Response to Periodic and Non-periodic Loadings. Giacomo Boffi.

Transcription

1 Periodic and Non-periodic Dipartimento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano March 7, 218 Outline Undamped SDOF systems Damped SDOF systems Series Series of the Series Series of the Undamped SDOF systems Damped SDOF systems

2 A periodic loading is characterized by the identity p(t) = p(t + T ) where T is the period of the loading, and ω 1 = 2π T frequency. p is its principal Series Series of the p(t) p(t + T ) T t theorem asserts that periodic loadings can be represented by an infinite series of harmonic functions. E.g., for an antisymmetric periodic loading of period T we have a series composed of antisymmetric harmonic functions p(t) = p( t) = j=1 p j sin jω 1 t = j=1 p j sin ω j t (with ω j = j 2π T ). Series Series of the The steady-state response of a SDOF system for a harmonic loading p j (t) = p j sin ω j t is known; with β j = ω j /ω n the s-s response is: x j,s-s = p j k D(β j, ζ) sin(ω j t θ(β j, ζ)) = a j cos ω j t + b j sin ω j t. The response to an individual harmonic component can be interpreted as a term of another different series, that possibly represents the steady-state response of the dynamic system to p(t). It can be shown that, under very wide assumptions, the infinite series whose terms are the s-s responses to the harmonic components of p(t) is indeed the series representation of the SDOF steady-state response to p(t). Due to the asymptotic behaviour of D(β; ζ) (D goes to zero as β 2 for β 1) it is apparent that a good approximation to the steady-state response can be obtained using a limited number of low-frequency terms. Series Series of the

3 Series Using theorem any practical periodic loading can be expressed as a series of harmonic loading terms. Consider a loading of period, its series is given by p(t) = a + a j cos ω j t + b j sin ω j t, j=1 j=1 ω j = j ω 1 = j 2π, where the harmonic amplitude coefficients have expressions: Series Series of the a = 1 Tp p(t) dt, a j = 2 Tp p(t) cos ω j t dt, b j = 2 Tp p(t) sin ω j t dt, as, by orthogonality, o etc. p(t)cosω j dt = o a j cos 2 ω j t dt = 2 a j, etc Coefficients If p(t) has not an analytical representation and must be measured experimentally or computed numerically, we may assume that it is possible (a) to divide the period in N equal parts t = /N, (b) measure or compute p(t) at a discrete set of instants t 1, t 2,..., t N, with t m = m t, obtaining a discrete set of values p m, m = 1,..., N (note that p = p N by periodicity). Under these asssumptions the, e.g., cosine-wave amplitude coefficients can be approximated using the trapezoidal rule of integration (note that p = p N and a j 2 t N p m cos ω j t m m=1 = 2 N N p m cos(jω 1 m t) = 2 N m=1 N p m cos m=1 jm 2π N. Series Series of the Periodicity The coefficients of the Discrete Series are periodic, with period N. E.g., here it is how we can compute a j+n according to its definition: a j+n = 2 N = 2 N = 2 N = 2 N N p m cos m=1 N p m cos m=1 N m=1 N m=1 2(j + N)mπ N 2(jm + Nm)π N ( ) 2jmπ p m cos N + 2mπ p m cos 2 jm π N = a j Series Series of the

4 Exponential Form The series can also be written in terms of exponentials of imaginary argument, p(t) = P j exp iω j t j= where the complex amplitude coefficients are given by P j = 1 Tp p(t) exp iω j t dt, j =,..., +. For a sampled p m we can write, using the trapezoidal integration rule and substituting t m = m t = m /N, ω j = j 2π/ : Series Series of the P j 1 N N m=1 p m exp( i 2π j m N ). For sampled input also the coefficients of the exponential series are periodic, P j+n = P j. Undamped We have seen that the steady-state response to the jth sine-wave harmonic can be written as [ ] x j = b j 1 k 1 βj 2 sin ω j t, β j = ω j /ω n, analogously, for the jth cosine-wave harmonic, [ ] x j = a j 1 k 1 βj 2 cos ω j t. Finally, we write [ ] x(t) = 1 k a βj 2 (a j cos ω j t + b j sin ω j t). j=1 Series Series of the Damped In the case of a damped, we must substitute the steady state response for both the jth sine- and cosine-wave harmonic, x(t) = a k + 1 k j=1 +(1 β 2 j ) a j 2ζβ j b j (1 β 2 j )2 + (2ζβ j ) 2 cos ω j t+ + 1 k j=1 As usual, the exponential notation is neater, +2ζβ j a j + (1 β 2 j ) b j (1 β 2 j )2 + (2ζβ j ) 2 sin ω j t. Series Series of the x(t) = j= P j k exp iω j t (1 β 2 j ) + i (2ζβ j).

5 Example As an example, consider the loading p(t) = max{p sin 2πt, } p p max[sin(2 π t/ ),.] Series Series of the.5 p..5 T Example a = 1 Tp /2 a j = 2 Tp /2 b j = 2 Tp /2 p o sin 2πt dt = p π, p o sin 2πt p o sin 2πt cos 2πjt dt = sin 2πjt dt = { [ ] for j odd p 2 π for j even, 1 j { 2 p 2 for j = 1 for n > 1. Series Series of the Example cont. Assuming β 1 = 3/4, from p = p ( π 1 + π 2 sin ω 1t 2 3 cos 2ω 1t 2 15 cos 4ω 2t... ) with the dynamic amplifiction factors 1 16 D 1 = 1 (1 3 = 4 )2 7, etc, we have x(t) = p kπ D 2 = 1 1 (2 3 4 )2 = 4 5, D 4 = 1 1 (4 3 4 )2 = 1 8, D 6 =... ( 1 + 8π 7 sin ω 1t cos 2ω 1t + 1 ) 6 cos 4ω 1t +... Take note, these solutions are particular solutions! If your solution has to respect given initial conditions, you must consider also the homogeneous solution. Series Series of the

6 Example cont. x(t) k π / p o x i = Σ j=1,..,i a j cosω j t + b j sinω j t x x 1 x 2 x 4 Series Series of the t/ Outline of transform Extension of Series to non periodic functions in the Frequency Domain Extension of Series to non periodic functions in the Frequency Domain Undamped SDOF systems Damped SDOF systems Non periodic loadings It is possible to extend the analysis to non periodic loading. Let s start from the series representation of the load p(t), + p(t) = P r exp(iω r t), ω r = r ω, ω = 2π, introducing P(iω r ) = P r and substituting, p(t) = 1 + P(iω r ) exp(iω r t) = ω 2π + P(iω r ) exp(iω r t). Due to periodicity, we can modify the extremes of integration in the expression for the complex amplitudes, Extension of Series to non periodic functions in the Frequency Domain P(iω r ) = +Tp /2 /2 p(t) exp( iω r t) dt.

7 Non periodic loadings (2) If the loading period is extended to infinity to represent the non-periodicity of the loading ( ) then (a) the frequency increment becomes infinitesimal ( ω = 2π dω) and (b) the discrete frequency ω r becomes a continuous variable, ω. In the limit, for we can then write p(t) = 1 2π P(iω) = P(iω) exp(iωt) dω p(t) exp( iωt) dt, which are known as the inverse and the direct s, respectively, and are collectively known as the transform pair. Extension of Series to non periodic functions in the Frequency Domain SDOF In analogy to what we have seen for periodic loads, the response of a damped SDOF system can be written in terms of H(iω), the complex frequency response function, x(t) = 1 H(iω) P(iω) exp iωt dt, where 2π H(iω) = 1 [ ] 1 = 1 k (1 β 2 ) + i(2ζβ) k [ ] (1 β 2 ) i(2ζβ), β = ω. (1 β 2 ) 2 + (2ζβ) 2 ω n To obtain the response through frequency, you should evaluate the above integral, but analytical integration is not always possible, and when it is possible, it is usually very difficult, implying contour integration in the complex plane (for an example, see Example E6-3 in Clough Penzien). Extension of Series to non periodic functions in the Frequency Domain Outline of the Discrete The Fast The Fast Undamped SDOF systems Damped SDOF systems

8 Discrete To overcome the analytical difficulties associated with the inverse transform, one can use appropriate numerical methods, leading to good approximations. Consider a loading of finite period, divided into N equal intervals t = /N, and the set of values p s = p(t s ) = p(s t). We can approximate the complex amplitude coefficients with a sum, P r = 1 Tp p(t) exp( iω r t) dt, 1 N t ( ) N 1 t p s exp( iω r t s ) = 1 N 1 N s= that, by trapezoidal rule, is s= p s exp( i 2πrs N ). The Fast Discrete (2) In the last two passages we have used the relations p N = p, exp(iω r t N ) = exp(ir ω ) = exp(ir2π) = exp(i) ω r t s = r ω s t = rs 2π N = 2π rs N. Take note that the discrete function exp( i 2πrs ), defined for integer r, s is N periodic with period N, implying that the complex amplitude coefficients are themselves periodic with period N. P r+n = P r Starting in the time with N distinct complex numbers, p s, we have found that in the frequency our load is described by N distinct complex numbers, P r, so that we can say that our function is described by the same amount of information in both s. The Fast Only N/2 distinct frequencies ( N 1 = +N/2 N/2 ) contribute to the load representation, what if the frequency content of the loading has contributions from frequencies higher than ω N/2? What happens is aliasing, i.e., the upper frequencies contributions are mapped to contributions of lesser frequency sin(21 * (2π)/ * s /N), N=2, s=,..,2 sin(22 * (2π)/ * s /N), N=2, s=,..,2 The Fast 1/4 See the plot above: the contributions from the high frequency sines, when sampled, are indistinguishable from the contributions from lower frequency components, i.e., are aliased to lower frequencies!

9 (2) The maximum frequency that can be described in the DFT is called the Nyquist frequency, ω Ny = 1 2π 2 t. It is usual in signal analysis to remove the signal s higher frequency components preprocessing the signal with a filter or a digital filter. It is worth noting that the resolution of the DFT in the frequency for a given sampling rate is proportional to the number of samples, i.e., to the duration of the sample. The Fast The Fast The operation count in a DFT is in the order of N 2. A Fast is an algorithm that reduces the number of arithmetic operations needed to compute a DFT. The first and simpler FFT algorithm is the Decimation in Time algorithm by Cooley and Tukey (1965). The algorithm introduced by Cooley and Tukey is quite complex because it allows to proceed without additional memory, we will describe a different algorithm, that is based on the same principles but requires additional memory and it s rather simpler than the original one. The Fast Decimation in Time DFT For simplicity, assume that N is even and split the DFT summation in two separate sums, with even and odd indices X r = = N 1 s= N/2 1 q= x s e 2πi N sr, r =,..., N 1 N/2 1 x 2q e 2πi N (2q)r + q= x 2q+1 e 2πi N (2q+1)r. Collecting e 2πi N r in the second term and letting 2q N = q N/2, we have X r = N/2 1 q= x 2q e 2πi N/2 qr + e 2πi N/2 1 N r q= x 2q+1 e 2πi N/2 qr, i.e., we have two DFT s of length N/2. The operations count is just 2(N/2) 2 = N 2 /2, but we have to combine these two halves in the full DFT. The Fast

10 Decimation in Time DFT Say that X r = E r + e 2πi N r O r where E r and O r are the even and odd half-dft s, of which we computed only coefficients from to N/2 1. To get the full sequence we have to note that 1. the E and O DFT s are periodic with period N/2, and 2. exp( 2πi(r + N/2)/N) = e πi exp( 2πir/N) = exp( 2πir/N), so that we can write { E r + exp( 2πir/N)O r X r = E r N/2 exp( 2πir/N)O r N/2 if r < N/2, if r N/2. The Fast The algorithm that was outlined can be applied to the computation of each of the half-dft s when N/2 were even, so that the operation count goes to N 2 /4. If N/4 were even... Pseudocode for CT algorithm def fft2(x, N): if N = 1 then Y = X else Y = fft2(x, N/2) Y1 = fft2(x1, N/2) for k = to N/2-1 Y_k = Y_k + exp(2 pi i k/n) Y1_k Y_(k+N/2) = Y_k - exp(2 pi i k/n) Y1_k endfor endif return Y The Fast from cmath i m p o r t exp, p i d e f d _ f f t ( x, n ) : """ D i r e c t f f t o f x, a l i s t o f n=2 m complex v a l u e s """ r e t u r n f f t ( x, n, [ exp ( 2 p i 1 j k /n ) f o r k i n r a n g e ( n / 2 ) ] ) d e f i _ f f t ( x, n ) : """ I n v e r s e f f t o f x, a l i s t o f n=2 m complex v a l u e s """ t r a n s f o r m = f f t ( x, n, [ exp (+2 p i 1 j k /n ) f o r k i n r a n g e ( n / 2 ) ] ) ] r e t u r n [ x /n f o r x i n t r a n s f o r m ] d e f f f t ( x, n, tw ) : """ D e c i m a t i o n i n Time FFT, t o be c a l l e d by d _ f f t and i _ f f t. x i s t h e s i g n a l t o t r a n s f o r m, a l i s t o f complex v a l u e s n i s i t s l e n g t h, r e s u l t s a r e u n d e f i n e d i f n i s n ot a power o f 2 tw i s a l i s t o f t w i d d l e f a c t o r s, precomputed by t h e c a l l e r r e t u r n s a l i s t o f complex v a l u e s, t o be n o r m a l i z e d i n c a s e o f an i n v e r s e t r a n s f o r m """ i f n == 1 : r e t u r n x # bottom r e a c h e d, DFT o f a l e n g t h 1 v e c x i s x # c a l l f f t w i t h t h e even and t h e odd c o e f f i c i e n t s i n x # t h e r e s u l t s a r e t h e so c a l l e d even and odd DFT s e, o = f f t ( x [ : : 2 ], n /2, tw [ : : 2 ] ), f f t ( x [ 1 : : 2 ], n /2, tw [ : : 2 ] ) The Fast # a s s e m b l e t h e p a r t i a l r e s u l t s : # 1 s t h a l f o f f u l l DFT i s put i n e ven DFT, 2 nd h a l f i n odd DFT f o r k i n r a n g e ( n / 2 ) : e [ k ], o [ k ] = e [ k]+tw [ k ] o [ k ], e [ k] tw [ k ] o [ k ] # c o n c a t e n a t e t h e two h a l v e s o f t h e DFT and r e t u r n t o c a l l e r r e t u r n e + o

11 Dynamic (1) To evaluate the dynamic response of a linear SDOF system in the frequency, use the inverse DFT, N 1 x s = V r exp(i r= 2π rs ), s =, 1,..., N 1 N where V r = H r P r. P r are the discrete complex amplitude coefficients computed using the direct DFT, and H r is the discretization of the complex frequency response function, that for viscous damping is H r = 1 k [ 1 (1 β 2 r ) + i(2ζβ r ) ] = 1 k [ (1 β 2 r ) i(2ζβ r ) (1 β 2 r ) 2 + (2ζβ r ) 2 ], β r = ω r ω n. The Fast while for hysteretic damping it is H r = 1 [ ] 1 k (1 βr 2 = 1 ) + i(2ζ) k [ ] (1 β 2 r ) i(2ζ) (1 βr 2 ) 2 + (2ζ) 2. Dynamic (2) Some word of caution... If you re going to approach the application of the complex frequency response function without proper concern, you re likely to be hurt. Let s say ω = 1., N = 32, ω n = 3.5 and r = 3, what do you think it is the value of β 3? If you are thinking β 3 = 3 ω/ω n = 3/ you re wrong! { r ω r N/2 Due to aliasing, ω r = (r N) ω r > N/2, note that in the upper part of the DFT the coefficients correspond to negative frequencies and, staying within our example, it is β 3 = (3 32) 1/ If N is even, P N/2 is the coefficient corresponding to the Nyquist frequency, if N is odd P N 1 corresponds to the largest positive frequency, 2 while P N+1 corresponds to the largest negative frequency. 2 The Fast Loading Undamped SDOF systems Damped SDOF systems

12 a short duration load An approximate procedure to evaluate the maximum displacement for a short impulse loading is based on the impulse-momentum relationship, m ẋ = t [p(t) kx(t)] dt. When one notes that, for small t, the displacement is of the order of t 2 while the velocity is in the order of t, it is apparent that the kx term may be dropped from the above expression, i.e., m ẋ t p(t) dt. a short duration load Using the previous approximation, the velocity at time t is ẋ(t ) = 1 m t p(t) dt, and considering again a negligibly small displacement at the end of the loading, x(t ), one has x(t t ) 1 t p(t) dt sin ω n (t t ). mω n Please note that the above equation is exact for an infinitesimal impulse loading. dx(t τ) = p(τ) dτ mω n sin ω n (t τ), t > τ, Undamped SDOF For an, the impulse-momentum is exactly p(τ) dτ and the response is dx(t τ) = p(τ) dτ mω n sin ω n (t τ), t > τ, and to evaluate the response at time t one has simply to sum all the infinitesimal contributions for τ < t, x(t) = 1 mω n t p(τ) sin ω n (t τ) dτ, t >. This relation is known as the Duhamel integral, and tacitly depends on initial rest conditions for the system. Jean-Marie Constant Duhamel (Saint-Malo, 5 February 1797 Paris, 29 April 1872)

13 Damped SDOF The derivation of the equation of motion for a generic load is analogous to what we have seen for undamped SDOF, the infinitesimal contribution to the response at time t of the load at time τ is dx(t) = p(τ) mω D dτ sin ω D (t τ) exp( ζω n (t τ)) t τ and integrating all infinitesimal contributions one has x(t) = 1 t p(τ) sin ω D (t τ) exp( ζω n (t τ)) dτ, t. mω D Evaluation, undamped Using the trig identity sin(ω n t ω n τ) = sin ω n t cos ω n τ cos ω n t sin ω n τ the Duhamel integral is rewritten as t x(t) = p(τ) cos ω nτ dτ sin ω n t mω n = A(t) sin ω n t B(t) cos ω n t where { A(t) = 1 t mω n p(τ) cos ω nτ dτ B(t) = 1 t mω n p(τ) sin ω nτ dτ t p(τ) sin ω nτ dτ cos ω n t mω n Undamped SDOF systems Damped SDOF systems Numerical evaluation, undamped Usual numerical procedures can be applied to the evaluation of A and B, e.g., using the trapezoidal rule, one can have, with A n = A(n τ), y n = p(n τ) cos(n τ) and z n = p(n τ) sin(n τ) we can write A n+1 = A n + τ (y n + y n+1 ), 2mω n B n+1 = B n + τ 2mω n (z n + z n+1 ). Undamped SDOF systems Damped SDOF systems

14 Evaluation, damped For a damped system, it can be shown that with x(t) = A(t) sin ω D t B(t) cos ω D t A(t) = exp ζω nt mω D B(t) = exp ζω nt mω D t t p(τ) exp ζω n τ cos ω D τ dτ, p(τ) exp ζω n τ sin ω D τ dτ. Undamped SDOF systems Damped SDOF systems Numerical evaluation, damped Numerically, using e.g. Simpson integration rule and y n = p(n τ) cos ω D τ, A n+2 = A n exp( 2ζω n τ)+ τ 3mω D [y n exp( 2ζω n τ) + 4y n+1 exp( ζω n τ) + y n+2 ] (You can write a similar relationship for B n+2 ) n =, 2, 4, Undamped SDOF systems Damped SDOF systems Transfer Functions The response of a linear SDOF system to arbitrary loading can be evaluated by a convolution integral in the time, x(t) = t p(τ) h(t τ) dτ, with the unit impulse response function h(t) = 1 mω D exp( ζω n t) sin(ω D t), or through the frequency using the integral x(t) = H(ω)P(ω) exp(iωt) dω, where H(ω) is the complex frequency response function.

15 Transfer Functions These response functions, or transfer functions, are connected by the direct and inverse transforms: H(ω) = h(t) = 1 2π h(t) exp( iωt) dt, H(ω) exp(iωt) dω. Relationship of transfer functions We write the response and its transform: x(t) = X (ω) = t p(τ)h(t τ) dτ = [ t t p(τ)h(t τ) dτ ] p(τ)h(t τ) dτ exp( iωt) dt the lower limit of integration in the first equation was changed from to because p(τ) = for τ <, and since h(t τ) = for τ > t, the upper limit of the second integral in the second equation can be changed from t to +, +s +s X (ω) = lim p(τ)h(t τ) exp( iωt) dt dτ s s s Relationship of transfer functions Introducing a new variable θ = t τ we have +s +s τ X (ω) = lim p(τ) exp( iωτ) dτ h(θ) exp( iωθ) dθ s s s τ with lim s τ =, we finally have s X (ω) = = P(ω) p(τ) exp( iωτ) dτ h(θ) exp( iωθ) dθ h(θ) exp( iωθ) dθ where we have recognized that the first integral is the transform of p(t).

16 Relationship of transfer functions Our last relation was X (ω) = P(ω) h(θ) exp( iωθ) dθ but X (ω) = H(ω)P(ω), so that, noting that in the above equation the last integral is just the transform of h(θ), we may conclude that, effectively, H(ω) and h(t) form a transform pair.