Abstract
This paper presents a quasi-static model of high-speed ball bearing, and the analytical expressions of dynamic stiffness are obtained. The action mechanism of centrifugal force and gyroscopic moment and its effect on kinematic and mechanical characteristics of bearing are studied. Results show that with the increase in rotational speed, centrifugal force increases continuously, while gyroscopic moment rapidly first and then slightly. With the increase in inertial force, the inner-ring contact angles and the outer-ring contact force increase, the outer-ring contact angles and inner-ring contact force decrease, and bearing stiffness will increase in all directions, while axial load will reduce the effect of inertial force. Radial force will increase the non-uniformity of load distribution, making the difference of load carried by each ball larger, thus aggravating the instability of the system. Compared with gyroscopic moment, centrifugal force is the main factor, while ignoring both inertial forces, rotational speed no longer affects mechanical characteristics of bearings. This study provides theoretical support for the dynamic analysis of high-speed bearing rotor system.
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Acknowledgements
We would like to express our appreciation to Key Project of Guangdong Education Department of China (2018KZDXM075), Program for Innovative Research Team in University of Guangdong Education Department of China (2018KCXTD032) and National Natural Science Foundation of China (U1710119) for supporting this research.
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Appendices
Appendix
Ball contact stiffness matrix
\({\partial {{\varvec{Q}}}_{j} } /{\partial {{\varvec{u}}}_{j} }\) is the contact stiffness matrix:
When \((z_{pj} ,r_{pj} )\) is selected as the reference point,
Here, let:
There are
Combined with the bearing’s force at high speed, according to the ball’s force balance condition formula (20), let:
There are
Then,
Write for
Taking the rolling element j as the research object, the parameters of the above equation are deduced, and the angle marker j is ignored here.
1. Solve for \(\frac{\partial (Q_{\mathrm{i}} \cos \alpha _{\mathrm{i}} )}{\partial v_{r} }\)
In combination with the \(\delta _{ij} \) and \(l_{ij} \), according to the chain rule.
(a). The partial derivative of equation \(\delta _{ij} \) with respect to \(l_{\mathrm{i}} \) can be obtained as
(b). The partial derivative of equation \(l_{ij} \) with respect to \(v_{r} \) can be obtained as
(c). Take the derivative of equation \(\delta _{\mathrm{i}} \) , then
Then,
Finally, the results were processed as
Similarly, the derivation results of other similar formulas can be obtained
2. Solve for \(\frac{\partial (M_{\mathrm{g}} \sin \alpha _{\mathrm{i}} )}{\partial v_{r} }\)
According to Formulas (4) and (5), let
where \(A=\frac{\cos \alpha _{\mathrm{e}} +\tan \beta \sin \alpha _{\mathrm{e}} }{1+\gamma '\cos \alpha _{\mathrm{e}} }\), \(B=\frac{\cos \alpha _{\mathrm{i}} +\tan \beta \sin \alpha _{\mathrm{i}} }{1-\gamma '\cos \alpha _{\mathrm{i}} }\).
Substitute (59) and (60) into (2), and get
(a) Solve for \(\frac{\partial W_{\mathrm{b}} }{\partial v_{r} }\)
where
(b) Solve for \(\frac{\partial W_{\mathrm{m}} }{\partial v_{r} }\)
In the same way,
(c) Solve for \(\frac{\partial \tan \beta }{\partial v_{r} }\)
3. Simplify and solve for \(\frac{\partial F_{\mathrm{c}} }{\partial v_{r} }\) and \(\frac{\partial F_{\mathrm{c}} }{\partial v_{z} }\)
And similarly, we can solve for \(\frac{\partial (M_{\mathrm{g}} \sin \alpha _{\mathrm{i}} )}{\partial v_{r} }\), \(\frac{\partial (M_{\mathrm{g}} \sin \alpha _{\mathrm{i}} )}{\partial v_{z} }\), \(\frac{\partial (M_{\mathrm{g}} \sin \alpha _{\mathrm{e}} )}{\partial v_{r} }\), \(\frac{\partial (M_{\mathrm{g}} \sin \alpha _{\mathrm{e}} )}{\partial v_{z} }\), \(\frac{\partial (M_{\mathrm{g}} \cos \alpha _{\mathrm{i}} )}{\partial v_{r} }\), \(\frac{\partial (M_{\mathrm{g}} \cos \alpha _{\mathrm{i}} )}{\partial v_{z} }\), \(\frac{\partial (M_{\mathrm{g}} \cos \alpha _{\mathrm{e}} )}{\partial v_{r} }\) and\(_{\mathrm { }}\frac{\partial (M_{\mathrm{g}} \cos \alpha _{\mathrm{e}} )}{\partial v_{z} }\).
The derivative of Eq. (36) with respect to \( {\tilde{{{\varvec{u}}}}}_{j} \) can be obtained as
Since \({{\varvec{B}}}_{\mathrm{i}j} ={{\varvec{B}}}_{i} \left( {{\tilde{{\varvec{w}}}}_{j} } \right) \), \({{\varvec{B}}}_{\mathrm{e}j} ={{\varvec{B}}}_{\mathrm{e}} \left( {\tilde{{\varvec{v}}}}_{j} \right) \), \({{\varvec{B}}}_{\mathrm{c}j} ={{\varvec{B}}}_{\mathrm{c}} \left( {\tilde{{\varvec{v}}}}_{j} \right) \), \({{\varvec{B}}}_{M\text{ i }j} ={{\varvec{B}}}_{M\text{ i }} \left( {\tilde{{\varvec{v}}}}_{j} \right) \), \({{\varvec{B}}}_{M\text{ e }j} ={{\varvec{B}}}_{M\text{ e }} \left( {\tilde{{\varvec{v}}}}_{j} \right) {\tilde{{\varvec{v}}}}_{j} ={\tilde{{\varvec{v}}}}\left( {{\tilde{{\varvec{u}}}}_{j} } \right) \), \({\tilde{{\varvec{w}}}}_{j} = {{\tilde{{\varvec{u}}}}}_{j} -{\tilde{{\varvec{v}}}}_{j} \), the following equation can be obtained by solving the partial derivative.
Substitute expressions (72)–(76) into Formula (71), then:
According to expression (14):
According to the expressions (25), (31) and (34), there is:
Substituting expression (78) into (79), it is found that:
Due to:
By substituting expression (78) into (81), we get:
After \({{\varvec{J}}}_{B_{\mathrm{i}j} } \), \({{\varvec{J}}}_{B_{\mathrm{e}j} } \), \({{\varvec{J}}}_{B_{\mathrm{c}j} } \), \({{\varvec{J}}}_{B_{M\text{ i }j} } \), \({{\varvec{J}}}_{B_{M\text{ e }j} } \) are substituted into the above equations, the contact stiffness matrix can be obtained by substituting Eqs. (80) and (82) into Formula (29).
The above is the analytical expression of bearing stiffness.
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Cheng, H., Zhang, Y., Lu, W. et al. Research on mechanical characteristics of fault-free bearings based on centrifugal force and gyroscopic moment. Arch Appl Mech 90, 2157–2184 (2020). https://doi.org/10.1007/s00419-020-01714-2
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DOI: https://doi.org/10.1007/s00419-020-01714-2