Abstract
The rotordynamics of a double-helical gear transmission system is investigated. The equation of motion of the system with bearing and gyroscopic effect is derived by using the finite element method, in which Timoshenko beam finite element is used to represent the shaft, a rigid mass for the gear. Natural frequencies, mode shapes and Campbell diagrams are illustrated to indicate the effects of gear input speed and time varying mesh stiffness. Besides, effects of mesh stiffness on the critical speed of the gear transmission system are analyzed. The numerical results show that the axial force has significant influence on the natural frequency and the mode shape of the double-helical gear transmission system, for which the mix whirling motion dominates the natural characteristics. There are two higher critical speed curves which increase with the mesh stiffness, but one of them is related to the gyroscopic effect.
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The authors gratefully acknowledge the support of the National Science Foundation of China (NSFC) through Grants Nos. 51305462 and 51275530.
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Appendices
Appendix 1
Mass and stiffness matrices for the shaft beam element are listed as follows:
1.1 Mass matrix \( {\mathbf{M}}^{e} \)
The mass matrix consists of three parts, which can be represented as
-
1.
Translational mass matrix \( {\mathbf{M}}_{T}^{e} \)
$$ {\mathbf{M}}_{T}^{e} = {\mathbf{M}}_{T1}^{e} + \phi {\mathbf{M}}_{T2}^{e} + \phi^{2} {\mathbf{M}}_{T3}^{e} $$(40)Here,
$$ {\mathbf{M}}_{T1}^{e} = m^{T} \left({\begin{array}{*{20}c} {312} & 0 & 0 & 0 & {44L} & 0 & {108} & 0 & 0 & 0 & {- 26L} & 0 \\ 0 & {312} & 0 & {- 44L} & 0 & 0 & 0 & {108} & 0 & {26L} & 0 & 0 \\ 0 & 0 & {280} & 0 & 0 & 0 & 0 & 0 & {140} & 0 & 0 & 0 \\ 0 & {- 44L} & 0 & {8L^{2}} & 0 & 0 & 0 & {- 26L} & 0 & {- 6L^{2}} & 0 & 0 \\ {44L} & 0 & 0 & 0 & {8L^{2}} & 0 & {26L} & 0 & 0 & 0 & {- 6L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {108} & 0 & 0 & 0 & {26L} & 0 & {312} & 0 & 0 & 0 & {- 44L} & 0 \\ 0 & {108} & 0 & {- 26L} & 0 & 0 & 0 & {312} & 0 & {44L} & 0 & 0 \\ 0 & 0 & {140} & 0 & 0 & 0 & 0 & 0 & {280} & 0 & 0 & 0 \\ 0 & {26{\mkern 1mu} L} & 0 & {- 6L^{2}} & 0 & 0 & 0 & {44L} & 0 & {8L^{2}} & 0 & 0 \\ {- 26L} & 0 & 0 & 0 & {- 6L^{2}} & 0 & {- 44L} & 0 & 0 & 0 & {8L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}} \right) $$(41)$$ \frac{{{\mathbf{M}}_{T2}^{e}}}{{m^{T}}} = \left({\begin{array}{*{20}c} {588} & 0 & 0 & 0 & {77L} & 0 & {252} & 0 & 0 & 0 & {- 63L} & 0 \\ 0 & {588} & 0 & {- 77L} & 0 & 0 & 0 & {252} & 0 & {63L} & 0 & 0 \\ 0 & 0 & {560} & 0 & 0 & 0 & 0 & 0 & {280} & 0 & 0 & 0 \\ 0 & {- 77L} & 0 & {14L^{2}} & 0 & 0 & 0 & {- 63L} & 0 & {- 14L^{2}} & 0 & 0 \\ {77L} & 0 & 0 & 0 & {14L^{2}} & 0 & {63L} & 0 & 0 & 0 & {- 14L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {252} & 0 & 0 & 0 & {63L} & 0 & {588} & 0 & 0 & 0 & {- 77L} & 0 \\ 0 & {252} & 0 & {- 63L} & 0 & 0 & 0 & {588} & 0 & {77L} & 0 & 0 \\ 0 & 0 & {280} & 0 & 0 & 0 & 0 & 0 & {560} & 0 & 0 & 0 \\ 0 & {63L} & 0 & {- 14L^{2}} & 0 & 0 & 0 & {77L} & 0 & {14L^{2}} & 0 & 0 \\ {- 63L} & 0 & 0 & 0 & {- 14L^{2}} & 0 & {- 77L} & 0 & 0 & 0 & {14L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}} \right) $$(42)$$ {\mathbf{M}}_{T3}^{e} = m^{T} \left({\begin{array}{*{20}c} {280} & 0 & 0 & 0 & {35L} & 0 & {140} & 0 & 0 & 0 & {- 35L} & 0 \\ 0 & {280} & 0 & {- 35L} & 0 & 0 & 0 & {140} & 0 & {35{\mkern 1mu} L} & 0 & 0 \\ 0 & 0 & {280} & 0 & 0 & 0 & 0 & 0 & {140} & 0 & 0 & 0 \\ 0 & {- 35L} & 0 & {7L^{2}} & 0 & 0 & 0 & {- 35L} & 0 & {- 7{\mkern 1mu} L^{2}} & 0 & 0 \\ {35L} & 0 & 0 & 0 & {7L^{2}} & 0 & {35L} & 0 & 0 & 0 & {- 7L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {140} & 0 & 0 & 0 & {35L} & 0 & {280} & 0 & 0 & 0 & {- 35L} & 0 \\ 0 & {140} & 0 & {- 35L} & 0 & 0 & 0 & {280} & 0 & {35L} & 0 & 0 \\ 0 & 0 & {140} & 0 & 0 & 0 & 0 & 0 & {280} & 0 & 0 & 0 \\ 0 & {35L} & 0 & {- 7L^{2}} & 0 & 0 & 0 & {35L} & 0 & {7L^{2}} & 0 & 0 \\ {- 35L} & 0 & 0 & 0 & {- 7L^{2}} & 0 & {- 35L} & 0 & 0 & 0 & {7L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}} \right) $$(43)and
$$ m^{T} = \frac{\rho AL}{{840\left({1 + \phi} \right)^{2}}} $$(44) -
2.
Rotational mass matrix \( {\mathbf{M}}_{R}^{e} \)
$$ {\mathbf{M}}_{R}^{e} = {\mathbf{M}}_{R1}^{e} + \phi {\mathbf{M}}_{R2}^{e} + \phi^{2} {\mathbf{M}}_{R3}^{e} $$(45)Here,
$$ {\mathbf{M}}_{R1}^{e} = m^{R} \left({\begin{array}{*{20}c} {36} & 0 & 0 & 0 & {3L} & 0 & {- 36} & 0 & 0 & 0 & {3L} & 0 \\ 0 & {36} & 0 & {- 3L} & 0 & 0 & 0 & {- 36} & 0 & {- 3L} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & {- 3L} & 0 & {4L^{2}} & 0 & 0 & 0 & {3L} & 0 & {- L^{2}} & 0 & 0 \\ {3L} & 0 & 0 & 0 & {4L^{2}} & 0 & {- 3L} & 0 & 0 & 0 & {- L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {- 36} & 0 & 0 & 0 & {- 3L} & 0 & {36} & 0 & 0 & 0 & {- 3L} & 0 \\ 0 & {- 36} & 0 & {3L} & 0 & 0 & 0 & {36} & 0 & {3L} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & {- 3L} & 0 & {- L^{2}} & 0 & 0 & 0 & {3L} & 0 & {4L^{2}} & 0 & 0 \\ {3L} & 0 & 0 & 0 & {- L^{2}} & 0 & {- 3L} & 0 & 0 & 0 & {4L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}} \right) $$(46)$$ {\mathbf{M}}_{R2}^{e} = m^{R} \left({\begin{array}{*{20}c} 0 & 0 & 0 & 0 & {- 15L} & 0 & 0 & 0 & 0 & 0 & {- 15L} & 0 \\ 0 & 0 & 0 & {15L} & 0 & 0 & 0 & 0 & 0 & {15L} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & {15L} & 0 & {5L^{2}} & 0 & 0 & 0 & {- 15L} & 0 & {- 5L^{2}} & 0 & 0 \\ {- 15L} & 0 & 0 & 0 & {5L^{2}} & 0 & {15L} & 0 & 0 & 0 & {- 5L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & {15L} & 0 & 0 & 0 & 0 & 0 & {15L} & 0 \\ 0 & 0 & 0 & {- 15L} & 0 & 0 & 0 & 0 & 0 & {- 15L} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & {15L} & 0 & {- 5L^{2}} & 0 & 0 & 0 & {- 15L} & 0 & {5L^{2}} & 0 & 0 \\ {- 15L} & 0 & 0 & 0 & {- 5L^{2}} & 0 & {15L} & 0 & 0 & 0 & {5L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}} \right) $$(47)$$ {\mathbf{M}}_{R3}^{e} = m^{R} \left({\begin{array}{*{20}c} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {10L^{2}} & 0 & 0 & 0 & 0 & 0 & {5L^{2}} & 0 & 0 \\ 0 & 0 & 0 & 0 & {10L^{2}} & 0 & 0 & 0 & 0 & 0 & {5L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & {5L^{2}} & 0 & 0 & 0 & 0 & 0 & {10L^{2}} & 0 & 0 \\ 0 & 0 & 0 & 0 & {5L^{2}} & 0 & 0 & 0 & 0 & 0 & {10L^{2}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}} \right) $$(48)where
$$ m^{R} = \frac{{\rho I_{ds}}}{{30L\left({1 + \phi} \right)^{2}}} $$(49) -
3.
Torsional mass matrix \( {\mathbf{M}}_{R}^{e} \)
$$ {\mathbf{M}}_{R}^{e} = \frac{{\rho I_{pe} l}}{6}\left({\begin{array}{*{20}c} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{array}} \right) $$(50)
1.2 Gyroscopic matrix \( {\mathbf{G}}^{e} \)
1.3 Stiffness matrix
The stiffness matrix of the beam element is written as
where
and
Appendix 2
2.1 Coefficients of mesh stiffness and damping
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Chen, S., Tang, J., Li, Y. et al. Rotordynamics analysis of a double-helical gear transmission system. Meccanica 51, 251–268 (2016). https://doi.org/10.1007/s11012-015-0194-0
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DOI: https://doi.org/10.1007/s11012-015-0194-0