An asymmetric quasi-zero stiffness vibration isolator with long stroke and large bearing capacity

Abstract

A novel passive asymmetric quasi-zero stiffness vibration isolator (AQZS-VI) comprising two linear springs acting in parallel with one negative stiffness element (NSE) is proposed, of which the NSE is mainly constructed by the combination of cantilever plate spring and L-shaped lever (CPS-LSL). The static model of the isolator is deduced considering the geometrical nonlinearity of the NSE and the bending deformation of plate spring. The nonlinear stiffness properties of the CPS-LSL and the AQZS-VI, as well as the nonlinear damping properties of the AQZS-VI, are discussed. The absolute displacement transmissibility of the AQZS-VI under base displacement excitation is obtained using harmonic balance method, and the effects of different excitation amplitudes and damping factors on the vibration isolation performance are analyzed. Better than other quasi-zero stiffness vibration isolators (QZS-VI) whose NSEs do not provide supporting force at zero stiffness point, the NSE of the AQZS-VI provides more supporting force than the parallel connected linear springs, which is very beneficial for improving the bearing capacity of the isolator. Compared with a typical symmetric QZS-VI with same damping property, the AQZS-VI has longer stroke with low stiffness and lower peak value of displacement transmissibility. The prototype experiments indicate that the AQZS-VI outperforms the linear counterpart with much smaller starting frequency of vibration isolation and lower displacement transmissibility. The proposed AQZS-VI has great potential for applying in various engineering practices with superior vibration isolation performance.

Appendices

Appendix A

$${f}_{f1}=2{f}_{c1}$$

(A.1)

$${f}_{f2}=4\frac{{\tilde{u }}^{2}}{{\delta }^{2}-{\tilde{u }}^{2}}{f}_{c2}$$

(A.2)

$${f}_{f3}=\frac{2}{\delta }\left({\cos}{\beta }_{c}-{\sin}{\beta }_{c}\frac{\tilde{u }}{\sqrt{{\delta }^{2}-{\tilde{u }}^{2}}}\right){f}_{c3}$$

(A.3)

$${f}_{f4}=2\frac{{\delta }^{2}}{{\delta }^{2}-{\tilde{u }}^{2}}{f}_{c4}$$

(A.4)

$${\varepsilon }_{0}={\left.4{f}_{c2}\frac{{\tilde{u }}^{2}}{{\delta }^{2}-{\tilde{u }}^{2}}\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.5)

$${\varepsilon }_{1}={\left.8{f}_{c2}\frac{{\delta }^{2}\tilde{u }}{{{(\delta }^{2}-{\tilde{u }}^{2})}^{2}}\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.6)

$${\varepsilon }_{2}={\left.4{f}_{c2}\frac{{\delta }^{4}+3{\delta }^{2}{\tilde{u }}^{2}}{{{(\delta }^{2}-{\tilde{u }}^{2})}^{3}}\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.7)

$${\varepsilon }_{3}={\left.2{f}_{c3}\frac{1}{\delta }\left({\cos}{\beta }_{c}-{\sin}{\beta }_{c}\frac{\tilde{u }}{\sqrt{{\delta }^{2}-{\tilde{u }}^{2}}}\right)\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.8)

$${\varepsilon }_{4}={\left.-2{f}_{c3}\frac{1}{\delta }\left[\left({\sin}{\beta }_{c}+{\cos}{\beta }_{c}\frac{\tilde{u }}{\sqrt{{\delta }^{2}-{\tilde{u }}^{2}}}\right)\frac{d{\beta }_{c}}{{\mathrm{d}}\tilde{u }}+{\sin}{\beta }_{c}{\delta }^{2}{\left({\delta }^{2}-{\tilde{u }}^{2}\right)}^{-\frac{3}{2}}\right]\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.9)

$${\varepsilon }_{5}={\left.-{f}_{c3}\frac{1}{\delta }\left[\left({\cos}{\beta }_{c}-{\sin}{\beta }_{c}\frac{\tilde{u }}{\Delta }\right){\left(\frac{d{\beta }_{c}}{{\mathrm{d}}\tilde{u }}\right)}^{2}+2{\cos}{\beta }_{c}{\delta }^{2}{\Delta }^{-3}\frac{d{\beta }_{c}}{{\mathrm{d}}\tilde{u }}+\left({\sin}{\beta }_{c}+{\cos}{\beta }_{c}\frac{\tilde{u }}{\Delta }\right)\frac{{d}^{2}{\beta }_{c}}{d{\tilde{u }}^{2}}+3{\sin}{\beta }_{c}{\delta }^{2}{\Delta }^{-5}\tilde{u }\right]\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.10)

where $\Delta =\sqrt{{\delta }^{2}-\tilde{u}^{2}}$.

$${\varepsilon }_{6}={\left.2{f}_{c4}\frac{{\delta }^{2}}{{\delta }^{2}-{\tilde{u }}^{2}}\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.11)

$${\varepsilon }_{7}={\left.4{f}_{c4}\frac{{\delta }^{2}\tilde{u }}{{{(\delta }^{2}-{\tilde{u }}^{2})}^{2}}\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.12)

$${\varepsilon }_{8}={\left.2{f}_{c4}\frac{{\delta }^{4}+3{\delta }^{2}{\tilde{u }}^{2}}{{{(\delta }^{2}-{\tilde{u }}^{2})}^{3}}\right|}_{\tilde{u }=\tilde{u}_{0}}$$

(A.13)

Appendix B

The equations for $a$, ${a}_{0}$ and $\varphi$ obtained by applying HBM are listed as below.

$$\left[\frac{8}{\pi }{\xi }_{c}\left({\mu }_{10}+{\mu }_{11}{a}_{0}+{\mu }_{12}{{a}_{0}}^{2}+\frac{1}{3}{\mu }_{12}{\mathrm{a}}^{2}\right)+2\mathrm{a\Omega }{\xi }_{v}\right]\mathrm{sin\varphi }-[\left({\varphi }_{1}+2{\varphi }_{2}{a}_{0}+3{\varphi }_{3}{{a}_{0}}^{2}+4{{{\varphi }_{4}a}_{0}}^{3}+{{5{\varphi }_{5}a}_{0}}^{4})\mathrm{a}+\frac{3}{4}\left({\varphi }_{3}+4{{\varphi }_{4}a}_{0}+10{\varphi }_{5}{{a}_{0}}^{2}\right){\mathrm{a}}^{3}+\frac{5}{8}{{\varphi }_{5}\mathrm{a}}^{5}-\mathrm{a}{\Omega }^{2}\right]\mathrm{cos\varphi }+{\Omega }^{2}{z}_{0}=0$$

(B.1)

$$\left[\frac{8}{\pi }{\xi }_{c}\left({\mu }_{10}+{\mu }_{11}{\mathrm{a}}_{0}+{\mu }_{12}{{\mathrm{a}}_{0}}^{2}+\frac{1}{3}{\mu }_{12}{\mathrm{a}}^{2}\right)+2\mathrm{a\Omega }{\xi }_{v}\right]\mathrm{cos\varphi }+[({\varphi }_{1}+2{\varphi }_{2}{a}_{0}+3{\varphi }_{3}{{a}_{0}}^{2}+4{{{\varphi }_{4}a}_{0}}^{3}+{{5{\varphi }_{5}a}_{0}}^{4})a+\frac{3}{4}\left({\varphi }_{3}+4{{\varphi }_{4}a}_{0}+10{\varphi }_{5}{{a}_{0}}^{2}\right){\mathrm{a}}^{3}+\frac{5}{8}{{\varphi }_{5}\mathrm{a}}^{5}-\mathrm{a}{\Omega }^{2}]\mathrm{sin\varphi }=0$$

(B.2)

$${\varphi }_{1}{a}_{0}+{\varphi }_{2}{{a}_{0}}^{2}+{{{\varphi }_{3}a}_{0}}^{3}+{\varphi }_{4}{{a}_{0}}^{4}+{\varphi }_{5}{{a}_{0}}^{5}+\frac{1}{2}({\varphi }_{2}+3{\varphi }_{3}{a}_{0}+6{\varphi }_{4}{{a}_{0}}^{2}+10{\varphi }_{5}{{a}_{0}}^{3}){a}^{2}+\frac{3}{8}\left({\varphi }_{4}+5{{\varphi }_{5}a}_{0}\right){a}^{4}=0$$

(B.3)