nonlinear identification method on simulated FRF and Transmissibility data. Numerical values are taken from reference [7] for consistency. The equation of motion of the forceexcited system (FRF) is (11) where the parameters are described in Tab.1. Eqn.(8) has been solved by direct integration using the built-in Matlab solver ODE45 for each case at different frequencies of excitation (separated by a step of 0.005Hz). The FRF has then been computed by simulating the operation of a Frequency Response Analyser (FRA) that is by computing the ratio between the Fourier coefficients of the response and the force at the excitation frequency. In addition, several levels of excitation force have been used. Similarly, the transmissibility (base-excitation) simulations derive from the integration of the equation of motion (12) where z x y = − is the relative displacement between the mass and the base and Yis the amplitude of the base displacement. After the integration the absolute displacement has been computed before calculating the absolute transmissibility which is needed to perform the nonlinear identification with CONCERTO. Table 1: Type and values of nonlinearities used for the numerical simulations. For the base-excitation simulations the stiffness and damping are functions of the relative displacement z As an example of the numerical simulation, Fig.2 (a,b) shows the FRF and transmissibility functions for the SDOF system with a combined hardening cubic stiffness and Coulomb damping. It should be pointed out that in Table 1 only the mass displacement x has been used. This is correct in the case of force excitation (FRF) but for the case of base-excitation the mount’s properties are function of the effective amplitude of deformation which is the relative displacement z. 483 0 ( ) ( ) sin( ) c k mxcx fX kx fX F tω + + + + = ɺ ɺɺ ɺ 2 ( ) ( ) sin( ) c k mzcz fZ kz fZ mY t ω ω + + + + = ɺ ɺɺ ɺ Mass m = 1.5 kg Damping Coefficient c = 0.8 Ns/m Stiffness k = 6000 N/m Nonlinearity Damping fc Stiffness fk Values Cubic stiffness (hardening) 0 3 k nl f k x = 6 7 nl k e = Nm-3 Quadratic Damping c nl f c x x = ɺ ɺ 0 8 nl c = N s 2m-2 Coulomb Damping ( ) c f f F sign x = ɺ 0 Ff = 0.85 Quadratic Damping + Cubic stiffness c nl f c x x = ɺ ɺ 3 k nl f k x = 6 7 nl k e = Nm-3 8 nl c = N s 2m-2
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