1 Introduction

Power loss in a gear drive system includes windage loss, oil loss, bearing power loss, gear meshing power loss, and so forth, of which the gear meshing power loss is a major constituent [1]. Many domestic and foreign scholars have proposed several theoretical models to predict the gear meshing power loss. To improve calculation efficiency, the normal load model of the tooth surface and friction coefficient model is included in the meshing power loss model.

At present, the friction coefficient model of the tooth surface mainly includes: the Coulomb friction coefficient model, the Buckingham [2] friction coefficient semi-empirical formula, the Benedict and Kelly [3] friction coefficient model, and the Xu [4] EHL friction coefficient model.

Several techniques have been implemented to deal with the normal load of the tooth surface. Liang [5], Marimuthu and Muthuveerappan [6] carried out load-bearing contact analysis of internal and external meshing gear pairs based on the finite element principle. The method assumes that the load is evenly distributed on both surfaces of the double-tooth meshing and static load is used as a normal load for the tooth surface. Kolivand et al. [7] proposed a gear meshing power loss model by combining the load-bearing contact model and the hybrid elastohydrodynamic lubrication (EHL) model. Lin et al. [

$$ {\mathbf{X}} = \left\{ {x_{1} ,y_{1} ,z_{1} ,\theta _{1} ,x_{2} ,y_{2} ,z_{2} ,\theta _{2} ,x_{3} ,y_{3} ,z_{3} ,\theta _{3} ,x_{4} ,y_{4} ,z_{4} ,\theta _{4} } \right\}^{T} $$
(1)

Then, the relative displacements δ12 and δ34 of the double helical gear along the mesh line can be expressed as

$$ \left\{ \begin{gathered} \delta _{{12}} = (y_{1} + r_{{b1}} \theta _{1} - y_{2} + r_{{b2}} \theta _{2} )\cos \beta _{b} - (z_{1} - z_{2} )\sin \beta _{b} - e_{1} \hfill \\ \delta _{{34}} = (y_{3} + r_{{b3}} \theta _{3} - y_{4} + r_{{b4}} \theta _{4} )\cos \beta _{b} + (z_{3} - z_{4} )\sin \beta _{b} - e_{2} \hfill \\ \end{gathered} \right. $$
(2)

In the above formula, rb1, rb2, rb3, and rb4 are the base radii of each gear, while e1 and e2 are the gear meshing errors. According to the gear manufacturing accuracy level, the error is expressed as a simple harmonic function [17].

$$ e(t) = e_{a} \cos (\omega t + \varphi ) $$
(3)

where ea is the meshing error amplitude, ω is the gear mesh engagement frequency in rad/s, and φ is the meshing phase angle.

Therefore, the elastic meshing force Flkm, Frkm and the viscous meshing force Flcm, Frkm of the left and right gear pairs can be expressed as

$$ \left\{ \begin{gathered} F_{{km}}^{l} = k_{m}^{l} \delta _{{12}} \begin{array}{*{20}c} {} & {F_{{km}}^{r} = k_{m}^{r} \delta _{{34}} } \\ \end{array} \hfill \\ F_{{cm}}^{l} = c_{m}^{l} \dot{\delta }_{{12}} \begin{array}{*{20}c} {} & {F_{{cm}}^{r} = c_{m}^{r} \dot{\delta }_{{34}} } \\ \end{array} \hfill \\ \end{gathered} \right. $$
(4)

where Flkm, Frkm, Flcm, and Frcm are, respectively, the meshing stiffness and meshing dam** of the left and right gear pairs. The meshing stiffness value is obtained by TCA.

Based on the above analysis, the dynamic differential equation of the external meshing double helical gear pair is given by:

$$ \left\{ \begin{gathered} m_{1} \ddot{x}_{1} + k_{{1x}} x_{1} + k_{{13x}} (x_{1} - x_{3} ) + c_{{1x}} \dot{x}_{1} + c_{{13x}} (\dot{x}_{1} - \dot{x}_{3} ) = F_{f}^{l} \hfill \\ m_{1} \ddot{y}_{1} + k_{{1y}} y_{1} + k_{{13y}} (y_{1} - y_{3} ) + c_{{1y}} \dot{y}_{1} + c_{{13y}} (\dot{y}_{1} - \dot{y}_{3} ) = - (F_{{km}}^{l} + F_{{cm}}^{l} )\cos \beta _{b} \hfill \\ m_{1} \ddot{z}_{1} + k_{{13z}} (z_{1} - z_{3} ) + c_{{13z}} (\dot{z}_{1} - \dot{z}_{3} ) = (F_{{km}}^{l} + F_{{cm}}^{l} )\sin \beta _{b} \hfill \\ I_{1} \ddot{\theta }_{1} = - (F_{{km}}^{l} + F_{{cm}}^{l} )\cos \beta _{b} r_{{b1}} - \frac{{T_{{in}} }}{2} + T_{{f1}} - k_{{J13}} (\theta _{1} - \theta _{3} ) - c_{{J13}} (\dot{\theta }_{1} - \dot{\theta }_{3} ) \hfill \\ m_{2} \ddot{x}_{2} + k_{{2x}} x_{2} + k_{{24x}} (x_{2} - x_{4} ) + c_{{2x}} \dot{x}_{2} + c_{{24x}} (\dot{x}_{2} - \dot{x}_{4} ) = - F_{f}^{l} \hfill \\ m_{2} \ddot{y}_{2} + k_{{2y}} y_{2} + k_{{24y}} (y_{2} - y_{4} ) + c_{{2y}} \dot{y}_{2} + c_{{24y}} (\dot{y}_{2} - \dot{y}_{4} ) = (F_{{km}}^{l} + F_{{cm}}^{l} )\cos \beta _{b} \hfill \\ m_{2} \ddot{z}_{2} + k_{{2z}} z_{2} + k_{{24z}} (z_{2} - z_{4} ) + c_{{2z}} \dot{z}_{2} + c_{{24z}} (\dot{z}_{2} - \dot{z}_{4} ) = - (F_{{km}}^{l} + F_{{cm}}^{l} )\sin \beta _{b} \hfill \\ I_{2} \ddot{\theta }_{2} = - (F_{{km}}^{l} + F_{{cm}}^{l} )\cos \beta _{b} r_{{b2}} - \frac{{T_{{out}} }}{2} + T_{{f2}} - k_{{J24}} (\theta _{2} - \theta _{4} ) - c_{{J24}} (\dot{\theta }_{2} - \dot{\theta }_{4} ) \hfill \\ m_{3} \ddot{x}_{3} + k_{{3x}} x_{3} + k_{{13x}} (x_{3} - x_{1} ) + c_{{3x}} \dot{x}_{3} + c_{{13x}} (\dot{x}_{3} - \dot{x}_{1} ) = F_{f}^{r} \hfill \\ m_{3} \ddot{y}_{3} + k_{{3y}} y_{3} + k_{{13y}} (y_{3} - y_{1} ) + c_{{3y}} \dot{y}_{3} + c_{{13y}} (\dot{y}_{3} - \dot{y}_{1} ) = - (F_{{km}}^{r} + F_{{cm}}^{r} )\cos \beta _{b} \hfill \\ m_{3} \ddot{z}_{3} + k_{{13z}} (z_{3} - z_{1} ) + c_{{13z}} (\dot{z}_{3} - \dot{z}_{1} ) = - (F_{{km}}^{r} + F_{{cm}}^{r} )\sin \beta _{b} \hfill \\ I_{3} \ddot{\theta }_{3} = - (F_{{km}}^{r} + F_{{cm}}^{r} )\cos \beta _{b} r_{{b3}} - \frac{{T_{{in}} }}{2} + T_{{f3}} - k_{{J13}} (\theta _{3} - \theta _{1} ) - c_{{J13}} (\dot{\theta }_{3} - \dot{\theta }_{1} ) \hfill \\ m_{4} \ddot{x}_{4} + k_{{4x}} x_{4} + k_{{24x}} (x_{4} - x_{2} ) + c_{{4x}} \dot{x}_{4} + c_{{24x}} (\dot{x}_{4} - \dot{x}_{2} ) = - F_{f}^{l} \hfill \\ m_{4} \ddot{y}_{4} + k_{{4y}} y_{4} + k_{{24y}} (y_{4} - y_{2} ) + c_{{4y}} \dot{y}_{4} + c_{{24y}} (\dot{y}_{4} - \dot{y}_{2} ) = (F_{{km}}^{r} + F_{{cm}}^{r} )\cos \beta _{b} \hfill \\ m_{4} \ddot{z}_{4} + k_{{4z}} z_{4} + k_{{24z}} (z_{4} - z_{2} ) + c_{{4z}} \dot{z}_{4} + c_{{24z}} (\dot{z}_{4} - \dot{z}_{2} ) = (F_{{km}}^{r} + F_{{cm}}^{r} )\sin \beta _{b} \hfill \\ I_{4} \ddot{\theta }_{4} = - (F_{{km}}^{r} + F_{{cm}}^{r} )\cos \beta _{b} r_{{b4}} - \frac{{T_{{out}} }}{2} + T_{{f4}} - k_{{J24}} (\theta _{4} - \theta _{2} ) - c_{{J24}} (\dot{\theta }_{4} - \dot{\theta }_{2} ) \hfill \\ \end{gathered} \right. $$
(5)

where mi and Ii (i = 1,2,3,4) are the mass and moment of inertia of gear i, respectively; Flf and Frf are the tooth surface friction forces of the left and right gear pairs, respectively; and Tf1, Tf2, Tf3, and Tf4 are the friction moments of the gear surface friction force on each gear.

The mass matrix M, stiffness matrix K, dam** matrix C, and excitation force matrix F can be extracted using the above formula. Since the dam** coefficient is difficult to determine, Rayleigh dam** is used in this study. The Rayleigh dam** coefficients α and β are used as reference [32].

$$ {\mathbf{C}} = \alpha {\mathbf{M}} + \beta {\mathbf{K}} $$
(6)

The differential equations can be arranged as a matrix:

$$ {\mathbf{M\ddot{X}}} + {\mathbf{KX}} + {\mathbf{C\dot{X}}} = {\mathbf{F}} $$
(7)

After eliminating the displacement of the rigid body, the displacement and velocity vectors are obtained using the Runge–Kutta method and then inserted into Eq. (4) to determine the dynamic meshing force.

2.2 Calculation of friction force and sliding friction power loss

From the dynamic equation in Sect. 2.1, it can be observed that the tooth surface meshing force and friction force are unknown, while the tooth surface friction force and friction coefficient are all dependent on the meshing force. When applying the Runge–Kutta method, the general processing method is to use the meshing force obtained from the solution of the previous loop to calculate the tooth surface friction force. The friction force is then used for the current cycle calculation in an iterative process until the two curves achieve reasonable stability.

Owing to the huge amount of finite element calculations, the meshing implemented for TCA should not be too dense. Therefore, the step size of the Runge–Kutta method will be much smaller than the step length of TCA regardless of whether a fixed-step or variable-step solver is used. To solve the data mismatch problem, it is considered that the load distribution in the vicinity of the discrete contact line has the same tendency since the discrete contact line load distribution curve is non-dimensionalized. According to the load of the current meshing position and the length of the contact line, the dimensionless load distribution curve is dimensioned such that the load distribution curve of the current meshing position is determined.

Assuming that the total meshing force is Fm at a certain time t, the expression for the friction force Fs on a single contact line is given as:

$$ F_{s} (t) = \int_{l} {\text{sgn} [v_{s} (u,t)]\mu (u,t)q_{L} (t)q_{C} (u,t)F_{m} du} $$
(8)

where qL(t) is the tooth-to-tooth load distribution rate at time t; qC(u, t) is the tooth surface load distribution function obtained from the loaded TCA (LTCA) model; u is the positional argument, which indicates the length from the starting point of the contact line; l is the contact line length of the current meshing position; vs(u, t) is the relative sliding speed of the tooth surface; and μ(u, t) is the friction coefficient. The above parameters are all functions of the position and time. The friction coefficient model uses the friction coefficient calculation model proposed by Xu [4] based on the theory of elastohydrodynamic lubrication. The term sgn() is a signum function whose expression is:

$$ \text{sgn} (x) = \left\{ \begin{gathered} 1 \hfill \\ 0 \hfill \\ - 1 \hfill \\ \end{gathered} \right.\begin{array}{*{20}c} {} & \begin{gathered} x > 0 \hfill \\ x = 0 \hfill \\ x < 0 \hfill \\ \end{gathered} \\ \end{array} $$
(9)

The friction torques of the single contact line on the driving wheel and the driven wheel are Ts1 and Ts2, respectively, and are given as:

$$ \left\{ \begin{gathered} T_{{s1}} (t) = \int_{l} {\text{sgn} [v_{s} (u,t)]\mu (u,t)q_{L} (t)q_{C} (u,t)F_{m} R_{{t1}} (u,t)du} \hfill \\ T_{{s2}} (t) = \int_{l} {\text{sgn} [v_{s} (u,t)]\mu (u,t)q_{L} (t)q_{C} (u,t)F_{m} R_{{t2}} (u,t)du} \hfill \\ \end{gathered} \right. $$
(10)

where Rt1(u, t) and Rt2(u, t) are, respectively, the force arm of the friction force on the driving wheel and the driven wheel at any position of the contact line at time t.

If the friction force curves and friction torque curves of adjacent contact lines differ in time by a meshing period Tm, then the total friction force and friction torque of the driving wheel and the driven wheel are:

$$ \left\{ \begin{gathered} F_{f} = \sum\limits_{{i = - N}}^{N} {F_{s} } (t + iT_{m} ) \hfill \\ T_{{f1}} = \sum\limits_{{i = - N}}^{N} {T_{{s1}} } (t + iT_{m} ) \hfill \\ T_{{f2}} = \sum\limits_{{i = - N}}^{N} {T_{{s2}} } (t + iT_{m} ) \hfill \\ \end{gathered} \right. $$
(11)

where N is the maximum number of teeth engaged in meshing at the same time.

The friction force at each point of the contact line and the relative sliding speed are different. The instantaneous power loss calculation integrates the power loss at each point along the contact line. The formula for the power loss of a single contact line is as follows:

$$ P_{s} (t) = \int_{l} {\mu (u,t)q_{L} (t)q_{C} (u,t)F_{m} v_{s} (u,t)du} $$
(12)

Similarly, the total power loss at any time t is:

$$ P_{f} = \sum\limits_{{i = - N}}^{N} {P_{s} } (t + iT_{m} ) $$
(13)

The average power loss is obtained by integrating the instantaneous power loss along the time axis and then averaging as given below:

$$ P = \frac{1}{{T_{m} }}\int_{{t_{1} }}^{{t_{1} + T_{m} }} {P_{f} (t)dt} $$
(14)

The average meshing efficiency can be expressed as:

$$ \eta = \left( {1 - \frac{P}{{P_{{in}} }}} \right) \times 100\% $$
(15)

where Pin is the input power.