Cutting torque and tangential cutting force coefficient identification from spindle motor current

Abstract

This article presents an enhanced methodology for cutting torque prediction from the spindle motor current, readily available in modern machine tool controllers. This methodology includes the development of the spindle power model which takes into account all mechanical and electrical power losses in a spindle motor for high-speed milling. The predicted cutting torque is further used to identify tangential cutting force coefficients in order to predict accurately the cutting forces and chatter-free regions for milling process planning purposes. The developed model is compared with other studies available in the literature, and it demonstrates significant improvements in terms of the completeness and accuracy achieved. The developed model is also validated experimentally, and the obtained results show good compliance between the predicted and the measured cutting torque. The developed enhanced procedure is very appealing for industrial implementation for cutting torque/force monitoring and tangential cutting force coefficient identification.

Appendix

1.1 Calculation of dynamic values of bearing contact angles

In Fig. 11a, free-state (nominal) value of the bearing contact angle (α _f) is illustrated. This condition refers to when the bearing is immobile and there is no preload. The outer (inner) contact angle is the one that the line connecting the contact point A on the outer ring (B on the inner ring) and the center of the ball forms with the normal direction to the bearing axis.

Fig. 11

Free-state, static (a) and dynamic (b) values of the bearing contact angle. Redrawn from [15]

In free-state conditions, both inner and outer contact angles have the same value, equal to α _f. The free-state angle of the bearing, a _f, is defined by the following trigonometric expression:

$$ \cos {\alpha_{{\text f} }} = \frac{{t - \frac{{{P_d}}}{2}}}{t} $$

(5)

where:

• t is the distance between the centers of the curvature of the inner and the outer raceway grooves, which is equal to:

  $$ t = {R_o} + {R_i} - {d_b} $$

  (6)

• R _o and R _i are, respectively, the radii of the outer and the inner raceway groove curvatures.

• P _d is the diametric clearance, which is, by definition [14]

  $$ {P_{{\text d} }} = {d_{{Bo}}} - {d_{{Bi}}} - 2{d_b} = 2\left( {{b_1} + {b_2}} \right) $$

  (7)

• d _Bo and d _Bi are the outer and inner ring raceway contact diameters (in millimeters), respectively

• d _b is the ball diameter (in millimeters)

• b ₁ and b ₂ are clearances between the ball and the raceways (in millimeters)

Figure 11a also depicts the condition where the bearing is still immobile, but there is a static preload, F _Bp, in the axial direction. An assumption is made that this load is equally distributed on each of the balls, where N _b stands for their number. Under the preload, the inner ring moves with respect to the outer ring in the direction of the bearing axis, X. Both contact points and the center of the ball are still collinear, but the angle (α _st) that this line forms with the direction normal to the bearing axis (Y) is greater than the free-state angle, α _f.

The static value of the bearing contact angle, α _st, can be calculated using the following expression [14]:

$$ \frac{{{F_{{Bp}}}}}{{{N_b}{K_a}d_{{\text b} }^{{2n - 1}}}} = \sin {\alpha_{{{\text st} }}}{\left( {\frac{{\cos {\alpha_f}}}{{\cos {\alpha_{{st}}}}} - 1} \right)^n} $$

(8)

where K _a is the axial deflection constant (in megapascals) that can be obtained from the corresponding experimentally determined diagrams [14] as a function of the total curvature B

$$ B = \frac{{{R_o}}}{{{d_b}}} + \frac{{{R_i}}}{{{d_b}}} - 1 $$

(9)

Equation 8 can be solved numerically by the Newton–Raphson method. The equation to be satisfied iteratively is

$$ \alpha_{\text{st}}^{\prime } = {\alpha_{{st}}} + \frac{{\frac{{{F_{{Bp}}}}}{{{N_b}{K_a}d_{{\text b} }^{{2n - 1}}}} - \sin {\alpha_{{st}}}{{\left( {\frac{{\cos {\alpha_f}}}{{\cos {\alpha_{{st}}}}} - 1} \right)}^n}}}{{\cos {\alpha_{{st}}}{{\left( {\frac{{\cos {\alpha_f}}}{{\cos {\alpha_{{st}}}}} - 1} \right)}^n} + n{{\tan }^2}{\alpha_{{st}}}{{\left( {\frac{{\cos {\alpha_f}}}{{\cos {\alpha_{{st}}}}} - 1} \right)}^{{n - 1}}}\cos {\alpha_f}}} $$

(10)

which is satisfied when $\alpha_{\text{st}}^{\prime } - {\alpha_{{st}}}$ is zero.

The main case that we are interested in is related to the dynamic state, when the bearing rotates under the load. The load on a bearing consists of an axial force, F _Sa, equally distributed on each of N _b balls and radial forces, F _ro and F _ri, that act on a ball through the outer and inner rings, respectively (Fig. 11b). The assumptions made are that the bearing has a constant rotational speed around the X-axis, the motion is without vibrations, and no sliding motion is supposed to appear at points A and B [15].

Balls follow the cage orbital motion around the X-axis. Due to friction forces in contact points, each ball also rotates around its own axis. This leads to the occurrence of the gyroscopic moments on the balls (Fig. 12a). Because of the inertial force (centrifugal force, F _C, in Fig. 12a), the outer and inner contact angles have now two different values, α _o and α _i.

Fig. 12

Free body diagram of an angular contact ball bearing. Redrawn from [14]

In order to calculate the bearing contact angles, a free body diagram is constructed in Fig. 12a. At the contact points A and B, the outer and inner contact actions are supposed to be reduced to the resultant forces F _o and F _i (acting on each ball). These forces are decomposed, respectively, into normal loads, W _o and W _i, and tangential loads, T _o and T _i. The result of the centrifugal force, F _C, is directed along the Y-axis and is applied at point C (center of the ball).

The assumption that the tangential forces compensate the gyroscopic moment allows for a simplification of the free body diagram [15]. Thus, we get the system of forces represented in Fig. 12b, which will help us determine the set of equations necessary for the calculation of contact angles.The sum of force projections in direction Y′ is

$$ {F_{{ri}}} - {F_{{ro}}} + {F_{{\text C} }} = 0{ } $$

(11)

From Fig. 12b, we also get

$$ \tan {\alpha_i} = \frac{{\frac{{{F_{{Sa}}}}}{{{N_b}}}}}{{{F_{{ri}}}}} \Rightarrow {F_{{ri}}} = \frac{{{F_{{Sa}}}}}{{{N_b}\tan {\alpha_{{\text i} }}}} $$

(12)

$$ \tan {\alpha_o} = \frac{{\frac{{{F_{{Sa}}}}}{{{N_b}}}}}{{{F_{{ro}}}}} \Rightarrow {F_{{ro}}} = \frac{{{F_{{Sa}}}}}{{{N_b}\tan {\alpha_o}}} $$

(13)

By introducing Eqs. 12 and 13 into Eq. 11, we get

$$ {F_{{ri}}} - {F_{{ro}}} + {F_C} = \frac{{{F_{{Sa}}}}}{{{N_{{\text b} }}\tan {\alpha_i}}} - \frac{{{F_{{{\text Sa} }}}}}{{{N_b}\tan {\alpha_o}}} + {F_C} = 0{ } \Rightarrow $$

$$ \frac{1}{{\tan {\alpha_{{\text o} }}}} - \frac{1}{{\tan {\alpha_i}}} = \frac{{{F_C} \cdot {N_b}}}{{{F_{{Sa}}}}} $$

(14)

where:

• F _C is the centrifugal force (in newton) which can be calculated as [19] (p. 175)

  $$ {F_C} = \frac{{\pi {\rho_b}d_{{\text b} }^3\omega_{{\text C} }^2{d_{{{\text Bm} }}}}}{{12}} $$

  (15)

• ρ _b is the mass density of balls/rollers (in kilograms per cubic meter)

• ω _C is the angular velocity of the cage (radian per second)

  $$ {\omega_C} = \frac{{2\pi {n_C}}}{{60}} = \frac{{\pi {n_C}}}{{30}} $$

  (16)

• n _C (1/s) is the rotational speed of the cage (orbital ball/roller speed) [14]

  $$ {n_C} = \frac{1}{2}\left[ {{n_i}\left( {1 - {\gamma_{{\text b} }}} \right) + {n_o}\left( {1 + {\gamma_{{\text b} }}} \right)} \right] $$

  (17)

• n _o and n _i are the rotational speeds of the outer and inner rings, respectively

• γ _b is a geometrical parameter

  $$ {\gamma_b} = \frac{{{d_b}\cos {\alpha_{{{\text st} }}}}}{{{d_{{Bm}}}}} $$

  (18)

For machine tools, n _o = 0 and Eq. 17 becomes

$$ {n_C} = \frac{{{n_{{\text i} }}}}{2}\left( {1 - {\gamma_b}} \right) = {n_C} = \frac{{{n_{{\text M} }}}}{2}\left( {1 - {\gamma_{{\text b} }}} \right) $$

(19)

where n _M is the number of rotations of the spindle shaft.

Bossmanns [19] makes an assumption that the changes of the contact angles in the outer and inner contact points follow the rules:

$$ {\alpha_o} = {\alpha_{{{\text st} }}} + \Delta {\alpha_o} $$

(20)

$$ {\alpha_i} = {\alpha_{{st}}} + \Delta {\alpha_i} $$

(21)

$$ \Delta {\alpha_o} = - \Delta {\alpha_i} $$

(22)

which leads to the conclusion

$$ {\alpha_{{\text o} }} + {\alpha_i} = 2{\alpha_{{st}}} $$

(23)

Equations 10 and 23 constitute a sufficient system for the calculation of the dynamic load contact angles, which can be solved by numerical iteration.