Inerter-Based Isolation System

Abstract

This chapter is concerned with the problem of analysis and optimization of the inerter-based isolators based on a “uni-axial” single-degree-of-freedom isolation system. In the first part, in order to gain an in-depth understanding of inerter from the prospective of vibration, the frequency responses of both parallel-connected and series-connected inerters are analyzed. In the second part, three other inerter-based isolators are introduced and the tuning procedures in both the $H_\infty$ optimization and the $H_2$ optimization are proposed in an analytical manner. The achieved $H_2$ and $H_\infty$ performance of the inerter-based isolators is superior to that achieved by the traditional dynamic vibration absorber (DVA) when the same inertance-to-mass (or mass) ratio is considered. Moreover, the inerter-based isolators have two unique properties, which are more attractive than the traditional DVA: first, the inertance-to-mass ratio of the inerter-based isolators can easily be larger than the mass ratio of the traditional DVA without increasing the physical mass of the whole system; second, there is no need to mount an additional mass on the object to be isolated.

Appendix

Proof of Proposition 3.1

Observing Fig. 3.6, it is shown that the curve horizontally passing through P indicates the optimal damping. This optimal damping can be obtained by solving the following equation:

$$\begin{aligned} \left. \frac{\mathrm {\partial } \mu ^2}{\mathrm {\partial } q^2}\right| _{q=q_P}=0. \end{aligned}$$

(3.46)

Denote $\mu =\sqrt{\frac{n}{m}}$, where $n=\delta ^2q^2+4(1-\delta q^2)^2\zeta ^2$, $m=\delta ^2(1-q^2)^2q^2+4(1-(1+\delta )q^2)^2\zeta ^2$. Equation (3.46) can be written in another form as

$$ n^\prime m-m^\prime n=0, $$

where $n^\prime =\mathrm {\partial } n/ \mathrm {\partial } q^2$ and $m^\prime =\mathrm {\partial } m/ \mathrm {\partial } q^2$. For the invariant point P,

$$ \frac{n}{m}=\frac{1}{(1-q^2)^2}=\frac{(1-\delta q^2)^2}{(1-(1+\delta )q^2)^2}, $$

therefore,

$$ (1-q^2)^2 n^\prime -m^\prime =0. $$

Since

$$ n^\prime =-8(1-\delta q^2)\delta \zeta ^2+\delta ^2, $$

$$ m^\prime =-8(1-(1+\delta )q^2)(\delta +1)\zeta ^2+\delta ^2(1-q^2)(1-3q^2), $$

after substituting $q_P$ into (3.14), one obtains

$$ \zeta _{opt}=\frac{1}{2}\sqrt{\delta (1+\delta -\sqrt{1+\delta ^2})}. $$

Proof of Proposition 3.2

Denote

$$ A=4\lambda ^2(1-\delta q^2)^2q^2,~B=(1-\delta (1+\lambda )q^2)^2, $$

$$ C=4\lambda ^2(1-(1+\delta )q^2)^2q^2,~D=(1-(\delta +1+\delta \lambda )q^2+\delta \lambda q^4)^2. $$

Then, $\mu$ in (3.17) can be rewritten as

$$\begin{aligned} \mu =\sqrt{\frac{A\zeta ^2+B}{C\zeta ^2+D}}. \end{aligned}$$

(3.47)

To find the invariant points which are independent of damping, it requires

$$\begin{aligned} \frac{A}{C}=\frac{B}{D}, \end{aligned}$$

that is,

$$ \frac{1-\delta q^2}{1-(1+\delta )q^2}=\pm \frac{1-\delta (1+\lambda )q^2}{1-(\delta +1+\delta \lambda )q^2+\delta \lambda q^4}. $$

With the plus sign, after cross-multiplication, one obtains $\delta ^2 \lambda q^6=0,$ which leads to the trivial solution $q=0$. With the minus sign, after simple calculation, one obtains

$$\begin{aligned} \delta ^2\lambda q^6-2\delta (\lambda +\delta +1+\delta \lambda )q^4+2(2\delta +1+\delta \lambda )q^2-2=0, \end{aligned}$$

(3.48)

which is a cubic form in $q^2$. Therefore, there are three invariant points for the configuration C3.

Denoting these three invariant points as P, Q, and R ($q_P<q_Q<q_R$), separately, one obtains

$$\begin{aligned} q^2_P+q^2_Q+q^2_R= & {} \frac{2}{\delta \lambda }(\lambda +\delta +1+\lambda \delta ),\end{aligned}$$

(3.49)

$$\begin{aligned} q^2_Pq^2_Qq^2_R= & {} \frac{2}{\delta ^2\lambda },\end{aligned}$$

(3.50)

$$\begin{aligned} q^2_Pq^2_Q+q^2_Pq^2_R+q^2_Qq^2_R= & {} \frac{2}{\delta ^2\lambda }(2\delta +1+\delta \lambda ). \end{aligned}$$

(3.51)

Since at points P and Q, the values of $\mu$ are independent of $\zeta$, then in the case of $\zeta =\infty$, one obtains

$$ \left| \frac{1-\delta q^2_P}{1-(1+\delta )q^2_P}\right| =\left| \frac{1-\delta q^2_Q}{1-(1+\delta )q^2_Q}\right| . $$

It can be checked that

$$ \frac{1-\delta q^2_P}{1-(1+\delta )q^2_P}>0,~\frac{1-\delta q^2_Q}{1-(1+\delta )q^2_Q}<0. $$

Then, one obtains

$$ \frac{1-\delta q^2_P}{1-(1+\delta )q^2_P}=-\frac{1-\delta q^2_Q}{1-(1+\delta )q^2_Q}. $$

After cross-multiplication and simplification, one obtains

$$\begin{aligned} 2\delta (1+\delta )q^2_Pq^2_Q-(q^2_P+q^2_Q)(1+2\delta )+2=0. \end{aligned}$$

(3.52)

Substituting (3.50) and (3.51) into (3.52), one can obtain a quadratic equation with respect to $q^2_R$ as

$$\begin{aligned} \delta \lambda (1+2\delta )q^4_R-2(\lambda +2\delta \lambda +3\delta +2\delta ^2+1+2\lambda \delta ^2)q^2_R+4(1+\delta )=0. \end{aligned}$$

(3.53)

Note that $q_R$ is the same solution as both (3.48) and (3.53) for the same $\delta$ and $\lambda$. Solving $\lambda$ from (3.48) and (3.53), separately, one obtains

$$\begin{aligned} \lambda= & {} \frac{2(q^4_R\delta (1+\delta )-(1+2\delta )q^2_R+1)}{\delta q^2_R(q^4_R\delta -2(\delta +1)q^2_R+2)},\end{aligned}$$

(3.54)

$$\begin{aligned} \lambda= & {} \frac{2((1+2\delta )(1+\delta )q^2_R-2(1+\delta ))}{q^2_R(\delta (1+2\delta )q^2_R-2(1+2\delta +2\delta ^2))}. \end{aligned}$$

(3.55)

Equating the solutions and simplifying the results, one obtains

$$\begin{aligned} \delta q^4_R-(2+3\delta )q^2_R+2=0. \end{aligned}$$

(3.56)

Then, one obtains $q^2_R$ as shown in (3.18).

From (3.18), it is easy to show that $q^2_R\ge 3$, which is relatively large compared with the natural frequency. This can explain why only invariant points P and Q are involved in the $H_\infty$ tuning of C3.

In this way, the optimal $\lambda$ can be obtained by substituting $q^2_R$ in (3.18) into (3.54) or (3.55). After obtaining $\lambda$, all the three invariant points can be obtained by solving

$$ q^4-\left( \frac{2}{\delta \lambda }(1+\lambda +\delta +\lambda \delta )-q^2_R\right) q^2+\frac{2}{\delta ^2\lambda q^2_R}=0, $$

which is obtained from (3.50) and (3.51).

The procedure of calculating the optimal damping ratio $\zeta$ is similar to the procedure in appendix, where the optimal $\zeta$ makes the gradients at invariant points P and Q zero. After calculation and simplification, one obtains (3.21). Taking an average of $\zeta ^2_P$ and $\zeta ^2_Q$, one obtains the optimal $\zeta _{opt}$ as in (3.20).

Proof of Proposition 3.3

Denote

$$ A=4(1-\delta (1+\lambda )q^2)^2, B=\delta ^2q^2,$$

$$ C=4(1-(1+\delta +\delta \lambda )q^2+\delta \lambda q^4)^2, D=\delta ^2(1-q^2)^2q^2, $$

and $\mu$ in (3.23) can be rewritten as

$$\begin{aligned} \mu =\sqrt{\frac{A\zeta ^2+B}{C\zeta ^2+D}}. \end{aligned}$$

(3.57)

To find the invariant points which are independent of damping, it requires

$$ \frac{A}{C}=\frac{B}{D}, $$

that is,

$$ \frac{1-\delta (1+\lambda )q^2}{1-(1+\delta +\delta \lambda )q^2+\delta \lambda q^4}=\pm \frac{1}{1-q^2}. $$

Again, with the plus sign, one obtains the trivial solution zero, and with the minus sign, one obtains

$$\begin{aligned} \delta (1+2\lambda )q^4-2(1+\delta +\delta \lambda )q^2+2=0. \end{aligned}$$

(3.58)

Then, one obtains the two invariant points P and Q ($q_P<q_Q$) as

$$\begin{aligned} q^2_{P,Q}=\frac{1+\delta +\delta \lambda \pm \sqrt{(1+\delta +\delta \lambda )^2-2\delta (1+2\lambda )}}{\delta (1+2\lambda )}. \end{aligned}$$

(3.59)

Letting the ordinates at invariant points P and Q equal, one has

$$ \left| \frac{1}{1-q^2_P}\right| =\left| \frac{1}{1-q^2_Q}\right| . $$

It can be checked that $\frac{1}{1-q^2_P}>0$ and $\frac{1}{1-q^2_Q}<0$. Then, one obtains

$$ \frac{1}{1-q^2_P}=-\frac{1}{1-q^2_Q}. $$

After cross-multiplication and simplification, one has

$$\begin{aligned} q^2_P+q^2_Q=2. \end{aligned}$$

(3.60)

Considering (3.58), one obtains

$$ \frac{2(1+\delta +\delta \lambda )}{\delta (1+2\lambda )}=2, $$

which leads to (3.24).

Similar to the method in appendix, the optimal $\zeta$ can be obtained by making $\mu$ to have zero gradients at invariant points P and Q. After calculation and simplification, one obtains

$$ \zeta ^2_{P,Q}=\frac{q^2_{P,Q} \delta ^2}{4\left( 1-\delta (1+\lambda )q^2_{P,Q}\right) \left( 1+2\delta +2\delta \lambda -\delta (1+3\lambda )q^2_{P,Q}\right) }. $$

After substituting (3.59) and (3.24), one obtains (3.26) and (3.27).

Taking an average of $\zeta ^2_p$ and $\zeta ^2_Q$, one obtains the optimal $\zeta _{opt}$ as in (3.25).

Proof of Proposition 3.4

Denote

$$ A=4(\lambda +1)^2q^2,B=(1-\delta (1+\lambda )q^2)^2,$$

$$ C=4(\lambda +1-\lambda q^2)^2q^2,D=(1-(1+\delta +\delta \lambda )q^2+\lambda \delta q^4)^2. $$

Then, $\mu$ in (3.28) can be rewritten as

$$\begin{aligned} \mu =\sqrt{\frac{A\zeta ^2+B}{C\zeta ^2+D}}. \end{aligned}$$

(3.61)

To find the invariant points which are independent of damping, it requires

$$ \frac{A}{C}=\frac{B}{D}, $$

that is,

$$ \frac{\lambda +1}{\lambda +1-\lambda q^2}=\pm \frac{1-\delta (1+\lambda )q^2}{1-(1+\delta +\delta \lambda )q^2+\delta \lambda q^4}. $$

Similarly, with plus sign, one obtains the trivial solution zero, and with minus sign, one obtains

$$\begin{aligned} 2\delta \lambda (\lambda +1)q^4-\left( 1+2\lambda +2\delta (1+\lambda )^2\right) q^2+2(\lambda +1)=0. \end{aligned}$$

(3.62)

Thus, one obtains the two invariant points P and Q ($q_P<q_Q$) as in (3.32).

Letting the ordinates at invariant points P and Q equal, one has

$$ \left| \frac{\lambda +1}{\lambda +1-\lambda q^2_P}\right| =\left| \frac{\lambda +1}{\lambda +1-\lambda q^2_Q}\right| . $$

It can be checked that $\frac{\lambda +1}{\lambda +1-\lambda q^2_P}>0$ and $\frac{\lambda +1}{\lambda +1-\lambda q^2_Q}<0$. Then, one obtains

$$ \frac{\lambda +1}{\lambda +1-\lambda q^2_P}=-\frac{\lambda +1}{\lambda +1-\lambda q^2_Q}. $$

After cross-multiplication and simplification, one has

$$ q^2_P+q^2_Q=\frac{2(\lambda +1)}{\lambda }. $$

Comparing with (3.62), one obtains

$$ \frac{1+2\lambda +2\delta (1+\lambda )^2}{2\delta \lambda (\lambda +1)}=\frac{2(\lambda +1)}{\lambda }, $$

which leads to

$$ 2\delta \lambda ^2-2(1-2\delta )\lambda +2\delta -1=0. $$

It can be checked that this equation has real solutions if and only if

$$ \delta \le 1/2. $$

Under this condition, the optimal $\lambda$ can be obtained as in (3.29).

Note that if $\delta =\frac{1}{2}$, from (3.29), one has $\lambda =0$ or $k=\infty$. In this case, C5 reduces to C1. Thus, the more reasonable assumption is $\delta <\frac{1}{2}$ rather than $\delta \le \frac{1}{2}$.

Similarly, the optimal $\zeta$ can be obtained by making $\mu$ to have zero gradients at invariant points P and Q. After calculation and simplification, one obtains $\zeta ^2_P$ and $\zeta ^2_Q$ as in (3.31).

Taking an average of $\zeta ^2_p$ and $\zeta ^2_Q$, one obtains the optimal $\zeta _{opt}$ as in (3.30).