Nonlinear response analysis for a dual-rotor system with a breathing transverse crack in the hollow shaft

Abstract

This paper focuses on the nonlinear response characteristics of a dual-rotor system with a breathing transverse crack in the hollow shaft of the high-pressure rotor (rotor 1). A finite element model of the system is set up, and the motion equations of the system are formulated, in which the unbalance excitations of the rotor 1 and rotor 2 (low-pressure rotor) and the time-varying stiffness of the cracked shaft are considered. By using the harmonic balance method, the motion equations are analytically solved to obtain the dynamic responses of the two rotors. Accordingly, the effects of the crack depth and location on the vibration amplitudes are discussed in detail. The results indicate that when a transverse crack appears, it may bring super-harmonic responses to the rotor system, and the resonance peaks at the second, third and even fourth subcritical whirling speeds of the two rotors can be observed. The deeper the crack is, the larger the resonances amplitudes are, especially when the crack is located in the middle of the shaft or around the disks. In addition, the super-harmonic responses of rotor 1 where the crack located, can also be observed in rotor 2, which means that the crack signals can be detected in the entire system. Moreover, the numerical computations are carried out by using the Newmark-$\beta$ method, which shows great agreement with the previous analytical results. The results obtained in this paper will contribute to the modeling and the fault diagnosis of dual-rotor systems with hollow-shaft crack.

Appendix

Finite element matrices:

$$\begin{aligned} \mathbf{K}_T^\mathrm{es}= & {} \frac{EI}{L^{3}}\left[ {{\begin{array}{cccccccc} {K_{B1} }&{} &{} &{} &{} &{} &{} &{} \\ 0&{} {K_{B1} }&{} &{} &{} &{} &{} &{} \\ 0&{} {-K_{B4} }&{} {K_{B2} }&{} &{} &{} \mathrm{symm}&{} &{} \\ {K_{B4} }&{} 0&{} 0&{} {K_{B2} }&{} &{} &{} &{} \\ {-K_{B1} }&{} 0&{} 0&{} {-K_{B4} }&{} {K_{B1} }&{} &{} &{} \\ 0&{} {-K_{B1} }&{} {K_{B4} }&{} 0&{} 0&{} {K_{B1} }&{} &{} \\ 0&{} {-K_{B4} }&{} {K_{B3} }&{} 0&{} 0&{} {K_{B4} }&{} {K_{B2} }&{} \\ {K_{B4} }&{} 0&{} 0&{} {K_{B3} }&{} {-K_{B4} }&{} 0&{} 0&{} {K_{B2} } \\ \end{array} }} \right] , \\ K_{B1}= & {} 12/(1+\varphi _s );K_{B2} =L^{2}(4+\varphi _s )/(1+\varphi _s );K_{B3} =L^{2}(2-\varphi _s )/(1+\varphi _s ); \\ K_{B4}= & {} 6L/(1+\varphi _s );\varphi _s =12\mathrm{EI}/(\mathrm{GAL}^{2}). \\ \mathbf{M}_T^\mathrm{es}= & {} \frac{\rho L}{(1+\varphi _s )^{2}}\left[ {{\begin{array}{cccccccc} {M_{T1} }&{} &{} &{} &{} &{} &{} &{} \\ 0&{} {M_{T1} }&{} &{} &{} &{} &{} &{} \\ 0&{} {-M_{T4} }&{} {M_{T2} }&{} &{} &{} \mathrm{symm}&{} &{} \\ {M_{T4} }&{} 0&{} 0&{} {M_{T2} }&{} &{} &{} &{} \\ {M_{T3} }&{} 0&{} 0&{} {M_{T5} }&{} {M_{T1} }&{} &{} &{} \\ 0&{} {M_{T3} }&{} {-M_{T5} }&{} 0&{} 0&{} {M_{T1} }&{} &{} \\ 0&{} {M_{T5} }&{} {M_{T6} }&{} 0&{} 0&{} {M_{T4} }&{} {M_{T2} }&{} \\ {-M_{T5} }&{} 0&{} 0&{} {M_{T6} }&{} {-M_{T4} }&{} 0&{} 0&{} {M_{T2} } \\ \end{array} }} \right] , \\ M_{T1}= & {} 13/35+7/10\varphi _s +1/3\varphi _s^2 ;M_{T2} =(1/105+1/60\varphi _s +1/120\varphi _s^2 )L^{2}; \\ M_{T3}= & {} 9/70+3/10\varphi _s +1/6\varphi _s^2 ;M_{T4} =(11/210+11/120\varphi _s +1/24\varphi _s^2 )L; \end{aligned}$$

$$\begin{aligned} M_{T5}= & {} (13/420+3/40\varphi _s +1/24\varphi _s^2 )L;M_{T6} =-(1/140+1/60\varphi _s +1/120\varphi _s^2 )L^{2}. \\ \mathbf{M}_R^\mathrm{es}= & {} \frac{\rho L}{(1+\varphi _s )^{2}}\left( \frac{r_\rho }{L} \right) ^{2}\left[ {{\begin{array}{cccccccc} {M_{R1} }&{} &{} &{} &{} &{} &{} &{} \\ 0&{} {M_{R1} }&{} &{} &{} &{} &{} &{} \\ 0&{} {-M_{R4} }&{} {M_{R2} }&{} &{} &{} \mathrm{symm}&{} &{} \\ {M_{R4} }&{} 0&{} 0&{} {M_{R2} }&{} &{} &{} &{} \\ {-M_{R1} }&{} 0&{} 0&{} {-M_{R4} }&{} {M_{R1} }&{} &{} &{} \\ 0&{} {-M_{R1} }&{} {M_{R4} }&{} 0&{} 0&{} {M_{R1} }&{} &{} \\ 0&{} {-M_{R4} }&{} {M_{R3} }&{} 0&{} 0&{} {M_{R4} }&{} {M_{R2} }&{} \\ {M_{R4} }&{} 0&{} 0&{} {M_{R3} }&{} {-M_{R4} }&{} 0&{} 0&{} {M_{R2} } \\ \end{array} }} \right] , \\ M_{R1}= & {} 6/5;M_{R2} =(2/15+1/6\varphi _s +1/3\varphi _s^2 )L^{2};r_\rho =\sqrt{I^{e}/A}; \\ M_{R3}= & {} (-1/30-1/6\varphi _s +1/6\varphi _s^2 )L^{2};M_{R4} =(1/10-1/2\varphi _s )L. \\ \mathbf{G}^\mathrm{es}= & {} \frac{\rho r_\rho ^2 }{15L(1+\varphi _s )^{2}}\left[ {{\begin{array}{cccccccc} 0&{} &{} &{} &{} &{} &{} &{} \\ {G_1 }&{} 0&{} &{} &{} &{} &{} &{} \\ {-G_2 }&{} 0&{} 0&{} &{} &{} \mathrm{antisymm}&{} &{} \\ 0&{} {-G_2 }&{} {G_4 }&{} 0&{} &{} &{} &{} \\ 0&{} {G_1 }&{} {-G_2 }&{} 0&{} 0&{} &{} &{} \\ {-G_1 }&{} 0&{} 0&{} {-G_2 }&{} {G_1 }&{} 0&{} &{} \\ {-G_2 }&{} 0&{} 0&{} {G_3 }&{} {G_2 }&{} 0&{} 0&{} \\ 0&{} {-G_2 }&{} {-G_3 }&{} 0&{} 0&{} {G_2 }&{} {G_4 }&{} 0 \\ \end{array} }} \right] , \\ G_1= & {} 36;G_2 =3L-15L\varphi _s ;G_3 =L^{2}+5L^{2}\varphi _s -15L^{2}\varphi _s^2 );G_4 =4L^{2}+5L^{2}\varphi _s +10L^{2}\varphi _s^2 ). \end{aligned}$$