PSI - Issue 24

Vincenzo D’Addio et al. / Procedia Structural Integrity 24 (2019) 510–525 Author name / Structural Integrity Procedia 00 (2019) 000 – 000 = [ 1 e ( + 1 ) . . e ( + ) . . e ( + ) ] = { }e (6) where e is the common term for all the components; the response will be harmonic and have the same angular frequency ω: = { }e (7) Substituting in (3) we obtain: [( − 2 ) + iω( − Ω )]{ } = { } (8) In the ANSYS program solver, the direct method has been set that consists in inverting the matrix on the left side: { } = [( − 2 ) + iω( − Ω )] −1 { } ( 9) The frequency response plot related to a couple of transversal forces fixed in space (the frequency varying in the range [1.3, 3.8] times the shaft operating rotational frequency) is shown in Fig. 8, where in the vertical axis the dimensionless acceleration magnitude of the node corresponding to the rotor side support is reported . 519 10

a

1.3 1.6

1.9

2.2

2.6

2.9

3.2

3.5

3.8

Fig. 8. Harmonic response of the node at the rotor side support to a couple of transversal forces of constant direction, at the bearings (see nomenclature).

It is evident that peaks in the response occur when the frequency of the excitation reaches the natural whirl frequencies of the system; the nature of this excitation (fixed in the space) is able to excite both BW and FW motions and so two peaks are observed in that range of frequency. On the contrary, if the load had been unbalanced (rotating at the same frequency and direction with the rotor, 1X) only FW peaks would appear, since unbalance is unable to activate BW motions.