Dynamics of an electromagnetic vibro-impact nonlinear energy sink, applications in energy harvesting and vibration absorption

Abstract

The dynamics of an electromagnetic vibro-impact nonlinear energy sink (VINES) is investigated. The system consists of a linear oscillator equipped with multiple turns of coil, coupled to a magnet attachment that undergoes two-sided inelastic impacts. The dynamical model is formulated in a dimensionless formula and handled using a multiscale expansion method with respect to the small mass ratio parameter. Under zero-order approximation, the simple periodic solutions with two impacts per cycle, together with their bifurcations and stabilities, are solved, giving a slow invariant manifold (SIM) including a symmetric branch and an asymmetric branch. Then the harmonic forced dynamics is studied by considering the first-order equations, evidencing the rich response regimes depending on the forcing condition. The developed framework contains the dynamics of classical VINES systems as its limit case when electric damping $\beta =0$, and the obtained analytical results are in good agreement with numerical simulations. Finally, the developments are applied to the applications of vibration absorption and energy harvesting. The analytical predictions provide a deeper understanding to clearly explain the energy harvesting potential of the proposed electromagnetic VINES that is numerically observed in our previous investigation. In terms of vibration suppression, it is found that the proposed electromagnetic VINES can enhance the robustness of the TET efficiency by taking advantage of the electric damping, thus leading to a better performance than the conventional ones.

Appendices

Appendix A: Derivation of Eq. (26)

This appendix is devoted to the derivation of Eq. (26) by using the relationships in Eq. (24).

First of all, subtracting Eq. (24b) with Eq. (24a) yields

$$\begin{aligned} \begin{aligned} \cos (\tau _{0,k}+\psi ) -\cos (\tau _{0,k-1}+\psi )=\frac{N_{k-1}(a-b)}{(1+r) M}, \end{aligned}\nonumber \\ \end{aligned}$$

(A.1)

squaring Eqs. (A.1) and  (25) and adding both sides, one has

$$\begin{aligned} 2-2\cos T_u = \frac{N_{k-1}^2}{(1+r)^2 M^2}(a-b)^2 + \frac{(2-cN_{k-1})^2}{M^2},\nonumber \\ \end{aligned}$$

(A.2)

which is Eq. (26b).

On the other hand, from trigonometric identity, one can rewrite $\cos (\tau _{0,k}+\psi )$ as

$$\begin{aligned} \begin{aligned}&\cos (\tau _{0,k}+\psi )=\\&\cos (\tau _{0,k-1}+\psi +T_u)=\\&\cos (\tau _{0,k-1}+\psi )\cos T_u-\sin (\tau _{0,k-1}+\psi )\sin T_u, \end{aligned} \end{aligned}$$

(A.3)

which gives

$$\begin{aligned} \begin{aligned}&[\cos (\tau _{0,k-1}+\psi )\cos T_u-\cos (\tau _{0,k}+\psi )]^2=\\&\sin ^2(\tau _{0,k-1}+\psi )\sin ^2 T_u=\\&[1-\cos ^2(\tau _{0,k-1}+\psi )](1-\cos ^2 T_u). \end{aligned} \end{aligned}$$

(A.4)

Applying the relationships in Eqs. (24a) and  (24b) leads to

$$\begin{aligned} \frac{N_{k-1}^2}{(1+r)^2 M^2}(a^2+b^2-2ab\cos T_u)=1-\cos ^2T_u,\nonumber \\ \end{aligned}$$

(A.5)

which is Eq. (26a).

Appendix B: Entries in $T_1$ and $T_2$

The entries $\frac{\partial \tau _{0,k}}{\partial \tau _{0,k-1}}$, $\frac{\partial \tau _{0,k}}{\partial N_{k-1}}$, $\frac{\partial N_{k}}{\partial \tau _{0,k-1}}$, and $\frac{\partial N_{k}}{\partial N_{k-1}}$ in matrix $T_1$ are obtained by taking partial derivatives with respect to $\tau _{0,k-1}$ and $N_{k-1}$ for both sides of Eqs. (21a) and  (21b), and are calculated as,

$$\begin{aligned} \begin{aligned}&\frac{\partial \tau _{0,k}}{\partial \tau _{0,k-1}} = \frac{M\cos (\tau _{0,k-1}+\psi ) + N_{k-1}e^{-\beta T_u}}{M\cos (\tau _{0,k} + \psi )+N_{k-1}e^{-\beta T_u}}\\&\frac{\partial \tau _{0,k}}{\partial N_{k-1}} = \frac{e^{-\beta T_u}-1}{\beta M\cos (\tau _{0,k} + \psi )+\beta N_{k-1}e^{-\beta T_u}}\\&\frac{\partial N_{k}}{\partial \tau _{0,k-1}} = - r \beta N_{k-1} e^{-\beta T_u} +\\&[(1+r)M\sin (\tau _{0,k} + \psi )+r \beta N_{k-1}e^{-\beta T_u}]\frac{\partial \tau _{0,k}}{\partial \tau _{0,k-1}}\\&\frac{\partial N_{k}}{\partial N_{k-1}} = - r e^{-\beta T_u} +\\&[(1+r)M\sin (\tau _{0,k} + \psi )+r \beta N_{k-1}e^{-\beta T_u}]\frac{\partial \tau _{0,k}}{\partial N_{k-1}}\\ \end{aligned} \end{aligned}$$

(B.1)

Similar calculations for Eqs. (22a) and  (22b) gives,

$$\begin{aligned} \begin{aligned}&\frac{\partial \tau _{0,k+1}}{\partial \tau _{0,k}} = \frac{M\cos (\tau _{0,k}+\psi ) + N_{k}e^{-\beta T_d}}{M\cos (\tau _{0,k+1} + \psi )+N_{k}e^{-\beta T_d}}\\&\frac{\partial \tau _{0,k+1}}{\partial N_{k}} = \frac{e^{-\beta T_d}-1}{\beta M\cos (\tau _{0,k+1} + \psi )+\beta N_{k}e^{-\beta T_d}}\\&\frac{\partial N_{k+1}}{\partial \tau _{0,k}} = - r \beta N_{k} e^{-\beta T_d} +\\&[(1+r)M\sin (\tau _{0,k+1} + \psi )+r \beta N_{k}e^{-\beta T_d}]\frac{\partial \tau _{0,k+1}}{\partial \tau _{0,k}}\\&\frac{\partial N_{k+1}}{\partial N_{k}} = - r e^{-\beta T_d} +\\&[(1+r)M\sin (\tau _{0,k+1} + \psi )+r \beta N_{k}e^{-\beta T_d}]\frac{\partial \tau _{0,k+1}}{\partial N_{k}}\\ \end{aligned} \end{aligned}$$

(B.2)