency Response H(ω) Spectrum F(ω) Spectrum X(ω)=H(ω)F(ω) Frequency Domain

13 Random Vibration Very often the excitation does not follows a periodic pattern but a random nature, for example wind loads, an uneven road, etc. In this case the excitation cannot be predicted over time. In order to obtain the response statistical procedures must be used. The Power Spectral Density (PSD) which represents the energy content in a signal per hertz. It is usually regarded as the spectrum of a signal. If one has a random force F(t) the PSD is defined as: 1 S f = lim x T F( ω ) T is the time over F(t) is measured. The PSD states that the random process (time) is decomposed into the frequency domain, i.e. the Fourier Transform of the random proccess

14 Ideal white noise White noise is a random process with a constant PSD over a frequency range. When the frequency band is infinitely wide it is called ideal white noise. F(t) S f (ω) If the spectral density of a is wide compared with the centre frequency of the band, it is termed a wide-band process t If the frequency range is narrow compared with the centre frequency it is termed a narrowband process ω Fig..50. process.

15 Random Vibration: FRF The mean square force is: S x Considering a MKC system subjected to a random force excitation k m F(t) f = S ( f ω )dω The response PSD is: ( ω ) = H ( ω ) S ( f ω ) c x(t) H ( ω ) = where: 1 k ω m + iωc i.e. the FRF of the MKC system Sx(ω) c Damping usually light, sharp resonance peak ω ω

16 Band-limited white noise When there is a finite cut-off in the band of frequencies, i.e. the frequency distribution is approximately constant for a range of frequencies, it is called band limited white noise, or broadband noise. The response PSD has a sharp resonance peak, i.e. it is narrowband. Sx(ω) Broadband in Excitation c ω ω Narrowband out Response Mean square displacement: x = πs f kc ( ω n )

17 Example A single degree of freedom system is subjected to random base displacement excitation with a PSD S x ( ω ). Derive an approximate expression for the mean square value y ( t) assuming that damping is light and S x ( ω ) is broadband and includes the natural frequencies of the system. m y(t) Considering the FRF of the system the response PSD is given k c x(t) S y ( ω ) = H ( ω ) S ( x ω ) Harmonic motion assumed to find the FRF of the system: ( ) = Y X = 1 + iζ H ω 1 ω ω n ω ω n + iζ ω ω n

18 Example The mean square value of amplitude is defined by: Assuming low damping and that the excitation includes the natural frequency of the system: Considering the PSD of the response as: S y ( ) = S y ω y t ( )dω ( ω ) = H ( ω ) S ( x ω ) S x H ω ( ωω) ) ( ) ω) b S x cb ( ω n ) ω ω By approximating the numerator and the PSD of the excitation at their respective values at ω n Then the mean square value is: S y ( ωω) ) cb ω y ( t) = 1 ω ω n 1+ ζ ω ω n + ζ ω ω n S x ( ω )dω y ωc n ( t) = S x ( ω n ) ω 1 ω ω n 1 + ( ζ ) + ζ ω ω n dω

19 Example 0 1 ω ω n By definition we have: 1 d ω + ζ ω ω n ω n = π 4ζ Then in order to approximate to the FRF of the MKC system the term ( ) 1 + ζ is considered negligible Then the value of mean square amplitude is: y ( t) = πω n ζ S x ( ω n ) If the excitation of the base has a constant PSD in the frequency range from 0 to 10 Hz and zero at other frequencies and the mean square value of x is 4 mm. The system has a natural frequency of 4 Hz and damping of 5%. Estimate the mean square value of y S x ( ω ) And considering that: dω = ω n d ω ω n 0 x ( t) = S ( x ω )dω f

20 Example x ( t) = S x ω = 0 π S x S x = x ( t) 40π = 4mm 40π y ( t) = πω n ζ S x y y ( t) = πω n ζ S x ( t) = 8πmm

21 Shock Vibration Short, non-periodic excitation usually having high energy levels. Mathematically represented using pulse functions. Undamped (or lightly damped) MK system subjected to transient excitation. Damping effect is minimal, see later Generic Response Generic Excitation ν ( t) ξ( t) m!!x =!kx + F ( t) ( ) m!!x =!k " # x! u t $ % ( ) m! "!!z + u!! t x = u + z # $ = %kz General form of equation of motion!!! " n + v = # t ( ) Excitation can be base motion, base acceleration, force, torque, etc

22 Excitation-Response Pairs!!! " n + v = # t ( )

23 Forcing functions Ideal pulse forcing functions have a finite duration and can be symmetric or non symmetric. Rectangular Trapezoidal Half sine Versed Sine ( ) = ξ 0 t τ p ξ t 0 τ t ξ( t) = t ξ c τ ξ c t ξ c τ 0 t τ τ τ + τ r τ + τ t r ( ) = ξ p ξ t πt sin τ 0 τ t 0 t τ ξ( t) = ξ p πt 1 cos τ 0 t τ 0 τ t

24 Shock response MAXIMAX: Maximum response at ANY time RELATIVE: Difference between shock amplitude and shock response SHOCK PULSE RESIDUAL. Response after the shock has finished

25 Shock Response Solution Methods Usually the equation of motion is solved in two parts, during the pulse (forced vibration) and after the pulse when the system is on free vibration considering the initial conditions at the end of the shock. The common methods of solution are: Classic Method (Undetermined coefficients). Laplace Transforms. Convolution. Numerical Methods

26 Shock response: Effect of pulse duration Shock duration is relative to natural period, defined by the period ratio: The duration of the pulse has strong influence in the response. For short pulses the shape has little significance and the maximax occurs after the pulse For long pulses the shape is more important. When pulse is very long the response follows the pulse closely.

27 Shock Response Spectra Used to evaluate the severity of impacts, perform shock tests, and select isolators. Several MK systems with different natural period subjected to the same shock pulse. ###% &$ ' $# ###% ($ ' %# ###% )$ ' &# ###% *$ ' '# ###% +$ ' (# ###%,$ ' )# #$ #$ #$ #$ #$ #$!!"" The response of each system is obtained (either maximax, relative or residual)

28 Shock Response Spectra Normalised Amplitude ν ξ p Period Ratio τ / T ###% &$ ' $# ###% ($ ' %# ###% )$ ' &# ###% *$ ' '# ###% +$ ' (# ###%,$ ' )# #$ #$ #$ #$ #$ #$!!""

29 Short Pulses - Impulsive Very short duration compared with natural period, response is smaller than pulse amplitude. Maximax response after pulse. ν Impulsive or Isolatio ξ p τ T < 0.5 τ / T

30 Amplification Response is greater than the pulse i.e. amplified, especially when pulse duration and natural period are similar. Amplification 0.5 τ T ν ξ p τ / T

31 Long pulses, quasistatic Very long duration compared to natural period, response follows closely the shape of the pulse. Quasistatic τ T ν ξ p τ / T

32 SRS for different pulses Rectangular Half sine Versed sine Cycloidal

33 Example: Rectangular Pulse Obtain the response of a SDOF undamped system to a rectangular pulse force excitation. F( t) F 0 t 0 x( t) = F t The integral can be then expressed as: t 0 Several methods can be used. Using Convolution Integral: The Impulse Response Function is: 0sin( ω ( n t τ ))dτ mω n Performing the integral: x( t) = F ( ( )) mω cos ω n t τ n 0 t ( ) = F τ x t x( t) = F t 0 h t ( )h( t τ )dτ ( ) = 1 Evaluating the limits: ( ) mω n sin ω n t 1 cosω n t t 0 mω n ( ) The response during the pulse

34 Example: Rectangular Pulse For the response after the pulse, the limits change x( t) = F x( t) = F ( ) cos ω ( n t τ ) t 0 mω 0 n Residual Response: mω cosω n t t 0 n ( ) cosω n t Which can be further simplified as: x( t) = F sin ω nt 0 mω n using: sinω n t t 0 cos(a - B )- cos(a + B ) = sin A sin B Plots of the response for different pulse durations: τ T = 0.5 τ T = 0.5 τ T = 1

35 $! c Using Laplace transforms! (t ) = 0 # t #" % 'is giv This is the case of a symmetrical pulse. The&forcing function 0 " # t ( 0 # t #" % ) $!! (The t ) =forcing The equation of motion in general function is given'by: & " # t) (0 form is:!cfunction is0given # t #by " %[]: $ The Laplace transform of the forcing!!! c ()! (t ) = & ' " # t) (0 " n 1 The Laplace!transform of the forcing function is(a3) given 1by []: Using a table of Laplace transform # e s! staking V (s) " sv " stransform V + V ( s )of=the # (s) Laplace +v=# t!! n ( ) " (s )of= "the c pairs the transform rectangular s pulse The Laplace transform of the forcing function is is: given by []: 0 0 previous equation ( ) 1 (A3) s! s V ( s ) " sv0 " sv!0 + V ( s ) = # (s) 1 # e!n " stake = "directly is initially at rest and rearranging equation (A3):one can From the previous analysis equation (A4) an c () s s! Considering that the system is in order to obtain the solution. The solution by: 1 # eis given initially at rest and(a1) rearranging: Combining the transformed eq. of " (s ) = " c tem is initially at rest and rearranging equation (A3): motion with the transforms of the pulse! n" (s) (A4) V (the s ) =previous From analysis one can take directly equation (A4) ' s) and & # s +! n 1 e -1 v(t ) = " c! n L $ '! n " (s)! ( ) ( s s + ( s s + ( (A1)Vin( sorder to obtain the solution. The solution is given by: (A4) )= n n )" % From thes previous analysis one can take directly equation (A4) and +! n on of motion(a1) is given by thetoinverse transformation of solution equation is given by: in order obtain the solution. The (s ) for isthisgiven mationof!motion particular is given uation by the case inverse transformation of equation 1 by:-1 & e ' s) #

36 v( t) v(t) = Using Laplace transforms Using the table of Inverse Transforms and the shifting theorem [ 1$ cos( " t) $ ( 1$ cos" (! )] = # t $ c n Simplifying and rearranging the response is given by: ( )! c $% 1 " cos # n t & ' 0 ( t ( ) 6 4 *! c sin # n) - +,. / sin# * t " ) 4 3 $ - & 4 0 n +,. / 1 % ' ) ( t n

37 Referencias. Notas del curso Vibraciones Mecánicas, Ledezma-Ramirez DF, FIME-UANL. Notas del curso Fundamentals of Mechanical Vibrations, Mace B, Institute of Sound and Vibration Research, University of Southampton. Shock and Vibration Handbook, Harris, C. McGraw Hill, 5ta edición. Structural Vibration, Analysis and Damping, C.E. Beards, Arnold, Primera Edición. Mechanical Vibrations, S. Rao, Prentice Hall, 4ta edición.