© D. Zhao et al., Hosted by EDP Sciences 2021

1 Introduction

Electromagnetic harmonic drive is a kind of mechatronics drive device combining rotating magnetic field with harmonic drive. It is widely used in the driving mechanism of machine tools and instruments, and is suitable for submarine navigation, aerospace and transportation fields [1–5].

Herdeg proposed the electromagnetic harmonic drive, and successfully developed the experimental prototype of the drive with external magnetic poles [6]. Janes developed an electromagnetic harmonic drive system with built-in electromagnetic winding, which is compact in structure and can transmit more torque per unit space [7–9]. Rens developed a new type of permanent magnet harmonic drive prototype, which is suitable for the requirements of large transmission ratio [10,11]. Tjahjowidodo proposed a harmonic drive model using statistical measures of variation and analyzed its reliability under different conditions [12,13]. Reinhard investigated small harmonic gear drive and developed the harmonic reducer with metal gears which was used in robot driverfor semiconductor chip packaging [14,15]. Jose designed harmonic drive with low temperature magnetic superconductor to improve the service life [16]. Chigira proposed a harmonic drive with a stackable structure that is easy to assemble. By adjusting the structure of the magnetic gear, the maximum transfer torque can be increased [17]. Based on studies of the electromagnetic harmonic drive, Xu proposed an electromagnetic harmonic movable tooth drive system and investigated its output torque [18–20]. Afanas'evcalculated the electromagnetic moments of anelectromagnetic gear reducer by the energy method [21]. Ando proposed a new harmonic gear with stackable structure easy to assemble and studied the effect of stack on the maximum transmit torque by experiments [22]. Koji proposed a new magnetic harmonic gear which has the stackable structure in which the maximum transmission torque of the gear was improved approximately 5.96 times and the torque density was improved about 3.82 times [23]. Liu et al. investigated eccentric harmonic magnetic gear and presented an analytical method for predicting the distribution of magnetic field in the air gap of harmonic gear [24]. Jing proposed a new type of eccentric harmonic magnetic gear and calculated corresponding magnetic field and static torque with the finite element analysis [25]. With the development of MEMS technology, the size of the driving link is more and more limited, and the micron scale drive technology is urgently needed [26].

Therefore, the Authors propose a micro electromagnetic harmonic drive system. It has advantages such as small volume, light weight, simple and compact structure, large speed ratio and small inertia, etc. Stator components with electrode segments and special piezoelectric ceramic materials are not needed, which is more conducive to the miniaturization.

In the drive system, the electromagnetic force has a decisive role for its operation behavior. However, with further reducing the size of the system, the effect of molecular force is becoming more and more significant. The Van der Waals force is an important molecular force, will significantly influence the dynamic performance of the micro electromagnetic harmonic drive system.

In this paper, considering Van der Waals force, dynamics equation of the flexible ring for the micro drive system is deduced and resolved. Using the equations, effects of the molecule force on the natural frequencies and vibration modes of the drive system are investigated. Results show that considering molecule force, natural frequencies of the flexible ring are reduced and its vibration modes are changed. The research is useful in design of the dynamics performance for the micro electromagnetic harmonic drive system.

2 Structure and operation principle

Figure 1 shows a micro electromagnetic harmonic drive system. It consists of micro flexible ring and stator. When the external magnetic field is applied sequentially, the rotating magnetic field will cause the flexible wheel to undergo periodic elastic deformation, thus driving the associated supporting shaft to rotate.

Here, r is radius of the flexible ring, l is the length of the flexible ring, t₀ is the clearance between the flexible ring and stator. The stator material is no magnetic and the flexible ring material is magnetic. The flexible ring under electromagnetic force and molecule force is given in Figure 2. In electromagnetic field, distributed electromagnetic force occurs on the flexible ring. q_re is electromagnetic force per unit length. Meanwhile, molecule force occurs at the angle range $\left[-\frac{\pi }{4},\frac{\pi }{4}\right].q_{r3}$. is Van der Waals force per unit length.

Fig. 1 An electromagnetic micro harmonic drive system. 1 Cover; 2 flexible ring; 3 stator. (a) Section diagram (b) 3D model.

Fig. 2 Electromagnetic and molecular forces on the flexible ring. (a) Flexible ring and stator (b) electromagnetic force (c) molecular force.

3 Static displacement of the flexible ring

The electromagnetic force per unit length on the flexible wheel is$$q_{re}=\frac{n_B^2{\mu }_0{R}_I^{4}l}{8{(R_I^2+x_B^2)}^3}{I}^2\cos \theta$$(1)

where μ₀ −vacuum permeability, μ₀ = 4π × 10⁻⁷N ⋅ A⁻²;

I −current in magnetic coil;

n_B −number of turns per unit length;

R_I −magnetic coil radius;

x_B −distance from the flexible ring to the magnetic coil plane;

_θ –position angle of the flexible ring.

Equation (1) can be simplified to$$q_{re}=q_{r0}\cos \theta$$(2) where $q_{r0}=\frac{n_B^2{\mu }_0{R}_I^{4}l}{8{(R_I^2+x_B^2)}^3}{I}^2$

The van der Waals force per unit length is [27]$$q_{r3}=\frac{Al}{6\pi {(t_0-u_0)}^3}$$(3)

Where A is Hamaker constant, A = 6.58 × 10⁻²⁰ J_.

The total force per unit length on the flexible ring is$$q_{r\text{s}}=q_{re}+q_{r3}$$(4)

The distribution of forces on the flexible wheel is as follows: $$\{\begin{array}{l}{q}_{rs}=q_{r0}\cos \theta +q_{r3}=\frac{n_B^2{\mu }_0{R}_I^{4}l}{8{(R_I^2+x_B^2)}^3}{I}^2\cos \theta +q_{r3}\left(-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}\right)\hfill \\ q_{rs}=q_{r0}\cos \theta \left(\frac{\pi }{4}\le \theta \le \frac{\pi }{2},-\frac{\pi }{2}\le \theta \le -\frac{\pi }{4}\right)\hfill \\ q_{rs}=-q_{r0}\cos \theta +q_{r3}=-\frac{n_B^2{\mu }_0{R}_I^{4}l}{8{(R_I^2+x_B^2)}^3}{I}^2\cos \theta +q_{r3}\left(\frac{3\pi }{4}\le \theta \le \frac{5\pi }{4}\right)\hfill \\ q_{rs}=-q_{r0}\cos \theta \left(\frac{\pi }{2}\le \theta \le \frac{3\pi }{4},\frac{4\pi }{4}\le \theta \le \frac{3\pi }{2}\right)\hfill \end{array}$$(5)

This load is expanded as a Fourier series$$q_{rs}=q_0+{\displaystyle \sum _1^{\infty }}{q}_k\cos k\theta$$(6)

The coefficients of the series satisfy the following conditions $$\begin{array}{l}{\int }_{\begin{array}{c}\hfill 0\hfill \end{array}}^{2\pi }\left[q_0+{\displaystyle \sum _1^{\infty }}{q}_k\cos k\theta \right]d\theta ={\int }_{-\frac{\pi }{\text{4}}}^{\frac{\pi }{\text{4}}}\left(q_{r0}\cos \theta +q_{r3}\right)d\theta +{\int }_{\frac{\pi }{\text{4}}}^{\frac{\pi }{\text{2}}}\left(q_{r0}\cos \theta \right)d\theta \\ \text{+}{\int }_{\frac{\text{3}\pi }{\text{4}}}^{\frac{\text{5}\pi }{\text{4}}}\left(-q_{r0}\cos \theta +q_{r3}\right)d\theta +{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{4}}\left(-q_{r0}\cos \theta \right)d\theta \\ \text{+}{\int }_{\frac{\text{5}\pi }{\text{4}}}^{\frac{\text{3}\pi }{\text{2}}}\left(-q_{r0}\cos \theta \right)d\theta +{\int }_{-\frac{\pi }{2}}^{-\frac{\pi }{4}}\left(q_{r0}\cos \theta \right)d\theta \end{array}$$(7)

Thus$$q_0=\frac{2q_{r0}}{\pi }+\frac{q_{r3}}{\text{2}}$$(8)

To determine the coefficient q_k, multiply the above integral equation by cos kθ, the following relation is obtained $$\begin{array}{l}{\int }_{\begin{array}{c}\hfill 0\hfill \end{array}}^{2\pi }\left[q_0+{\displaystyle \sum _1^{\infty }}{q}_k\cos k\theta \right]\cos k\theta d\theta ={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\left(q_{r0}\cos \theta +q_{r3}\right)\cos k\theta d\theta +{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\left(q_{r0}\cos \theta \right)\cos k\theta d\theta \\ +{\int }_{\frac{3\pi }{4}}^{\frac{5\pi }{4}}\left(-q_{r0}\cos \theta +q_{r3}\right)\cos k\theta d\theta +{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{4}}\left(-q_{r0}\cos \theta \right)\cos k\theta d\theta \\ +{\int }_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}\left(-q_{r0}\cos \theta \right)\cos k\theta d\theta +{\int }_{-\frac{\pi }{2}}^{-\frac{\pi }{4}}\left(q_{r0}\cos \theta \right)\cos k\theta d\theta \end{array}$$(9)

In order to make the load symmetric with respect to the vertical axis and the horizontal axis, only take the coefficient of the cosine series to be even, yields$$q_k=\frac{2q_{r0}}{\pi }\left[\frac{\sin \frac{\left(k+1\right)\pi }{2}}{k+1}+\frac{\sin \frac{\left(k-1\right)\pi }{2}}{k-1}\right]\text{+}\frac{\text{4}{q}_{r3}}{\pi k}\sin \frac{k\pi }{\text{4}}\left(k\ne 1\right)$$(10)

Equation (10) can be changed into following form$$q_{rs}=\frac{2q_{r0}}{\pi }+\frac{q_{r3}}{\text{2}}+{\displaystyle \sum _{k=2,4,6,\cdot \cdot \cdot }}\left\{\frac{2q_{r0}}{\pi }\left[\frac{\sin \frac{\left(k+1\right)\pi }{2}}{k+1}+\frac{\sin \frac{\left(k-1\right)\pi }{2}}{k-1}\right]\text{+}\frac{\text{4}{q}_{r3}}{\pi k}\sin \frac{k\pi }{\text{4}}\right\}\cos k\theta$$(11)

Substituting q_rs into the dynamics equation of the flexible wheel:$$\frac{{\partial }^{5}u}{\partial {\theta }^5}+2\frac{{\partial }^{3}u}{\partial {\theta }^3}+\frac{\partial u}{\partial \theta }=\frac{r^4}{E{I}_x}\frac{\partial q_{rs}}{\partial \theta }-\frac{r^4\rho A}{E{I}_x}\frac{\partial {\displaystyle \stackrel{\mathrm{}}{u}}}{\partial \theta }$$(12)This is a static solution, so the derivative of the displacement with respect to time is zero, that is, $\partial {\displaystyle \stackrel{\mathrm{}}{u}}/\partial \theta =0$, yields$$\begin{array}{l}\frac{{\partial }^{5}u}{\partial {\theta }^5}+2\frac{{\partial }^{3}u}{\partial {\theta }^3}+\frac{\partial u}{\partial \theta }\\ =-\frac{r^4}{E{I}_x\pi }{\displaystyle \sum _{k=2,4,6,\cdot \cdot \cdot }}\left\{2q_{r0}\left[\frac{k\sin \frac{\left(k+1\right)\pi }{2}}{k+1}+\frac{k\sin \frac{\left(k-1\right)\pi }{2}}{k-1}\right]\text{+4}{q}_{r3}\sin \frac{k\pi }{\text{4}}\right\}\sin k\theta \end{array}$$(13)

Write the solution of equation (13) as a series as follows$$u={\displaystyle \sum _{k=2,4,6,\cdot \cdot \cdot }}{C}_k\cos k\theta$$(14)

Substituting equation (14) into equation (13) to obtain$$\begin{array}{l}{\displaystyle \sum _{k=2,4,6,\cdot \cdot \cdot }}{C}_k\left(-k^5+2k^3-k\right)\sin k\theta \\ =-\frac{r^4}{E{I}_x}{\displaystyle \sum _{k=2,4,6,\cdot \cdot \cdot }}\left\{\frac{2q_{r0}k}{\pi }\left[\frac{\sin \frac{\left(k+1\right)\pi }{2}}{k+1}+\frac{\sin \frac{\left(k-1\right)\pi }{2}}{k-1}\right]+\frac{4q_{r3}}{\pi }\sin \frac{k\pi }{\text{4}}\right\}\sin k\theta \end{array}$$(15)

On both sides of equation (15), if the corresponding terms of the same k value are equal, then$$C_k=\frac{r^4}{E{I}_x}\frac{\frac{2q_{r0}k}{\pi }\left[\frac{\sin \frac{\left(k+1\right)\pi }{2}}{k+1}+\frac{\sin \frac{\left(k-1\right)\pi }{2}}{k-1}\right]+\frac{4q_{rn}}{\pi }\sin \frac{k\pi }{\text{4}}}{k{\left(k^2-1\right)}^2}$$(16)

So $$u=\frac{r^4}{\pi E{I}_x}{\displaystyle \sum _{k=2,4,6,\cdot \cdot \cdot }}\left[2q_{r0}\frac{\left(k-1\right)\sin \frac{\left(k+1\right)\pi }{2}+\left(k+1\right)\sin \frac{\left(k-1\right)\pi }{2}}{{\left(k^2-1\right)}^3}+4q_{rn}\frac{\sin \frac{k\pi }{\text{4}}}{k{\left(k^2-1\right)}^2}\right]\cos k\theta$$(17)

Letting u₀ be equal to the average value of the displacement u at the angle range $\left[-\frac{\pi }{\text{4}}\le \theta \le \frac{\pi }{\text{4}}\right]$, using equation (17), u₀ can be obtained $$\begin{array}{l}{u}_0=\frac{2}{\pi }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}ud\theta \\ =\frac{8r^4}{{\pi }^{2}E{I}_x}{\displaystyle \sum _{k=\mathrm{2,4,6},\cdot \cdot \cdot }}\left[q_{r0}\frac{\left(k-1\right)\sin \frac{\left(k+1\right)\pi }{2}+\left(k+1\right)\sin \frac{\left(k-1\right)\pi }{2}}{k{\left(k^2-1\right)}^3}+2q_{rn}\frac{\sin \frac{k\pi }{4}}{k^2{\left(k^2-1\right)}^2}\right]\sin \frac{k\pi }{4}\end{array}$$(18)

4 Solution of the dynamics equation for flexible ring

4.1 Mode function

Let the radial displacement u of the flexible ring be composed of static displacement u₀ and dynamic displacement Δu $$u=u_0+\Delta u$$(19)

The radial load q_r is composed of static load q_rs and dynamic load Δq_r $$q_r=q_{rs}+\Delta q_r$$(20)

Substituting equation (20) into equation (12), yields$$\frac{{\partial }^5\Delta u}{\partial {\theta }^5}+2\frac{{\partial }^3\Delta u}{\partial {\theta }^3}+\frac{\partial \Delta u}{\partial \theta }=\frac{r^4}{E{I}_x}\frac{\partial \Delta q_r}{\partial \theta }-\frac{r^4\rho A_s}{E{I}_x}\frac{\partial \Delta {\displaystyle \stackrel{\mathrm{}}{u}}}{\partial \theta }$$(21)

where $\Delta {\displaystyle \stackrel{\mathrm{}}{u}}$ is the second derivative of the dynamic displacement Δu with respect to time, E is the elastic modulus of the flexible ring material, ρ is density of the flexible ring, A_s is the section area of the flexible ring, I_x is the cross section modulus of the flexible ring.

Equation (21) can be changed into following form$$\frac{{\partial }^4\Delta u}{\partial {\theta }^4}+2\frac{{\partial }^2\Delta u}{\partial {\theta }^2}+\Delta u=\frac{r^4}{E{I}_x}\Delta q_r-\frac{r^4\rho A}{E{I}_x}\Delta {\displaystyle \stackrel{\mathrm{}}{u}}$$(22)

where Δq_r is the dynamic load, $\Delta q_r=\frac{d{q}_r}{du}\Delta u$, from equations (1) and (3), the dynamic load can be determined by:

• Only considering electromagnetic force:

$$\Delta q_{re}=\frac{d{q}_{re}}{du}\Delta u_e=0$$(23)

• Considering electromagnetic force and van der Waals force:

$$\{\begin{array}{l}\Delta q_{r3}=\frac{d{q}_{r3}}{du}\Delta u_3=\frac{Al}{2\pi {(t_0-u_0)}^4}\Delta u_3\left(0\le \theta \le \frac{\pi }{4}\right)\hfill \\ \Delta q_{r3}=0\left(\frac{\pi }{4}\le \theta \le \frac{\pi }{2}\right)\hfill \end{array}$$(24)

Substituting equations (23) and (24) into (22), the dynamics equation of the micro electromagnetic harmonic drive system is as follows:

• Only considering electromagnetic force:

$$\frac{{\partial }^4\Delta u_e}{\partial {\theta }^4}+2\frac{{\partial }^2\Delta u_e}{\partial {\theta }^2}+\Delta u_e=-\frac{r^4}{E{I}_x}\rho A_s\Delta {{\displaystyle \stackrel{\mathrm{}}{u}}}_e$$(25)

• Considering electromagnetic force and van der Waals force:

$$\{\begin{array}{l}\frac{{\partial }^4\Delta u_{13}}{\partial {\theta }^4}+2\frac{{\partial }^2\Delta u_{13}}{\partial {\theta }^2}+\Delta u_{13}=\frac{r^4}{E{I}_x}\frac{Al}{2\pi {(t_0-u_0)}^4}\Delta u_{13}-\frac{r^4}{E{I}_x}\rho A_s\Delta {{\displaystyle \stackrel{\mathrm{}}{u}}}_{13}\left(0\le \theta \le \frac{\pi }{4}\right)\hfill \\ \frac{{\partial }^4\Delta u_{23}}{\partial {\theta }^4}+2\frac{{\partial }^2\Delta u_{23}}{\partial {\theta }^2}+\Delta u_{23}=-\frac{r^4}{E{I}_x}\rho A_s\Delta {{\displaystyle \stackrel{\mathrm{}}{u}}}_{23}\left(\frac{\pi }{4}\le \theta \le \frac{\pi }{2}\right)\hfill \end{array}$$(26)

Letting$$\Delta u=\phi \left(\theta \right)q\left(t\right)$$(27)

Then

• Only considering electromagnetic force:

$$\frac{{\displaystyle \stackrel{\mathrm{}}{q}}\left(t\right)}{q\left(t\right)}=-\frac{{\phi }_e^{(4)}\left(\theta \right)+2{{\phi }^{″}}_e\left(\theta \right)+{\phi }_e\left(\theta \right)}{\frac{\rho A_s{r}^4}{E{I}_x}{\phi }_e\left(\theta \right)}$$(28)

From equation (28), it is given$${\displaystyle \stackrel{\mathrm{}}{q}}\left(t\right)+{\omega }^{2}q\left(t\right)=0$$(29) $${\phi }_e^{(4)}\left(\theta \right)+2{{\phi }^{″}}_e\left(\theta \right)+Q{\phi }_e\left(\theta \right)=0$$(30) where $Q=1-\frac{\rho A_s{r}^4}{E{I}_x}{\omega }^2$

Letting φ_e (θ) = e ^λθ and substituting it into (30), yields$${\lambda }^4+2{\lambda }^2+Q=0$$(31)

The four eigen values can be obtained as follows $\pm \sqrt{-1+\sqrt{1-Q}}$ and $\pm i\sqrt{1+\sqrt{1-Q}}$, and then the mode function can be given by$${\phi }_e\left(\theta \right)=B_1\cos k_1\theta +B_2\sin k_1\theta +B_{3}ch{k}_2\theta +B_{4}sh{k}_2\theta$$(32)where $k_1=\sqrt{1+\sqrt{1-Q}}$ and $k_2=\sqrt{-1+\sqrt{1-Q}}$

The integral constants B_j(j = 1, 2, 3, 4) and frequency equation can be determined by symmetry and continuity conditions of the flexible ring.

• Considering electromagnetic force and van der Waals force:

(a) at $0\le \theta \le \frac{\pi }{4}$

In a same manner, following equation can be given$$\frac{{\displaystyle \stackrel{\mathrm{}}{q}}\left(t\right)}{q\left(t\right)}=-\frac{{\phi }_{31}^{(4)}\left(\theta \right)+2{\phi }_{31}^{\text{'}\text{'}}\left(\theta \right)+P{\phi }_{31}\left(\theta \right)}{\frac{\rho A_s{r}^4}{E{I}_x}{\phi }_{31}\left(\theta \right)}$$(33)where $P=1-\frac{r^{4}Al}{2E{I}_x\pi {\left(t_0-u_0\right)}^4}$

From equation (33), it is obtained$${\phi }_{31}^{(4)}\left(\theta \right)+2{{\phi }^{″}}_{31}\left(\theta \right)+R{\phi }_{31}\left(\theta \right)=0$$(34) where $R=P-\frac{\rho A_s{r}^4}{E{I}_x}{\omega }^2$

Letting φ₃₁ (θ) = e ^λθ and substituting it into (34), yields$${\lambda }^4+2{\lambda }^2+R=0$$(35)

Thus, the mode function can be given as$${\phi }_{31}\left(\theta \right)=A_1\cos m_1\theta +A_2\sin m_1\theta +A_{3}ch{m}_2\theta +A_{4}sh{m}_2\theta$$(36) where $m_1=\sqrt{1+\sqrt{1-R}}$, $m_2=\sqrt{-1+\sqrt{1-R}}$

(b) at $\frac{\pi }{4}\le \theta \le \frac{\pi }{2}$

In a same manner, following equation can be given$$\frac{{\displaystyle \stackrel{\mathrm{}}{q}}\left(t\right)}{q\left(t\right)}=-\frac{{\phi }_{32}^{(4)}\left(\theta \right)+2{{\phi }^{″}}_{32}\left(\theta \right)+{\phi }_{32}\left(\theta \right)}{\frac{\rho A_s{r}^4}{E{I}_x}{\phi }_{32}\left(\theta \right)}$$(37)

From equation (37), it is obtained$${\phi }_{32}^{(4)}\left(\theta \right)+2{{\phi }^{″}}_{32}\left(\theta \right)+S{\phi }_{32}\left(\theta \right)=0$$(38) where $S=1-\frac{\rho A_s{r}^4}{E{I}_x}{\omega }^2$

Letting φ₃₂ (θ) = e ^λθ and substituting it into (38), yields$${\lambda }^4+2{\lambda }^2+S=0$$(39)

Thus, the mode function can be given as$${\phi }_{32}\left(\theta \right)=A_5\cos n_1\theta +A_6\sin n_1\theta +A_{7}ch{n}_2\theta +A_{8}sh{n}_2\theta$$(40)where $n_1=\sqrt{1+\sqrt{1-S}}$ and $n_2=\sqrt{-1+\sqrt{1-S}}.$

The integral constants A_j(j = 1, 2, 3, 4, 5, 6, 7, 8) and frequency equation can be determined by symmetry and continuity conditions of the flexible ring.

4.2 Natural frequencies

From symmetry and continuity conditions of the flexible ring, both angle ϕ and tangent displacement v are zero at θ = 0 and $\theta =\frac{\pi }{2}$, so$$\{\begin{array}{l}{\phi }^{\text{'}}{\left(\theta \right)}_{\theta =0}=0\\ {\phi }^{\text{'}}{\left(\theta \right)}_{\theta =\frac{\pi }{2}}=0\\ {\phi }^{(3)}{\left(\theta \right)}_{\theta =0}=0\\ {\phi }^{(3)}{\left(\theta \right)}_{\theta =\frac{\pi }{2}}=0\end{array}$$(41)

• Only considering electromagnetic force:

Substituting equation (41) into (32), yields$$\begin{array}{c}{B}_2{k}_1+B_4{k}_2=0\\ -B_1{k}_1\sin \left(\frac{\pi }{2}{k}_1\right)+B_2{k}_1\cos \left(\frac{\pi }{2}{k}_1\right)+B_3{k}_{2}sh\left(\frac{\pi }{2}{k}_2\right)+B_4{k}_{2}ch\left(\frac{\pi }{2}{k}_2\right)=0\\ -B_2{k}_1^3+B_4{k}_2^3=0\\ B_1{k}_1^3\sin \left(\frac{\pi }{2}{k}_1\right)-B_2{k}_1^3\cos \left(\frac{\pi }{2}{k}_1\right)+B_3{k}_2^{3}sh\left(\frac{\pi }{2}{k}_2\right)+B_4{k}_2^{3}ch\left(\frac{\pi }{2}{k}_2\right)=0\end{array}\}$$(42)

From equation (42), it can be known that B₂ = 0 and B₄ = 0, so equation (42) can be changed into following form$$C_e{X}_e=D_e$$(43)

where$$C_e=\left[\begin{array}{cc}\hfill -k_1\sin \left(\frac{\pi }{2}{k}_1\right)\hfill & \hfill k_{2}sh\left(\frac{\pi }{2}{k}_2\right)\hfill \\ \hfill k_1^3\sin \left(\frac{\pi }{2}{k}_1\right)\hfill & \hfill k_2^{3}sh\left(\frac{\pi }{2}{k}_2\right)\hfill \end{array}\right]$$ $$X_e={\left[\begin{array}{cc}\hfill B_1\hfill & \hfill B_3\hfill \end{array}\right]}^T{D}_e={\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \end{array}\right]}^T$$

If there are non-zero solutions for equation (43), the determinant of is zero, that is $$\left|\begin{array}{cc}\hfill -k_1\sin \left(\frac{\pi }{2}{k}_1\right)\hfill & \hfill k_{2}sh\left(\frac{\pi }{2}{k}_2\right)\hfill \\ \hfill k_1^3\sin \left(\frac{\pi }{2}{k}_1\right)\hfill & \hfill k_2^{3}sh\left(\frac{\pi }{2}{k}_2\right)\hfill \end{array}\right|=0$$(44)

• Considering electromagnetic force and van der Waals force:

From continuity conditions of the flexible ring, following relationship can be given at $\theta =\frac{\pi }{4}$ $$\{\begin{array}{l}{\phi }_{31}\left(\theta \right)={\phi }_{32}\left(\theta \right)\\ {\phi }_{31}^{\text{'}}\left(\theta \right)={\phi }_{32}^{\text{'}}\left(\theta \right)\\ {\phi }_{31}^{\text{'}\text{'}}\left(\theta \right)={\phi }_{32}^{\text{'}\text{'}}\left(\theta \right)\\ {\phi }_{31}^{(3)}\left(\theta \right)={\phi }_{32}^{(3)}\left(\theta \right)\end{array}$$(45)

Substituting equations (36) and (40) into (45), yields $$\begin{array}{c}\hfill A_2{m}_1+A_4{m}_2=0\hfill \\ \hfill -A_5{n}_1\sin \left(\frac{\pi }{2}{n}_1\right)+A_6{n}_1\cos \left(\frac{\pi }{2}{n}_1\right)+A_7{n}_{2}sh\left(\frac{\pi }{2}{n}_2\right)+A_8{n}_{2}ch\left(\frac{\pi }{2}{n}_2\right)=0\hfill \\ \hfill -A_2{m}_1^3+A_4{m}_2^3=0\hfill \\ \hfill A_5{n}_1^3\sin \left(\frac{\pi }{2}{n}_1\right)-A_6{n}_1^3\cos \left(\frac{\pi }{2}{n}_1\right)+A_7{n}_2^{3}sh\left(\frac{\pi }{2}{n}_2\right)+A_8{n}_2^{3}ch\left(\frac{\pi }{2}{n}_2\right)=0\hfill \\ \hfill \begin{array}{l}{A}_1\cos \left(\frac{\pi }{4}{m}_1\right)+A_2\sin \left(\frac{\pi }{4}{m}_1\right)+A_{3}ch\left(\frac{\pi }{4}{m}_2\right)+A_{4}sh\left(\frac{\pi }{4}{m}_2\right)-A_5\cos \left(\frac{\pi }{4}{n}_1\right)\\ -A_6\sin \left(\frac{\pi }{4}{n}_1\right)-A_{7}ch\left(\frac{\pi }{4}{n}_2\right)-A_{8}sh\left(\frac{\pi }{4}{n}_2\right)=0\end{array}\hfill \\ \hfill \begin{array}{l}-A_1{m}_1\sin \left(\frac{\pi }{4}{m}_1\right)+A_2{m}_1\cos \left(\frac{\pi }{4}{m}_1\right)+A_3{m}_{2}sh\left(\frac{\pi }{4}{m}_2\right)+A_4{m}_{2}ch\left(\frac{\pi }{4}{m}_2\right)\\ +A_5{n}_1\sin \left(\frac{\pi }{4}{n}_1\right)-A_6{n}_1\cos \left(\frac{\pi }{4}{n}_1\right)-A_7{n}_{2}sh\left(\frac{\pi }{4}{n}_2\right)-A_8{n}_{2}ch\left(\frac{\pi }{4}{n}_2\right)=0\end{array}\hfill \\ \hfill \begin{array}{l}-A_1{m}_1^2\cos \left(\frac{\pi }{4}{m}_1\right)-A_2{m}_1^2\sin \left(\frac{\pi }{4}{m}_1\right)+A_3{m}_2^{2}ch\left(\frac{\pi }{4}{m}_2\right)+A_4{m}_2^{2}sh\left(\frac{\pi }{4}{m}_2\right)\\ +A_5{n}_1^2\cos \left(\frac{\pi }{4}{n}_1\right)+A_6{n}_1^2\sin \left(\frac{\pi }{4}{n}_1\right)-A_7{n}_2^{2}ch\left(\frac{\pi }{4}{n}_2\right)-A_8{n}_2^{2}sh\left(\frac{\pi }{4}{n}_2\right)=0\end{array}\hfill \\ \hfill \begin{array}{l}{A}_1{m}_1^3\sin \left(\frac{\pi }{4}{m}_1\right)-A_2{m}_1^3\cos \left(\frac{\pi }{4}{m}_1\right)+A_3{m}_2^{3}sh\left(\frac{\pi }{4}{m}_2\right)+A_4{m}_2^{3}ch\left(\frac{\pi }{4}{m}_2\right)\\ -A_5{n}_1^3\sin \left(\frac{\pi }{4}{n}_1\right)+A_6{n}_1^3\cos \left(\frac{\pi }{4}{n}_1\right)-A_7{n}_2^{3}sh\left(\frac{\pi }{4}{n}_2\right)-A_8{n}_2^{3}ch\left(\frac{\pi }{4}{n}_2\right)=0\end{array}\hfill \end{array}\}$$(46)

From equation (46), it can be known that A₂ = 0 and A₄ = 0, so equation (46) can be changed into following form$$C_1{X}_1=D_1$$(47)

where $$C_1=\left[\begin{array}{cccccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -n_1\sin \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill n_1\cos \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill n_{2}sh\left(\frac{\pi }{2}{n}_2\right)\hfill & \hfill n_{2}ch\left(\frac{\pi }{2}{n}_2\right)\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill n_1^3\sin \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill -n_1^3\cos \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill n_2^{3}sh\left(\frac{\pi }{2}{n}_2\right)\hfill & \hfill n_2^{3}ch\left(\frac{\pi }{2}{n}_2\right)\hfill \\ \hfill \cos \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill ch\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill -\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -ch\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -sh\left(\frac{\pi }{4}{n}_2\right)\hfill \\ \hfill -m_1\sin \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill m_{2}sh\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill n_1\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_1\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_{2}sh\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -n_{2}ch\left(\frac{\pi }{4}{n}_2\right)\hfill \\ \hfill -m_1^2\cos \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill m_2^{2}ch\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill n_1^2\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill n_1^2\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_2^{2}ch\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -n_2^{2}sh\left(\frac{\pi }{4}{n}_2\right)\hfill \\ \hfill m_1^3\sin \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill m_2^{3}sh\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill -n_1^3\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill n_1^3\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_2^{3}sh\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -n_2^{3}ch\left(\frac{\pi }{4}{n}_2\right)\hfill \end{array}\right]$$

If there are non-zero solutions for equation (47), the corresponding determinant is zero, that is $$\left|\begin{array}{cccccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -n_1\sin \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill n_1\cos \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill n_{2}sh\left(\frac{\pi }{2}{n}_2\right)\hfill & \hfill n_{2}ch\left(\frac{\pi }{2}{n}_2\right)\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill n_1^3\sin \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill -n_1^3\cos \left(\frac{\pi }{2}{n}_1\right)\hfill & \hfill n_2^{3}sh\left(\frac{\pi }{2}{n}_2\right)\hfill & \hfill n_2^{3}ch\left(\frac{\pi }{2}{n}_2\right)\hfill \\ \hfill \cos \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill ch\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill -\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -ch\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -sh\left(\frac{\pi }{4}{n}_2\right)\hfill \\ \hfill -m_1\sin \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill m_{2}sh\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill n_1\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_1\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_{2}sh\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -n_{2}ch\left(\frac{\pi }{4}{n}_2\right)\hfill \\ \hfill -m_1^2\cos \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill m_2^{2}ch\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill n_1^2\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill n_1^2\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_2^{2}ch\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -n_2^{2}sh\left(\frac{\pi }{4}{n}_2\right)\hfill \\ \hfill m_1^3\sin \left(\frac{\pi }{4}{m}_1\right)\hfill & \hfill m_2^{3}sh\left(\frac{\pi }{4}{m}_2\right)\hfill & \hfill -n_1^3\sin \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill n_1^3\cos \left(\frac{\pi }{4}{n}_1\right)\hfill & \hfill -n_2^{3}sh\left(\frac{\pi }{4}{n}_2\right)\hfill & \hfill -n_2^{3}ch\left(\frac{\pi }{4}{n}_2\right)\hfill \end{array}\right|=0$$(48)

Using equations (42) and (46), the constants B_j and A_j of mode functions can be determined. Using equation. (44) and (48), natural frequencies of the flexible ring can be determined. Then, the primary mass M_pi of the first order can be calculated by$$M_{pi}={\int }_0^{\beta }\rho {\left[{\phi }_{1i}\left(\theta \right)\right]}^{2}Ard\theta +{\int }_{\beta }^{\pi /2}\rho {\left[{\phi }_{2i}\left(\theta \right)\right]}^{2}Ard\theta$$(49)

Multiplying φ_i(θ) by $M_{pi}^{-1/2}$, the normal mode function of the flexible ring is obtained.

5 Effects of friction and air damping

Under the electromagnetic force, the flexible wheel after deformation contacts with the outer ring stator at the angle range [− π/4, π/4]. Here, the friction force occurs. At other angle range, the air damping force is applied to the flexible wheel. Letting q_t denote friction force per unit length on the flexible wheel, q_p denote damping force per unit length on the flexible wheel.

From friction torque equation of the flexible wheel, the friction force q_t3considering electromagnetic force and Van der Walls force can be given by$$q_{t3}=\frac{\mu \left(q_{r0}{Y}_1+2q_{r3}{Y}_2\right)}{2Y_2}-\frac{\mu t_0\pi E{I}_x}{4r^4{Y}_2}=\frac{\mu q_{r0}{Y}_1}{2Y_2}-\frac{\mu t_0\pi E{I}_x}{4r^4{Y}_2}+\mu q_{r3}$$(50)

This friction force is expanded as a Fourier series and high order terms are neglected, the dynamic friction force can be obtained$$\Delta q_{t3}=\frac{\mu Al}{2\pi {(t_0-u_0)}^4}\Delta u$$(51)

The air damping force q_p per unit length on the flexible wheel is [28]$$q_p=\frac{\eta l^3}{{(t_0-u)}^3}\frac{\partial u}{\partial t}$$(52)where η is the air dynamic viscosity, η = 1.86 × 10⁻⁵ N ⋅ S ⋅ m⁻².

This air damping force is expanded as a Fourier series and high order terms are neglected, the dynamic air damping force can be obtained$$q_p=\frac{\eta l^3}{{(t_0-u_0)}^3}\frac{\partial \Delta u}{\partial t}$$(53)

Considering tangent friction force, the dynamics equation (12) of the flexible wheel is changed into following form$$\frac{{\partial }^5\Delta u}{\partial {\theta }^5}+2\frac{{\partial }^3\Delta u}{\partial {\theta }^3}+\frac{\partial \Delta u}{\partial \theta }=\frac{r^4}{E{I}_x}\left(\frac{\partial \Delta q_r}{\partial \theta }+\Delta q_t\right)-\frac{r^4\rho A_s}{E{I}_x}\frac{\partial \Delta {\displaystyle \stackrel{\mathrm{}}{u}}}{\partial \theta }$$(54)

The forces on the flexible wheel include radial one and tangent one, and the force distribution can be given by $$\{\begin{array}{c}\Delta q_{rp}=\frac{Al}{2\pi {(t_0-u_0)}^4}\Delta u_p\Delta q_t=\frac{\mu Al}{2\pi {(t_0-u_0)}^4}\Delta u_p\left(0\le \theta \le \frac{\pi }{4}\right)\hfill \\ \Delta q_{rp}=-\frac{\eta l^3}{{(t_0-u_0)}^3}\frac{\partial \Delta u_p}{\partial t}\Delta q_t=0\left(\frac{\pi }{4}\le \theta \le \frac{\pi }{2}\right)\hfill \end{array}$$(55)

Combining equation (54) with (55), yields $$\{\begin{array}{l}\frac{{\partial }^5\Delta u_{\text{1}p}}{\partial {\theta }^5}+2\frac{{\partial }^3\Delta u_{\text{1}p}}{\partial {\theta }^3}+\frac{\partial \Delta u_{\text{1}p}}{\partial \theta }\\ =\frac{r^4}{E{I}_x}\left(\frac{Al}{2\pi {(t_0-u_0)}^4}\frac{\partial \Delta u_{\text{1}p}}{\partial \theta }+\frac{\mu Al}{2\pi {(t_0-u_0)}^4}\Delta u_{\text{1}p}\right)-\frac{r^4\rho A_s}{E{I}_x}\frac{\partial \Delta {{\displaystyle \stackrel{\mathrm{}}{u}}}_{\text{1}p}}{\partial \theta }\left(0\le \theta \le \frac{\pi }{4}\right)\\ \frac{{\partial }^4\Delta u_{2p}}{\partial {\theta }^4}+2\frac{{\partial }^2\Delta u_{2p}}{\partial {\theta }^2}+\Delta u_{2p}=-\frac{r^4}{E{I}_x}\frac{\eta l^3}{{(t_0-u_0)}^3}\frac{\partial \Delta u_{2p}}{\partial t}-\frac{r^4}{E{I}_x}\rho A_s\Delta {{\displaystyle \stackrel{\mathrm{}}{u}}}_{2p}\left(\frac{\pi }{4}\le \theta \le \frac{\pi }{2}\right)\end{array}$$(56)

Letting$$\Delta u=\phi \left(\theta \right)q\left(t\right)$$(57)

Substituting equation (57) into (56), yields: $$\frac{{\displaystyle \stackrel{\mathrm{}}{q}}\left(t\right)}{q\left(t\right)}=-\frac{{\phi }_{p1}^{(5)}\left(\theta \right)+2{\phi }_{p1}^{(3)}\left(\theta \right)+\left(1-T\right){{\phi }^{\prime }}_{p1}\left(\theta \right)-\mu T{\phi }_{p1}\left(\theta \right)}{\frac{\rho A_s{r}^4}{E{I}_x}{{\phi }^{\prime }}_{p1}\left(\theta \right)}\left(0\le \theta \le \frac{\pi }{4}\right)$$(58-a) $$\frac{{\displaystyle \stackrel{\mathrm{}}{q}}\left(t\right)}{q\left(t\right)}+\frac{\eta l^3}{{(t_0-u_0)}^3\rho A_s}\frac{{\displaystyle \stackrel{}{q}}\left(t\right)}{q\left(t\right)}=-\frac{{\phi }_{p2}^{(4)}\left(\theta \right)+2{\phi }_{p2}^{\text{'}\text{'}}\left(\theta \right)+{\phi }_{p2}\left(\theta \right)}{\frac{\rho A_s{r}^4}{E{I}_x}{\phi }_{p2}\left(\theta \right)}\left(\frac{\pi }{4}\le \theta \le \frac{\pi }{2}\right)$$(58-b) where $T=\frac{r^4}{E{I}_x}\frac{Al}{2\pi {(t_0-u_0)}^4}$