C.-H. Lamarque. University of Lyon/ENTPE/LGCB & LTDS UMR CNRS 5513

Transcription

1 Nonlinear Dynamics of Smooth and Non-Smooth Systems with Application to Passive Controls 3rd Sperlonga Summer School on Mechanics and Engineering Sciences on Dynamics, Stability and Control of Flexible Structures Sperlonga (Italy), September 213 C.-H. Lamarque University of Lyon/ENTPE/LGCB & LTDS UMR CNRS 5513

2 1. Numerical simulations and question of identification 2. Identification process: Theoretical examples 3. Identification: Numerical examples

3 Introduction Numerical simulations and question of identification : Responses of some studied examples: Numerical examples, Transient, cycles, responses under loading solicitation Numerics... = "experimental" results Identification: Theoretical examples via modelling Assuming type for non smooth elements: E.g. dry friction, Saint-Venant Theoretical location of "non smooth" points of cycles Identification: Numerical examples For discrete data (numerical, experimental), how to process? Adapted numerica tools

4 Schedule 1. Numerical simulations and question of identification 2. Identification process: Theoretical examples 3. Identification: Numerical examples

5 Description of the mechanical systems We studied problems with the general mathematical formulation: Proposition: Under assumptions, for all ξ D( φ), there exists a unique function X in W 1,1 (, T, R p ) such that { Ẋ (t) + M φ(x (t)) G(t, X (t)) a.e. on ], T [, (1) X () = ξ. M symmetric definite positive matrix, φ maximal monotone... In every case, models are based on masses, springs (stiffness), dashpots (damping coefficient), Saint-Venant elements (threshold). These parameters have to be identified.

6 Thus, all the systems we have seen can be written under the form (1) and have a unique solution. We can observe that there are three classes of mechanical systems : in the first class, the function φ is a linear combinations of x i ; in the second class, the function φ is a sum of functions ψ [ ηi,η i ] (indicatrix function); in the third class, φ involves the both functions x and ψ [ ηi,η i ]. But a model must be given to be identified. Let us look at a few numerical results...

7 Numerical simulations We present some numerical simulations for : - the Prandtl model, - the Prandtl model with linear hardening, - the generalized Prandtl model with linear hardening - several viscoelastoplastic models, For all these numerical simulations, we solve the differential inclusion (1) by using the implicit Euler scheme We denote by X h the linear interpolation of the X n s. The function X h converges to the solution X of system (1) in C ([, T ], R p ).

8 Study of the rheological Prandtl models Let us consider the rheological Prandtl models: k α 1 1 k k α 2 2 k 1 α 1 m F k α 2 2 kn αn m F (a) kn αn The generalized Prandtl model (a) and the generalized Prandtl model + linear hardening (b). Displacement x and mass m. We choose m = 1. We study a harmonic forcing and other periodic forcings. (b)

9 Harmonic forcing For this section, we choose F (t) = f cos(ωt). (2) We first present essential difference between the Prandtl model and the Prandtl model with linear hardening : we compute x, y and z for the Prandtl model on the interval [, T ] with T = 2, F (t) = 2 cos(.1t), η = k = 1, x = y = u =, (3) and for the Prandtl model with linear hardening onl [, T ] with T = 3, F (t) = 2 cos(.1t), k = 1, η = k = 1, Sx = y = u =. (4) Results :

10 x(t) x (t) u(t) t t t x(t) x (t) u(t) t t t The functions x, y and u for the Prandtl model The functions x, y and u for the Prandtl model with linear hardening,

11 We can see that the amplitude of the functions x and ẋ is larger for the Prandtl model, which corresponds to k = than for Prandtl model with hardening, which is the same system with k. We sketch a physical explanation: If the St-Venant element slips, then and u εη, ẍ(t) + k m x(t) = f εkη cos(ωt) m m, (5) where ε { 1, 1}. If k =, and if the slip phase starts at t, then x(t) = 2f sin(ω(t + t mω 2 )) sin(ω(t t )) εkη 2m (t t ) 2 + ( ẋ(t ) f mω sin(ωt ) ) (6) (t t ) + x(t ). For ω 1 and t t, we have x(t) 2f mω 2 sin(ω(t t )); (7) thus, the approximate amplitude of x is roughly 2f = 4, (8) mω2

12 If k = 1 and if the slip phase starts at t, (5) implies x(t) = D cos(ω t + φ) + f εkη m(ω 2 cos(ωt) ω2 ) mω 2, (9) where ω = k /m ω and (D, φ) depends of (x(t ), ẋ(t )). Here, with ω =.1, the amplitude of the sinusoidal component of period ω of the function x is roughly as we can see on the right. - Coherent results. f m(ω 2 2, (1) ω2 ) - We start observing also "smoothness" of the solutions.

13 To examine "smoothness", see in next Fig. a) curve {x(t), F (t)} t [35,8] for the Prandtl model with F (t) = 2 cos(.1t), η = k = 1, and x = y = u =. We observe a limit cycle, but it gives no information on the physically relevant parameters of the system. If we plot as in Fig. b), {x(t), F (t) mẍ(t)} t [35,8], we observe that the slope of the oblique parts of the cycle is equal to k (a) (b) F(t) 5 1 F(t) mx (t) x(t) x(t) {x(t), F (t)} t [35,8] (a) and {x(t), F (t) mẍ(t)} t [35,8] (b) for the Prandtl model defined by F (t) = 2 cos(.1t), η = k = 1, and x = y = u =. These two figures differ only by the choice of coordinates.

14 t (a) F(t) mx (t) F(t) mx"(t) x(t) x(t) {x(t), F (t) mẍ(t), t} t [,18] for the Prandtl model, defined by F (t) = 2 cos(.1t), η = k = 1, x = y = u =. {x(t), F (t) mẍ(t)} t [5,2] for the generalized Prandtl model with linear hardening F (t) = cos(.5t), n = 1, k =, η 1 = k 1 = 1, (x = y = u,1 = ) We can observe (left Figure) transients followed by a periodic regime (t 15)of hysteresis cycles. Without transient (right Figure a), the cycle does not contain (, ) in the plane (x, F mẍ). Slopes and location of slopes changes...

15 F(t) mx (t) 2 F(t) mx (t) x(t) x(t) {x(t), F (t) mẍ(t)} t [75,1] for the generalized Prandtl rheological model with linear hardening, with mechanical parameters and F (t) = f cos(.5t), n = 5, m = 1, x = y =, i {1,..., 5}, u,i =, f = 6.6. {x(t), F (t) mẍ(t)} t [96,1] for the generalized Prandtl rheological model with linear hardening, defined by F (t) = 8H 1 (.5t), n = 3, k = 1/n, x = y =, i {1,..., 3}, k i = 1/3, η i = i, u,i =.

16 Viscoelastoplastic model Adding classical dissipation cẋ in our one dof model... 1 (a) (b) F(t) mx (t) F(t) mx (t) x(t) x(t) {x(t), F (t) mẍ(t)} t [4,2] for the viscoelastoplastic model defined by F (t) = cos(.5t), η = k = m = 1, x = y = u =, c = 1 (a) and c = 1 (b). These two figures differ only by (mathematical) value of damping c.

17 Schedule 1. Numerical simulations and question of identification 2. Identification process: Theoretical examples 3. Identification: Numerical examples

18 Process for identification To get such cycle or at least half cycle: Continuous data (theory), Discrete data (experiments), F known, To deal with damping... Find slope s changes (non smooth), Identify from a given model. Let us start first from theoretical point of view. Remark: - sometimes hysteresis limit cycles periodic but not convex. - elastic part may be nonlinear... - in both cases: Similar process works.

19 Most of the responses of the generalized Prandtl model to a sinusoidal forcing or to a periodic (but non harmonic) forcing tend to hysteresis limit cycles. We observe that these hysteresis cycles have a center of symmetry. Therefore, we study a loading phase corresponding to a half-cycle in the (x, F mẍ) plan : F(t)-mx"(t) A n+1 A n A n+2 A i+1 A i A 2 A 1 d 1 d i d n-1 d n x(t) Loading curve for the generalized Prandtl rheological model with linear hardening.

20 Let us denote by A 1, A 2,...,A n+1 and A n+2 the ends of segments which constitute the hysteresis half-cycle. For i {1,..., n}, d i denotes the difference between abscissa of A i+1 and A i (positive real). For i {1,..., n + 1}, p i denotes the slope of the segment [A i, A i+1 ]. F(t)-mx"(t) A n A n+1 A n+2 A i+1 A i A 2 A 1 d 1 d i d n-1 d n x(t)

21 If the forcing F is not constant on any open non empty subinterval of [, T ], then x has the same property; assume indeed that x is constant on (t 1, t 2 ); then ẋ and ẍ vanish on (t 1, t 2 ) and by uniqueness of the solution of { u i + β(u i /η i ), (11) u i (t 1 ) given, u i (t) is equal to u i (t 1 ) on (t 1, t 2 ). This implies that the expression F = mẍ + k x + n k i u i i=1 is constant over (t 1, t 2 ), which contradicts our assumption.

22 We are now able to describe the shape of the representation of a trajectory of the system in the x and F mẍ coordinates, in some special case. We assume that the η i = α i /k i are all distinct; otherwise two elements with identical ratio η i would enter the plasticity phase simultaneously. We reorder the indices so that η 1 < η 2 <... < η n 1 < η n. (12) Assume that F vanishes on no open non empty subinterval of [, T ], and the following properties: 1) u j () = η j, j = 1,..., n, 2) x is increasing on [, T ].

23 Then there exists a increasing sequence t 1 t 2... t n such that for some j {1,..., n} and we have moreover t 1 < t 2 <... < t j = t j+1 =... = T (13) x(t j ) x() = 2η j, (14) n F mẍ k + x is constant on ]t j, t j+1 ]. (15) i=j+1 k i

24 Observe first that x is strictly increasing on [, T ], thanks to our assumption on F. Let û j (t) = η j + x(t) x(), (16) and denote by ξ the inverse function of x and define { ξ (2η j + x()) if 2η j + x() < x(t ), t j = T if 2η j + x() x(t ). (17) We check immediately that the function u j (t) = is the (unique) solution of { {ûj (t) if t < t j, η j if t t j, u j + β(u j /η j ) ẋ, u j () = η j. (18) (19)

25 Therefore, on ]t j, t j+1 ], relation (15) holds. Moreover, relation (17) is equivalent to (14). We write equations (15) and (14) as j {1,..., n + 1}, p j = k + n k l, (2) i {1,..., n}, d i = 2η i. (21) On the other hand, we remark that the assumptions k, k i > and equation (2) imply equations (12) and (21) imply l=j p n+1 < p n <... < p 2 < p 1 ; (22) d 1 < d 2 <... < d n 1 < d n. (23)

26 From (2) and (21) we have a one to one correspondence between the parameters of generalized Prandtl model k i and η i and geometrical parameters p j and d j of the hysteresis cycle in the (x, F mẍ) plane. Thus, a partial identification of the model is possible. But we cannot infer from the values of (p i ) 1 i n+1 and (d i ) 1 i n the abscissa and the ordinate of the lower left point of the cycle. They depend on the initial conditions and on the forcing:

27 (a) (b) F(t) mx (t) F(t) mx (t) x(t) x(t) x 1 4 {x(t), F (t) mẍ(t)} t [5,2] for the generalized Prandtl rheological model with linear hardening, with F(t) = cos(.5t), n = 1, k =, η 1 = k 1 = 1, (x = y = u,1 = ) (a), (x = 1, y = 2, u,1 =.99) (b). These two figures differ only by the initial conditions.

28 There are more or less parameters to identify...: F(t) mx (t) 2 2 F(t) mx (t) x(t) x(t) {x(t), F (t) mẍ(t)} t [75,1] for the generalized Prandtl rheological model with linear hardening, with mechanical parameters and F (t) = f cos(.5t), n = 5, m = 1, x = y =, i {1,..., 5}, u,i =, f = phases... (n) {x(t), F (t) mẍ(t)} t [96,1] for the generalized Prandtl rheological model with linear hardening, defined by F (t) = 8H 1 (.5t), n = 3, k = 1/n, x = y =, i {1,..., 3}, k i = 1/3, η i = i, u,i =. 3 phases... (n)

29 Schedule 1. Numerical simulations and question of identification 2. Identification process: Theoretical examples 3. Identification: Numerical examples

30 Numerical identification of a model from its limit cycles If limit cycle or at least half cycle is known from continuous data, with no damping or known damping, F known: OK (calculation of parameters...) If limit cycle or at least half cycle is known from discrete data with no noise, high quality sampling, with no damping or known damping, F known (discrete): OK (calculation of parameters, mean squares,...). If limit cycle or at least half cycle is known from discrete data with poor sampling, noise etc. (real data), with no damping or known damping, F known (discrete): OK with wavelet analysis. To deal with damping...also wavelet analysis... Identification from a given model and high enough solicitation: n = 1 rather than n = 2 leads to an error. Unlocking only one Saint-Venant element rather than two because of too weak forcing F also leads to an error... obviously.

31 Damping with wavelets Continuous wavelet transform: f signal, g analyzing function (b, a) R R +, W g f (b, a) = 1 + a f (t)g( t b )dt, (24) a For f (t) = A exp( ct) cos(ωt + φ), focusing on b j = jt, j N, T = 2π/ω, one can prove c = 1 (m n)t log( W g f (nt, a) (25) (mt, a)), W g f for m, n integers, at the analyzing scale a > small enough. C.-H. LAMARQUE, S. PERNOT, A. CUER, Damping identification in multi-degree-of-freedom systems via a wavelet-logarithmic decrement- Part I: Theory, Journal of Sound and Vibration, 235 (3), , 2.

32 Multiresolution wavelet analysis: f signal, φ scaling function, φ jk (x) = r Z φ jk (x + r), (26) f f J = 2 J 1 k= s Jk φjk (27) For f (t) = A exp( ct) cos(ωt + φ), focusing at scale a = 2 ja on b k = k/2 ja and b l = l/2 ja, one can prove for k, l, j a integers. c = 2ja (k l) log( s j al s jak ), (28) C.-H. LAMARQUE, S. PERNOT, A. CUER, Damping identification in multi-degree-of-freedom systems via a wavelet-logarithmic decrement- Part I: Theory, Journal of Sound and Vibration, 235 (3), , 2.

33 Theoretical and signal processing validation: C.-H. LAMARQUE, S. PERNOT, A. CUER, Damping identification in multi-degree-of-freedom systems via a wavelet-logarithmic decrement- Part I: Theory, Journal of Sound and Vibration, 235 (3), , 2. Experimental validation: S. HANS, E. IBRAIM, S.PERNOT, C. BOUTIN, C.-H. LAMARQUE, Damping identification in multi-degree-of-freedom systems via a wavelet-logarithmic decrement- Part II: Study of a civil engineering building, Journal of Sound and Vibration, 235 (3), , 2. It permits to process identification of damping from more general signal, discrete, with noise, from experiments. Better than several classical algorithms (logarithmic decrement, mean squares methods, etc.). It includes filters...

34 Identification of non locally Lipschitz with damping mẍ + G(x) + G(ẋ) + i {1,..., n}, n k i u i = F, i=1 u i + β(u i /η i ) ẋ (29) where F is deterministic or stochastic. We choose n = 5, k =, m = 1, σ = 3, T = 1 and G(y) =.5y, G(x) =.5x 3. We focus on a monotonic phase of x assumed to be hold between and T. On the graph (x, F mẍ) t T = (X, G(X )) some regularity s phases can be observed. In fact, in the deterministic case, ẋ is a derivable function with respect to x and we observe only n discontinuities on the graph (X, G (X )). We can locate this discontinuities by using wavelet s tools.

35 Identification of non locally Lipschitz with damping Slope s detections in next figure for n = 5, k =, m = 1, σ = 3, T = 1 and the functions G 1 (x, y) = G(y) =.5y and G 2 (x, y) =. Horizontal one: we represent on the horizontal axis not the value of the discretization (x i ) 1 i N of x but the number i. Vertical axis = scales Color = highest amplitude of wavelets coefficients

36 Identification of non locally Lipschitz with damping x x 1 4 Amplitude of wavelet s coefficients versus displacement x. High amplitudes separation of phases lest square identification... Remark: Identification may be still difficult (external forcing, model of damping, choice of model)...