﻿ 偏心误差影响下的斜齿轮摩擦激励计算方法
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 浙江大学学报(工学版)  2017, Vol. 51 Issue (8): 1559-1567  DOI:10.3785/j.issn.1008-973X.2017.08.011 0

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dx.doi.org/10.3785/j.issn.1008-973X.2017.08.011
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LIU Wen, YANG Yun, LIN Teng-jiao, ZHANG Jin-hong. Method of calculating friction excitation of helical gear with geometric eccentricity[J]. Journal of Zhejiang University(Engineering Science), 2017, 51(8): 1559-1567.
dx.doi.org/10.3785/j.issn.1008-973X.2017.08.011
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### 文章历史

Method of calculating friction excitation of helical gear with geometric eccentricity
LIU Wen , YANG Yun , LIN Teng-jiao , ZHANG Jin-hong
State Key Laboratory of Mechanical Transmission, Chongqing University, Chongqing 400044, China
Abstract: In order to analyse the calculation method of friction excitation of helical gear pair with eccentricity error, the formulas of dynamic working pressure angle, transient pitch radius and transient transmission ratio were derived by establishing the geometrical model of a single helical gear pair. Describing the variational process of the contact line of a single tooth pair by piecewise function, the total contact line length of all meshing tooth pairs could be got by addition. The friction and frictional torque were obtained through cutting off the contact line into two segments with pitch line and compared with the results without considering eccentricity error. It shows that when considering the geometric eccentricity, the length of contact line, friction and frictional torque are more complicated, and the frequency spectrum peaks appear both at meshing frequency and shaft rotational frequency; and meanwhile, side frequency band occurs with the meshing frequency as central frequency and affects the stability of the transmission system.
Key words: helical gear pair    eccentricity error    the length of contact line    friction    frictional torque

1 考虑偏心误差时斜齿轮副几何模型

 图 1 含偏心误差的斜齿轮副几何模型 Fig. 1 Geometric model of helical gear with geometric eccentricity

 $\begin{array}{*{20}{c}} {{{O'}_1}\left( {{x_1},{y_1}} \right),{{O'}_2}\left( {{x_2},{y_2}} \right);}\\ {\left. \begin{array}{l} {x_1} = {e_1}\cos \left( {{\varphi _1} + {\omega _1}t} \right),\\ {y_1} = {e_1}\sin \left( {{\varphi _1} + {\omega _1}t} \right),\\ {x_2} = {e_2}\cos \left( {{\varphi _2} - {\omega _2}t} \right) + H,\\ {y_2} = {e_2}\sin \left( {{\varphi _2} - {\omega _2}t} \right). \end{array} \right\}} \end{array}$ (1)

 $l = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} .$ (2)

 $\psi = \arccos \left( {\frac{{a\cos {\alpha _{\rm{t}}}}}{l}} \right).$ (3)

 $\gamma = \arctan \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right).$ (4)

 ${{\alpha '}_{\rm{t}}} = \psi - \gamma .$ (5)

 $k = \frac{1}{{\tan {{\alpha '}_{\rm{t}}}}}.$ (6)

O1O2同时也是主、从动轮基圆圆心, 故两基圆方程可分别表示为

 $\left. \begin{array}{l} {\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = r_{{\rm{b1}}}^2\\ {\left( {x - {x_2}} \right)^2} + {\left( {y - {y_2}} \right)^2} = r_{{\rm{b2}}}^2 \end{array} \right\}.$ (7)

 $- \frac{{{x_{01}} - {x_1}}}{{{y_{01}} - {y_1}}} = - \frac{{{x_{02}} - {x_2}}}{{{y_{02}} - {y_2}}} = k.$ (8)

 $\begin{array}{l} {x_{01}} = {x_1} + \frac{k}{{\sqrt {{k^2} + 1} }}{r_{{\rm{b1}}}}\\ {y_{01}} = {y_1} - \frac{{{r_{{\rm{b1}}}}}}{{\sqrt {{k^2} + 1} }}\\ {x_{02}} = {x_2} - \frac{k}{{\sqrt {{k^2} + 1} }}{r_{{\rm{b2}}}}\\ {y_{02}} = {y_2} + \frac{{{r_{{\rm{b2}}}}}}{{\sqrt {{k^2} + 1} }}. \end{array}$ (9)

 $\frac{{y - {y_{02}}}}{{x - {x_{02}}}} = \frac{{{y_{02}} - {y_{01}}}}{{{x_{02}} - {x_{01}}}}.$ (10)

y=0即可得P点横坐标为

 ${x_P} = {x_{02}} - {y_{02}}\frac{{{x_{02}} - {x_{01}}}}{{{y_{02}} - {y_{01}}}}.$ (11)

 ${i_{12}} = \frac{{{\omega _1}}}{{{\omega _2}}} = \frac{{H - {x_P}}}{{{x_P}}}.$ (12)

 $v\left( t \right) = {\omega _1}{x_P}\cos \psi .$ (13)
2 时变接触线长度计算

 图 2 理想无误差斜齿轮单齿接触线长度 Fig. 2 Single-gear contact line length of helical gear without errors

 ${\varepsilon _\alpha } = \frac{{{L_{CD}}}}{{{P_{\rm{t}}}}},{\varepsilon _\beta } = \frac{{b\tan {\beta _b}}}{{{P_{\rm{t}}}}}.$ (14)

 图 3 偏心误差斜齿轮动态啮合线及动态接触线 Fig. 3 Dynamic mesh line and contact line of helical gear with geometric eccentricity

 图 4 2种不同重合度斜齿轮接触线位置 Fig. 4 Contact lines location of helical gear with two different contact ratios

 图 5 主从动轮中心距变大和变小时接触线长度示意图 Fig. 5 Contact line of pinion change with center distance become large or small

 ${L_i} = \left\{ \begin{array}{l} {D_i}/\sin {\beta _b},\\ \min \left( {b\tan {\beta _b},{L_{CD}}} \right)/\sin {\beta _b},\\ \left[ {{L_{CD}} - \left( {{D_i} - b\tan {\beta _b}} \right)} \right]/\sin {\beta _b},\\ 0. \end{array} \right.$ (15)

 $\left. \begin{array}{l} 0 \le {D_i} < \min \left( {b\tan {\beta _b},{L_{CD}}} \right)\\ \min \left( {b\tan {\beta _b},{L_{CD}}} \right) \le {D_i} < \max \left( {b\tan {\beta _b},{L_{CD}}} \right)\\ \max \left( {b\tan {\beta _b},{L_{CD}}} \right) \le {D_i} < \left( {{L_{CD}} + b\tan {\beta _b}} \right)\\ {D_i} \ge \left( {{L_{CD}} + b\tan {\beta _b}} \right) \end{array} \right\}$ (16)

 $L\left( t \right) = \sum\limits_{i = 0}^n {{L_i}\left( t \right)} .$ (17)
3 时变摩擦力和摩擦力矩计算

 ${F_{ij}} = \frac{F}{{L\left( t \right)}}{L_{ij}}.$ (18)
 图 6 含偏心误差斜齿轮接触线摩擦力方向和摩擦力臂 Fig. 6 Friction force direction and their corresponding moment arm of helical gear with geometric eccentricity

 ${r_1} = {x_P},{r_2} = L - {r_1}.$ (19)

 $\begin{array}{*{20}{c}} {{L_{CP}} = \sqrt {{{\left( {{x_{01}} - {x_P}} \right)}^2} + y_{01}^2} - {L_{{\rm{AC}}}}}\\ {{L_{PD}} = {L_{CD}} - {L_{CP}}.} \end{array}$ (20)

 ${F_{{\rm{f}}i}} = \mu \frac{F}{{L\left( t \right)}}\left( {{L_{i1}} - {L_{i2}}} \right).$ (21)

 $\left. \begin{array}{l} {L_{i1}} = 0,{L_{i2}} = {L_i},0 \le {D_i} < {L_{CP}};\\ {L_{i1}} = \left( {{D_i} - {L_{CP}}} \right)/\sin {\beta _b},{L_{i2}} = {L_i} - {L_{i1}},\\ {L_{CP}} \le {D_i} < {L_{CP}} + b\tan {\beta _b};\\ {L_{i1}} = {L_i},{L_{i2}} = 0,\\ {L_{CP}} + b\tan {\beta _b} \le {D_i} < {L_{CD}} + b\tan {\beta _b};\\ {L_{i1}} = 0,{L_{i2}} = 0,\\ {D_i} \le {L_{CD}} + b\tan {\beta _b}. \end{array} \right\}$ (22)

 ${F_{\rm{f}}} = \sum\limits_{i = 0}^n {{F_{{\rm{f}}i}}} .$ (23)

 $S = {r_1}\sin \psi .$ (24)

 ${T_{{\rm{f}}i}} = {F_{{\rm{f}}i2}}{S_{i2}} - {F_{{\rm{f}}i1}}{S_{i1}}.$ (25)

 $\left. \begin{array}{l} {S_{i1}} = 0,{S_{i2}} = S - {L_{CP}} + \frac{{{L_{i2}}\sin {\beta _b}}}{2},D < {D_i} < {L_{CP}};\\ {S_{i1}} = S + \frac{{{L_{i1}}\sin {\beta _b}}}{2},{S_{i2}} = S + \frac{{{L_{i2}}\sin {\beta _b}}}{2},\\ {L_{CP}} \le {D_i} < {L_{CP}} + b\tan {\beta _b};\\ {S_{i1}} = S + {L_{PD}} - \frac{{{L_{i1}}\sin {\beta _b}}}{2},{S_{i2}} = 0,\\ {L_{CP}} + b\tan {\beta _b} < {D_i} < {L_{CD}} + b\tan {\beta _b};\\ {S_{i1}} = 0,{S_{i2}} = 0,{D_i} \ge {L_{CD}} + b\tan {\beta _b}. \end{array} \right\}$ (26)

 ${T_{\rm{f}}} = \sum\limits_{i = 0}^n {{T_{{\rm{f}}i}}} .$ (27)
4 数值计算及分析

 $tan{\alpha _t} = \tan {\alpha _{\rm{n}}}\sec \beta ,$ (28)
 $\sin {\beta _b} = \tan \beta \cos {\alpha _{\rm{t}}}.$ (29)

 图 7 从动轮转速时域频域曲线 Fig. 7 Time-domain and frequency-domain of gear speed

 图 8 动态接触线总长度 Fig. 8 Total length of dynamic contact line

 图 9 不同几何中心距时齿轮副动态接触线总长度 Fig. 9 Total length of dynamic contact line with different center distance

 图 10 含不同偏心误差时频域曲线对比 Fig. 10 Compare with differences error in pinion or gear

 图 11 无误差单齿摩擦力曲线 Fig. 11 Friction force of single-gear without errors

 图 12 动态摩擦力曲线 Fig. 12 The curve of dynamic friction force

 图 13 动态摩擦力矩曲线 Fig. 13 The curve of dynamic friction torque

5 结论

(1) 考虑偏心误差的影响, 建立了斜齿轮副几何模型, 采用几何方法推导了含偏心误差的斜齿轮副动态啮合角、瞬时节圆半径以及瞬时传动比表达式, 并计算了主动轮恒定转速输入时, 从动轮转速变化规律.

(2) 通过将接触线划分为4个不同位置以分段函数形式表示各段接触线长度, 并进行叠加, 得出接触线总长度, 而后将各条接触线以节线为界分为两段, 分别计算各段摩擦力以及总摩擦力, 以各段接触线中点作为摩擦力的作用点计算各接触线摩擦力矩及总摩擦力矩.

(3) 对比分析含偏心误差与不含偏心误差的斜齿轮接触线长度、摩擦力及摩擦力矩的时域频域曲线, 结果表明：含偏心误差的斜齿轮接触线长度、摩擦力及摩擦力矩都比不含偏心误差时更加复杂, 其频域不仅在啮合频率及其倍频处出现峰值, 在输入及输出轴转频处也出现峰值, 同时还在啮合频率及其倍频两侧出现许多间隔为转频的边频, 影响了传动系统的平稳性.

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