2.1 Equivalent Dynamic Model
To study the variation in the dynamic characteristics of a vertical ball screw feed system under upward acceleration and deceleration, the structure of the system, as shown in Figure 1, is first described. The vertical ball screw feed system consists of a spindle system with a spindle unit, screw shaft, screwnut joints, upperend bearing joints, lowerend bearing joints, linear rolling guides, slider blocks, a nut bracket, and a servo motor. Compared with the stiffness in the two other directions (X and Y), the transmission stiffness in the Z direction is the lowest owing to the stiffness of the screwnut joints and bearing joints. The contact stiffness of the joints is affected not only by the inertial force, but also by the spindle system weight during acceleration and deceleration. In addition, the contact stiffness is also affected by the friction force between the guide and the slide. However, the friction is relatively small compared to the inertial force and spindle system weight, and can therefore be neglected.
The equivalent dynamic model of the vertical ball screw feed system is established and shown in Figure 2. In the dynamic model, the screw shaft, screwnut joints, and bearing joints are simplified as lumped spring elements. k_{ls}(z), k_{rs}(z) and C_{ls}, C_{rs} respectively represent the axial stiffness and damping of the screw shaft at both sides of the nut. k_{lb}(a), k_{rb}(a), k_{nut}(a) and C_{lb}, C_{rb}, and C_{nut} represent the equivalent axial stiffness and damping of the upper and lowerend bearing joints and screwnut joints, respectively. The components of the spindle system are modeled as an equivalent lumped mass m. α denotes the acceleration in the transmission direction, z the distance between the nut and the lowerend bearing joints, and L the distance between the upper and lowerend bearing joints. The influence of the servo stiffness is neglected.
2.2 Dynamic Equation Considering the Effect of Acceleration
To analyze the influence of the inertial force induced by acceleration and deceleration on the natural frequency of the system, a variablecoefficient dynamic equation of the feed system can be established based on the equivalent dynamic model and the D'Alembert principle, as follows:
$$ m\ddot{z} + c_{e} \dot{z} + k_{e} \left( a \right)z = 0, $$
(1)
where k_{e} is the total stiffness coefficient, which is determined by the acceleration of a. In this study, only the natural frequency of the ball screw feed system is considered. Therefore, the damping coefficient, c_{e}, is ignored.
2.3 Calculation of Stiffness Coefficient
The total transmission stiffness in the transmission direction comprises the stiffness of the screwnut joints, bearing joints, and screw shaft. Therefore, the stiffness of each component and the transmission stiffness need to be calculated.
2.3.1 Equivalent Axial Stiffness of the ScrewNut Joints
The force diagram of the gaskettype doublenut screwnut joints considering the spindle system weight and inertial force under upward acceleration and deceleration is shown in Figure 3. When the spindle system accelerates upward, the inertial force acts on the screwnut joints. The action direction of the force is downward, which is the same direction as that of the spindle system weight. When the spindle system decelerates upwards, the direction of the inertial force is upward, which is opposite to the direction of the spindle system weight. In addition, the preload produced by the gasket acting on nut A is downward, and it acts on nut B upward.

(1)
Upward acceleration
Under acceleration, the vertical ball screw feed system is affected by the weight of the spindle system, the inertial force due to acceleration, and the preload of the screwnut joints. The ball between the screw shaft and nut deforms elastically. As shown in Figure 3, the direction of the preload acting on nut A is the same as that of the spindle system weight and inertial force, while the direction of the preload acting on nut B is opposite to the direction of the spindle system weight and inertial force. According to the Hertz contact theory [29], the initial normal force and deformation of each ball in nuts A and B can be obtained using the following equations:
$$ \left\{ \begin{aligned} P_{A} & = \frac{P + m \cdot g}{{N \cdot \sin \alpha \cdot \cos \varphi }} = \frac{{P_{d} \cdot c_{p} + m \cdot g}}{{\left( {i\frac{{\uppi \cdot d_{s0} }}{{d_{sb} \cos \varphi }}} \right) \cdot \sin \alpha \cdot \cos \varphi }}, \\ \delta_{A} & = \left( {\frac{1}{{K_{h} }}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot P_{A}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} , \\ \end{aligned} \right. $$
(2)
$$ \left\{ \begin{aligned} P_{B} & = \frac{P  m \cdot g}{{N \cdot \sin \alpha \cdot \cos \varphi }} = \frac{{P_{d} \cdot c_{p}  m \cdot g}}{{\left( {i\frac{{\uppi \cdot d_{s0} }}{{d_{sb} \cos \varphi }}} \right) \cdot \sin \alpha \cdot \cos \varphi }}, \\ \delta_{B} & = \left( {\frac{1}{{K_{h} }}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot P_{B}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} , \\ \end{aligned} \right. $$
(3)
where d_{s0} and d_{sb} are the diameters of the screw and ball, respectively. α is the contact angle between the ball and the raceway. φ is the pitch angle of the screw. i is the number of rings in a single nut. d_{s0} and d_{sb} are the diameters of the screw shaft and ball, respectively. N is the number of balls in a single nut. P_{d} is the rated dynamic load of the screwnut joints. P is the preload of nut A and nut B. P_{A} and P_{B} are the initial normal forces in a single ball of nut A and nut B, respectively. c_{p} is the rated dynamic load factor. \(\delta_{A}\) and \(\delta_{B}\) are the normal deformations of a ball in nuts A and B, respectively. \(K_{h}\) is the Hertz contact coefficient, which mainly depends on the contact shape and the material properties of the screwnut joints [30].
When the spindle system accelerates upward in the Z direction with the magnitude a, the load of each ball in nuts A and B will change owing to the action of the inertial force. The normal force, deformation, and contact stiffness of each ball between the screw and nut A can be derived as follows:
$$ \left\{ \begin{aligned} P_{AT} & = P_{A} + P_{AG} = \frac{P + m \cdot g}{{N \cdot \sin \alpha \cdot \cos \varphi }} + \frac{m \cdot a}{{N \cdot \sin \alpha \cdot \cos \varphi }}, \\ \delta_{AT} & = \left( {\frac{1}{{K_{h} }}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot P_{AT}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} , \\ k_{nA} & = \frac{3}{2}K_{h} \cdot \delta_{AT}^{{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern\nulldelimiterspace} 2}}} = \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot P_{AT}^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} , \\ \end{aligned} \right. $$
(4)
where P_{AG} is the additional normal force on each ball in nut A generated by the inertial force; and P_{AT}, \(\delta_{AT}\), and \(k_{nA}\) are the normal force, contact deformation, and contact stiffness of each ball in nut A, respectively.
Based on the force decomposition, the equivalent axial stiffness of each ball in nut A and the stiffness of nut A can be calculated as
$$\left\{ \begin{aligned} k_{axA} & = k_{nA} \sin^{2} \alpha \cdot \cos^{2} \varphi , \\ K_{axA \uparrow } & = \frac{3}{2}k_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot \left( {\left( {P + m \cdot (g + a)} \right) \cdot \sin^{5} \alpha \cdot \cos^{5} \varphi } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \left( {i \cdot \frac{{\uppi d_{s0} }}{{d_{sb} \cos \varphi }}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} , \\ \end{aligned} \right.$$
(5)
where \(k_{axA}\) and \(K_{axA \uparrow }\) are the equivalent axial stiffness of a single ball in nut A and the equivalent axial stiffness of nut A, respectively.
Applying the boundary conditions, the elastic deformation of a single rolling ball in nut B can be obtained using deformation compatibility theory as
$$ \left\{ \begin{aligned} \delta_{BT} & = \delta_{B}  \delta_{BF} ,\left( {\delta_{BF} < \delta_{B} } \right), \\ \delta_{BT} & = 0,\quad \left( {\delta_{BF} \ge \delta_{B} } \right). \\ \end{aligned} \right. $$
(6)
If δ_{BF} ≥ δ_{B}, the load on each ball in nut B is equal to zero, and the corresponding contact stiffness will be abrupt. Therefore, the equivalent axial stiffness of nut B can be obtained as
$$K_{axB \uparrow } = \left\{ \begin{aligned} & \frac{3}{2}k_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot \left( {2 \cdot P^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {P + m \cdot (g + a)} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern\nulldelimiterspace} 2}}} \\ & \quad \cdot \left( {\sin^{5} \alpha \cdot \cos^{5} \varphi \cdot \left( {i \cdot \frac{{\uppi d_{s0} }}{{d_{sb} \cos \varphi }}} \right)^{2} } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} ,\;\left( {\delta_{BF} < \delta_{B} } \right), \\ & 0,\quad \left( {\delta_{BF} \ge {\kern 1pt} \delta_{B} } \right). \\ \end{aligned} \right.$$
(7)
The stiffness of the screwnut joints is mainly determined by \(K_{axA \uparrow }\) and \(K_{axB \uparrow }\) and can be expressed as
$$ K_{nut \uparrow } \left( a \right) = \max \left( {K_{axA \uparrow } ,K_{axB \uparrow } } \right) \cdot c_{wn} , $$
(8)
where c_{wn} is the weight coefficient of the equivalent axial stiffness of the screwnut joints and has the value of 0.9.

(2)
Upward deceleration
As shown in Figure 3, the principle of calculating the stiffness of nuts A and B under upward deceleration is the same as that under upward acceleration. Both calculations are based on the combination of Hertz contact theory and deformation compatibility theory. The only difference is that the direction of the inertial force acting on nuts A and B is switched. In this case, the equivalent axial stiffnesses of nuts A and B are calculated as
$$K_{axA \downarrow } = \left\{ \begin{aligned} & \frac{3}{2}k_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot \left( {\left( {P + m\left( {g  a} \right)} \right) \cdot \sin^{5} \alpha \cdot \cos^{5} \varphi } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \left( {i\frac{{\uppi d_{s0} }}{{d_{sb} \cos \varphi }}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} ,\;\left( {a \le g} \right), \\ & \frac{3}{2}k_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot \left( {2 \cdot P^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {P + m\left( {a  g} \right)} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern\nulldelimiterspace} 2}}} \\ & \quad \cdot \left( {\sin^{5} \alpha \cdot \cos^{5} \varphi \left( {i\frac{{\uppi d_{s0} }}{{d_{sb} \cos \varphi }}} \right)} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} ,{\kern 1pt} \;\left( {a > g} \right), \\ \end{aligned} \right.$$
(9)
$$K_{axB \downarrow } = \left\{ \begin{aligned} & \frac{3}{2}k_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot \left( {2 \cdot P^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {P + m\left( {g  a} \right)} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern\nulldelimiterspace} 2}}} \\ & \quad \cdot \left( {\sin^{5} \alpha \cdot \cos^{5} \varphi \left( {i\frac{{\uppi d_{s0} }}{{d_{sb} \cos \varphi }}} \right)^{2} } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} ,{\kern 1pt} \;\left( {a \le g} \right), \\ & \frac{3}{2}k_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot \left( {\left( {\left( {P + m\left( {a  g} \right)} \right) \cdot \sin^{5} \alpha \cdot \cos^{5} \varphi } \right)} \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \left( {i\frac{{\uppi d_{s0} }}{{d_{sb} \cos \varphi }}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} ,\;\left( {a > g} \right). \\ \end{aligned} \right.$$
(10)
The stiffness of the screwnut joints depends mainly on the values of \(K_{axA \downarrow }\) and \(K_{axB \downarrow }\) and can be expressed as
$$ K_{nut \downarrow } \left( a \right) = \max \left( {K_{axA \downarrow } ,K_{axB \downarrow } } \right) \cdot c_{wn} . $$
(11)
2.3.2 Equivalent Axial Stiffness of the Bearing Joints
In this study, the angular contact ball bearings at both ends have the same specifications and protrusions. The pretightening method is applied to eliminate the initial axial clearance. However, the configurations differ at the two ends. The configuration of the upperend bearing joints is backtoback (DB), as shown in Figure 4(a), while that in the lowerend bearing joints is facetoface(DF), as shown in Figure 4(b).

(1)
Upward acceleration
Figure 5(a)–(b) show the force analysis of the upper and lowerend bearing joints, respectively. As shown in Figure 5(a)–(b), there are three forces acting on the upper and lowerend bearing joints, namely, the screw tension force, spindle system weight, and inertial force. Under the action of the screwtension force, the clearance formed by the protrusion of bearings A and B, as well as that between bearings C and D, is eliminated. The other two forces act on the bearing joints as external forces. The direction of the spindle system weight is downward, and the direction of the inertial force is upward. Both forces are equally divided at the ends of the lead screw.
According to the force equilibrium in the vertical direction, the normal contact force of each ball in bearings A and D can be obtained as
$$ \begin{aligned} T_{D \uparrow } & = T_{A \uparrow } = \frac{{F_{as} + {{mg} \mathord{\left/ {\vphantom {{mg} 2}} \right. \kern\nulldelimiterspace} 2} + {{ma} \mathord{\left/ {\vphantom {{ma} 2}} \right. \kern\nulldelimiterspace} 2}}}{{n_{b} \sin \alpha_{b} }} \\ {\kern 1pt} & = \frac{{{M \mathord{\left/ {\vphantom {M {\left( {T_{C} \cdot d_{M} } \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{C} \cdot d_{M} } \right)}} + {{mg} \mathord{\left/ {\vphantom {{mg} 2}} \right. \kern\nulldelimiterspace} 2} + {{ma} \mathord{\left/ {\vphantom {{ma} 2}} \right. \kern\nulldelimiterspace} 2}}}{{n_{b} \sin \alpha_{b} }}, \\ \end{aligned} $$
(12)
where n_{b} is the number of balls in a single bearing, α_{b} is the contact angle of each ball in the upper and lowerend bearing joints, d_{M} is the diameter of the pretension nut, and M and T_{C} are the torque and torque coefficient of the pretension nut, respectively.
The inner and outer rings in a bearing are connected by spring elements distributed around the raceway, which provides the stiffness at both ends to sustain the ball screw. Based on the Hertz contact theory, the relationship between the contact force and the local deformation at the contact point is obtained. The normal contact stiffness of a single ball in bearings A and D can then be derived as
$$ k_{D \uparrow } = k_{A \uparrow } = \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} T_{A}^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} . $$
(13)
Therefore, the equivalent axial stiffness of the upperend bearing A and lowerend bearing D can be derived as
$$ \begin{aligned} K_{D \uparrow } \left( a \right) & = K_{A \uparrow } \left( a \right) = \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \left( {\frac{M}{{T_{c} \cdot d}} + \frac{m \cdot (a + g)}{2}} \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \sin^{{{5 \mathord{\left/ {\vphantom {5 3}} \right. \kern\nulldelimiterspace} 3}}} \alpha_{b} \cdot \left( {N_{b} \cdot n_{b} } \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot c_{wb} , \\ \end{aligned} $$
(14)
where c_{wb} is the weight coefficient of the equivalent axial stiffness of the bearing joints and has the value of 0.9.
Based on the principle of deformation compatibility, the normal contact force of each ball in bearing B at the upper end and bearing C at the lower end can be expressed as
$$ \begin{aligned} T_{C \uparrow } & = T_{B \uparrow } \\ & = \left\{ \begin{aligned} & \frac{{\left( {2\left( {{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}} + {{m \cdot (a + g)} \mathord{\left/ {\vphantom {{m \cdot (a + g)} 2}} \right. \kern\nulldelimiterspace} 2}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern\nulldelimiterspace} 2}}} }}{{n_{b} \sin \alpha_{b} }}, \\ & \quad \left( {\frac{m \cdot a}{2} < 1.83\frac{M}{{T_{c} \cdot d}}  \frac{m \cdot g}{2}} \right), \\ & 0,\quad \left( {\frac{m \cdot a}{2} \ge 1.83\frac{M}{{T_{c} \cdot d}}  \frac{m \cdot g}{2}} \right). \\ \end{aligned} \right. \\ \end{aligned} $$
(15)
Similarly, the axial contact stiffness of bearing B at the upper end and bearing C at the lower end can be expressed as
$$ \begin{aligned} K_{C \uparrow } \left( a \right) & = K_{B \uparrow } \left( a \right) \\ & = \left\{ \begin{aligned} & \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \left( {\left( {2\left( {\frac{M}{{T_{c} \cdot d}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {\frac{M}{{T_{c} \cdot d}} + \frac{m \cdot (a + g)}{2}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern\nulldelimiterspace} 2}}} } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \sin^{{{5 \mathord{\left/ {\vphantom {5 3}} \right. \kern\nulldelimiterspace} 3}}} \alpha_{b} \cdot \left( {N_{b} \cdot n_{b} } \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot c_{wb} ,{\kern 1pt} \;\left( {\frac{m \cdot (a + g)}{2} < 1.83\frac{M}{{T_{c} \cdot d}}} \right), \\ & 0,\quad \left( {\frac{m \cdot (a + g)}{2} \ge 1.83\frac{M}{{T_{c} \cdot d}}} \right). \\ \end{aligned} \right. \\ \end{aligned} $$
(16)
Therefore, when the spindle system accelerates upward, the stiffness of the upper end bearing joints is
$$ K_{U \uparrow } = \max \left( {K_{A \uparrow } ,K_{B \uparrow } } \right) \cdot c_{wb} . $$
(17)
The stiffness of the lowerend bearing joints is
$$ K_{L \uparrow } = \max (K_{C \uparrow } ,K_{D \uparrow } ) \cdot c_{wb} . $$
(18)

(2)
Upward deceleration
Figure 6(a)–(b) show the force analysis of the upper and lowerend bearing joints under upward deceleration. As shown in Figure 6(a)–(b), during upward deceleration, except for the downward direction of the inertial force, the other forces are the same as those under upward acceleration.
According to the force equilibrium of the single bearing in the vertical direction, the normal contact force of each ball in bearings A and D is calculated as
$$ \begin{aligned} T_{D \downarrow } & = T_{A \downarrow } \\ & = \left\{ \begin{aligned} & \frac{{{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}} + \left( {m \cdot g} \right)/2  \left( {m \cdot a} \right)/2}}{{n_{b} \sin \alpha_{b} }},\;\left( {\frac{ma}{2} < \frac{mg}{2}} \right), \\ & \frac{{\left( {2\left( {\left( {{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}}} \right)} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}} + {{\left( {m \cdot a} \right)} \mathord{\left/ {\vphantom {{\left( {m \cdot a} \right)} 2}} \right. \kern\nulldelimiterspace} 2}  {{\left( {m \cdot g} \right)} \mathord{\left/ {\vphantom {{\left( {m \cdot g} \right)} 2}} \right. \kern\nulldelimiterspace} 2}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern\nulldelimiterspace} 2}}} }}{{n_{b} \sin \alpha_{b} }}, \\ {\kern 1pt} & \quad {\kern 1pt} \left( {\frac{mg}{2} \le \frac{ma}{2} \le 1.83\left( {\frac{M}{{T_{c} \cdot d}}} \right) + \frac{mg}{2}} \right), \\ & 0,\quad \left( {\frac{ma}{2} > 1.83\left( {\frac{M}{{T_{c} \cdot d}}} \right) + \frac{mg}{2}} \right). \\ \end{aligned} \right. \\ \end{aligned} $$
(19)
Similarly, according to the Hertz contact theory, when the spindle system decelerates upwards with the acceleration magnitude of a, the equivalent axial stiffness for a single ball in bearings A and D can be derived as
$$ k_{D \downarrow } = k_{A \downarrow } = \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} T_{A}^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} . $$
(20)
Therefore, the equivalent axial stiffness for bearings A and D can be derived using
$$ \begin{aligned} K_{D \downarrow } \left( a \right) & = K_{A \downarrow } \left( a \right) \\ & = \left\{ \begin{aligned} & \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \left( {\frac{M}{{T_{c} \cdot d}} + \frac{m \cdot g}{2}  \frac{m \cdot a}{2}} \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot \sin^{{{5 \mathord{\left/ {\vphantom {5 3}} \right. \kern\nulldelimiterspace} 3}}} \alpha_{b} \cdot \left( {n_{b} } \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot c_{wb} , \\ & \quad \left( {\frac{m \cdot a}{2} < \frac{m \cdot g}{2}} \right), \\ & \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \left( {\left( {2\left( {\frac{M}{{T_{c} \cdot d}}{\kern 1pt} } \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {\frac{M}{{T_{c} \cdot d}} + \frac{m \cdot a}{2}  \frac{m \cdot g}{2}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern\nulldelimiterspace} 2}}} } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \sin^{{{5 \mathord{\left/ {\vphantom {5 3}} \right. \kern\nulldelimiterspace} 3}}} \alpha_{b} \cdot \left( {n_{b} } \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot c_{wb} ,{\kern 1pt} \\ {\kern 1pt} & \quad \left( {\frac{m \cdot g}{2} \le \frac{m \cdot a}{2} \le 1.83\left( {\frac{M}{{T_{c} \cdot d}}} \right) + \frac{m \cdot g}{2}} \right), \\ & 0,\quad \left( {\frac{m \cdot a}{2} > 1.83\left( {\frac{M}{{T_{c} \cdot d}}} \right) + \frac{m \cdot g}{2}} \right). \\ \end{aligned} \right. \\ \end{aligned} $$
(21)
Similarly, according to the theory of deformation compatibility, the normal contact force of a single ball in bearing C and bearing B is
$$ \begin{aligned} T_{C \downarrow } & = T_{B \downarrow } \\ & = \left\{ \begin{array} {*{20}ll} \frac{{\left( {2\left( {{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}} + \left( {m \cdot g} \right)/2  \left( {m \cdot a} \right)/2} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern\nulldelimiterspace} 2}}} }}{{n_{b} \sin \alpha_{b} }}, & \left( {\frac{ma}{2} < \frac{mg}{2}} \right), \\ \frac{{{M \mathord{\left/ {\vphantom {M {\left( {T_{c} \cdot d} \right)}}} \right. \kern\nulldelimiterspace} {\left( {T_{c} \cdot d} \right)}} + {{\left( {m \cdot a} \right)} \mathord{\left/ {\vphantom {{\left( {m \cdot a} \right)} 2}} \right. \kern\nulldelimiterspace} 2}  {{\left( {m \cdot g} \right)} \mathord{\left/ {\vphantom {{\left( {m \cdot g} \right)} 2}} \right. \kern\nulldelimiterspace} 2}}}{{n_{b} \sin \alpha_{b} }}{\kern 1pt}, & \left( {\frac{ma}{2} \ge \frac{mg}{2}} \right). \\ \end{array} \right. \\ \end{aligned} $$
(22)
The axial contact stiffness of bearing B and bearing C is expressed as
$$ \begin{aligned} K_{C \downarrow } \left( a \right) & = K_{B \downarrow } \left( a \right) \\ & = \left\{ \begin{aligned} & \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \left( {\left( {2\left( {\frac{M}{{T_{c} \cdot d}}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}}  \left( {\frac{M}{{T_{c} \cdot d}} + \frac{m \cdot g}{2}  \frac{m \cdot a}{2}} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} } \right)^{{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern\nulldelimiterspace} 2}}} } \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \sin^{{{5 \mathord{\left/ {\vphantom {5 3}} \right. \kern\nulldelimiterspace} 3}}} \alpha_{b} \cdot \left( {N_{b} \cdot n_{b} } \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot c_{wb} , \quad \left( {\frac{ma}{2} < \frac{mg}{2}} \right), \\ & \frac{3}{2}K_{h}^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \left( {\frac{M}{{T_{c} \cdot d}} + \frac{m \cdot a}{2}  \frac{m \cdot g}{2}} \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern\nulldelimiterspace} 3}}} \\ & \quad \cdot \sin^{{{5 \mathord{\left/ {\vphantom {5 3}} \right. \kern\nulldelimiterspace} 3}}} \alpha_{b} \cdot \left( {N_{b} \cdot n_{b} } \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern\nulldelimiterspace} 3}}} \cdot c_{wb} ,\quad \left( {\frac{ma}{2} \ge \frac{mg}{2}} \right). \\ \end{aligned} \right. \\ \end{aligned} $$
(23)
Therefore, when the spindle system decelerates upwards, the stiffness of the upperend bearing joints is
$$ K_{U \downarrow } = \max (K_{A \downarrow } ,K_{B \downarrow } ) \cdot c_{wb} . $$
(24)
The stiffness of the lowerend bearing joints is
$$ K_{L \downarrow } = \max \left( {K_{C \downarrow } ,K_{D \downarrow } } \right) \cdot c_{wb} . $$
(25)
2.3.3 Calculation of Axial Stiffness of the Ball Screw
In the axial fixedfixed support mode, the lead screw is divided into two sections by screwnut joints. The servo motor drives the rotation of the screw shaft, causing the upper section of the screw shaft to rotate. In addition, the upper and lower sections of the screw shaft are axially restrained by the bearings. However, the lengths of the upper and lower sections of the screw shaft change as the spindle system moves up and down. As a result, the axial tensile/compressive stiffness of the upper and lower sections and the axial torsional stiffness of the upper section of the screw shaft change. The expressions for the stiffness of the upper and lower sections of the screw shaft can be obtained as
$$ K_{us1} (z) = \frac{{4GE\uppi ^{3} d_{1}^{4} }}{{16G\uppi ^{2} d_{1}^{2} \left( {L  z} \right) + 32p^{2} E\left( {L  z} \right)}}, $$
(26)
$$ K_{ds1} (z) = \frac{{\uppi d_{1}^{2} E}}{4z}, $$
(27)
where \(K_{us1}\) is the axial stiffness of the upper section of the screw shaft, \(K_{ds1}\) is the axial stiffness of the lower section of the screw shaft, d_{1} is the bottom diameter of the screw shaft, E is the elastic modulus, G is the shear modulus, and p is the screw pitch.
2.3.4 Calculation of Transmission Stiffness
The transmission stiffness in the transmission direction consists of the axial stiffness of both the upper and lower sections of the screw shaft and the equivalent axial stiffness of the screwnut joints and bearing joints, as shown in Figure 2.
Therefore, the transmission stiffness under upward acceleration in the transmission direction can be calculated as
$$ \begin{aligned} k_{e \uparrow } \left( {a,z} \right) & = \\ & \quad \frac{{\left( {\frac{{K_{us} (z) \cdot K_{U \uparrow } (a)}}{{K_{us} (z) + K_{U \uparrow } (a)}} + \frac{{K_{ds} (z) \cdot K_{L \uparrow } (a)}}{{K_{ds} (z) + K_{L \uparrow } (a)}}} \right) \cdot K_{nut \uparrow } \left( a \right)}}{{\left( {\frac{{K_{us} (z) \cdot K_{U \uparrow } (a)}}{{K_{us} (z) + K_{U \uparrow } (a)}} + \frac{{K_{ds} (z) \cdot K_{L \uparrow } (a)}}{{K_{ds} (z) + K_{L \uparrow } (a)}}} \right) + K_{nut \uparrow } \left( a \right)}}. \\ \end{aligned} $$
(28)
Similarly, the transmission stiffness under upward deceleration in the transmission direction can be calculated as
$$ \begin{aligned} k_{e \downarrow } (a,z) & = \\ & \quad \frac{{\left( {\frac{{K_{us} (z) \cdot K_{U \downarrow } (a)}}{{K_{us} (z) + K_{U \downarrow } (a)}} + \frac{{K_{ds} (z) \cdot K_{L \downarrow } (a)}}{{K_{ds} (z) + K_{L \downarrow } (a)}}} \right) \cdot K_{nut \downarrow } (a)}}{{\left( {\frac{{K_{us} (z) \cdot K_{U \downarrow } (a)}}{{K_{us} (z) + K_{U \downarrow } (a)}} + \frac{{K_{ds} (z) \cdot K_{L \downarrow } (a)}}{{K_{ds} (z) + K_{L \downarrow } (a)}}} \right) + K_{nut \downarrow } (a)}}. \\ \end{aligned} $$
(29)