Design of a metastructure for vibration isolation with quasi-zero-stiffness characteristics using bistable curved beam

Abstract

This work designs and analyzes a metastructure-based vibration isolation model to improve small-scale equipment's isolation effectiveness under low-frequency excitations. The feature of the proposed model is the high static and low dynamic stiffness characteristics, also called quasi-zero-stiffness (QZS), possessed by the metastructure under vertical load. The metastructure consists of four parallelly arranged unit cells, and the QZS property is realized in each unit cell by the snap-through behavior of the cosine beam system and the bending-dominated behavior of semicircular arches. The static characteristics of the metastructure are studied analytically and numerically and validated with experimental results. Based on the static analysis results, the dynamic equation of the proposed metastructure is set up in the form of Duffing's equation. The harmonic balance method is used to calculate the frequency response and motion transmissibility of the metastructure at steady state for a harmonic load. The time and frequency responses under the sinusoidal base excitation are examined analytically and numerically, and their results are compared. The simulation results revealed that the proposed QZS metastructure obtains lower transmissibility and wider effective isolation range compared to the equivalent linear model. The parametric study shows that in the low-frequency excitation region, the motion transmissibility increases with decreasing damping ratio, whereas for the effective isolation range, the motion transmissibility increases with increasing damping ratio. Finally, stability analysis is performed to study the unstable region in the frequency response curve. The parametric study indicates that the unstable region reduces with the increase in damping ratio and remains unaffected with varying excitation amplitude.

Appendix

1.1 A. Derivation of Eq. (27)

Substituting the static equilibrium condition (x = 0) in Eq. (25) and equating k_QZS to zero

$$ k_{{{\text{QZS}}}} (x) = 8 + 4\mu - \frac{{4\mu \gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} $$

(25)

$$ 0 = 8 + 4\mu - \frac{{4\mu \gamma^{2} }}{{\left( {\gamma^{2} + 0} \right)^{3/2} }} $$

(63)

$$ 0 = 8 + 4\mu - \frac{{4\mu \gamma^{2} }}{{\gamma^{3} }} $$

(64)

$$ 0 = 8 + 4\mu \left( {1 - \frac{1}{\gamma }} \right) $$

(65)

$$ \mu \left( {1 - \frac{1}{\gamma }} \right) = - 2 $$

(66)

$$ \mu_{{{\text{QZS}}}} = \frac{ - 2\gamma }{{\gamma - 1}} = \frac{2\gamma }{{1 - \gamma }} $$

(67)

$$ {\text{or}},\quad 0 = 8 + 4\mu \left( {1 - \frac{1}{\gamma }} \right) $$

(65)

$$ \mu \left( {1 - \frac{1}{\gamma }} \right) = - 2 $$

(68)

$$ - \frac{1}{\gamma } = \frac{ - 2}{\mu } - 1 $$

(69)

$$ \frac{1}{\gamma } = \frac{2}{\mu } + 1 $$

(70)

$$ \gamma_{{{\text{QZS}}}} = \frac{\mu }{\mu + 2} $$

(71)

$$ \mu_{{{\text{QZS}}}} = \frac{2\gamma }{{1 - \gamma }}\quad \gamma_{{{\text{QZS}}}} = \frac{\mu }{\mu + 2} {.}$$

(27)

1.2 B. Dependence of geometrical parameter (γ) and stiffness parameter (µ)

Based on Eq. (25), a study is performed to show the variation of k_QZS with x for different values of γ and µ based on the three different conditions:

$$ k_{{{\text{QZS}}}} (x) = 8 + 4\mu - \frac{{4\mu \gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} $$

(25)

$$ \mu_{{{\text{QZS}}}} = \frac{2\gamma }{{1 - \gamma }}\quad \gamma_{{{\text{QZS}}}} = \frac{\mu }{\mu + 2} $$

(27)

B.i. Let’s consider $\mu$ as some constant value to analyze the dependency of $\gamma$ on Eq. (25).

Assume $\mu = 2$, then $\gamma_{{{\text{QZS}}}} = 0.5$ (from Eq. 27). Here, three different cases are discussed: (a) $\gamma = \gamma_{{{\text{QZS}}}} = 0.5$, (b) $\gamma = 0.6 > \gamma_{{{\text{QZS}}}}$, (c) $\gamma = 0.4 < \gamma_{{{\text{QZS}}}}$.

A curve is plotted in Fig. 31 based on Eq. (25) for comparing the three cases. It can be observed that for $\gamma = \gamma_{{{\text{QZS}}}}$, the equivalent stiffness is zero at the static equilibrium position (x = 0). For $\gamma < \gamma_{{{\text{QZS}}}}$, the equivalent stiffness becomes negative at static equilibrium position making the system unstable. For $\gamma > \gamma_{{{\text{QZS}}}}$, the equivalent stiffness becomes positive, and the QZS property is lost.

Fig. 31

Stiffness–displacement curve of metastructure based on Eq. (25) for varying γ, when μ = 2

It can be observed that the system is unstable for the condition $\gamma < \gamma_{{{\text{QZS}}}}$.

B.ii. Let’s consider $\gamma$ as some constant value to analyze the dependency of $\mu$ on Eq. (25).

Assume $\gamma = 0.5$, then $\mu_{{{\text{QZS}}}} = 2$ (from Eq. 27). Here, three different cases are discussed: (a) $\mu = \mu_{{{\text{QZS}}}} = 2$, (b) $\mu = 2.5 > \mu_{{{\text{QZS}}}}$, (c) $\mu = 1.5 < \mu_{{{\text{QZS}}}}$.

A curve is shown in Fig. 32 based on Eq. (25) for comparing the three cases. It can be observed that for $\mu = \mu_{{{\text{QZS}}}}$, the equivalent stiffness is zero at the static equilibrium position (x = 0). For $\mu > \mu_{{{\text{QZS}}}}$, the equivalent stiffness becomes negative at static equilibrium position making the system unstable. For $\mu < \mu_{{{\text{QZS}}}}$, the equivalent stiffness becomes positive, and the QZS property is lost.

Fig. 32

Stiffness–displacement curve of metastructure based on Eq. (25) for varying μ, when γ = 0.5

It can be observed that the system is unstable for the condition $\mu > \mu_{{{\text{QZS}}}}$.

B.iii. Equation (27) shows the relation between $\mu$ and $\gamma$, i.e.,

$$ \mu_{{{\text{QZS}}}} = \frac{2\gamma }{{1 - \gamma }}\quad \gamma_{{{\text{QZS}}}} = \frac{\mu }{\mu + 2} $$

(27)

From Fig. 5a and Eq. (18), it can be observed that

$$ \gamma = \frac{{L_{1} }}{{L_{0} }} = \cos \theta $$

(72)

As it is known that the range of cosine varies from 0 to 1, the extreme values of $\gamma$ are 0 and 1. By substituting these values of $\gamma$ in Eq. (27), we can observe that

$$ {\text{when }}\gamma_{\min } = 0,\mu_{{{\text{QZS}}}} = 0\,{\text{and}}\,{\text{when}}\gamma_{\max } = 1,\mu_{{{\text{QZS}}}} = {\text{undefined}} $$

(73)

Therefore, it can be seen that $\mu$ achieves singularity at $\gamma = 1$. The conditions shown in Eq. (73) are plotted in Fig. 33. It can be observed from the curve that for $\gamma_{\min } = 0$, the system behaves as an equivalent linear isolator with equivalent QZS equal to 8, and for $\gamma_{\max } = 1$, the value of equivalent QZS remains constant as zero throughout, which is practically difficult to achieve.

Fig. 33

Stiffness–displacement curve of metastructure based on Eq. (25) plotted for two different conditions obtained from Eq. (73) i.e., (i) $\gamma_{\min } = 0$ and (ii) $\gamma_{\max } = 1$

The limiting value obtained for $\gamma$ can be used to find the limiting conditions for $\mu$ as well; Eq. (27) leads to

$$ 0 < \frac{\mu }{\mu + 2} < 1 $$

(74)

From Eq. (74), it can be observed that

$$ \mu > 0 $$

(75)

As the stiffness cannot be negative, $\mu$ should always hold the condition mentioned in Eq. (75), and also $\mu \ne 0$, as it will lead the model to linear condition. A stiffness–displacement curve (shown in Fig. 

Fig. 34

Stiffness–displacement curve of metastructure based on Eq. (25) plotted for four different conditions of $\mu$ satisfying Eq. (27)

34) based on Eq. (25) is plotted for four different conditions of $\mu$ satisfying Eq. (27). It can be observed from the curve that $k_{{{\text{QZS}}}} = 0$ is achieved at static equilibrium position (x = 0) for all the four conditions of $\mu$, therefore satisfying the QZS criterion.

Hence, it can be observed from the above discussions that the system is unstable for $\gamma < \gamma_{{{\text{QZS}}}}$ and $\mu > \mu_{{{\text{QZS}}}}$. The metastructure will hold the QZS property when $0 < \gamma < 1$ and $\mu > 0$. The singularity for $\mu$ is achieved when $\gamma_{QZS} = 1$.

1.3 C. Derivation of Eq. (29)

$$ k_{{{\text{QZS}}}} (x) = 8 + 4\mu - \frac{{4\mu \gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} $$

(25)

Satisfying the condition $k_{{{\text{QZS}}}} \left( x \right) < 8$ from Eq. (28) and substituting $\mu_{{{\text{QZS}}}} = \frac{2\gamma }{{1 - \gamma }}$ from Eq. (27), Eq. (25) can be rewritten as

$$ k_{{{\text{QZS}}}} (x) = 8 + 4\mu - \frac{{4\mu \gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} < 8 $$

(76)

$$ 8 + 4\left( {\frac{2\gamma }{{1 - \gamma }}} \right) - \frac{{4\left( {\frac{2\gamma }{{1 - \gamma }}} \right)\gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} < 8 $$

(77)

$$ 4\left( {\frac{2\gamma }{{1 - \gamma }}} \right) - \frac{{4\left( {\frac{2\gamma }{{1 - \gamma }}} \right)\gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} < 0 $$

(78)

$$ 1 - \frac{{\gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} < 0 $$

(79)

$$ \frac{{\gamma^{2} }}{{\left( {\gamma^{2} + x^{2} } \right)^{3/2} }} > 1 $$

(80)

$$ \left( {\gamma^{2} + x^{2} } \right)^{3/2} < \gamma^{2} $$

(81)

$$ x^{2} < \gamma^{4/3} - \gamma^{2} $$

(82)

$$ x^{2} < \gamma^{4/3} - \gamma^{{\left( {4/3} \right) + \left( {2/3} \right)}} $$

(83)

$$ x^{2} < \gamma^{4/3} \left( {1 - \gamma^{2/3} } \right) $$

(84)

$$ \left| x \right| < \gamma^{2/3} \sqrt {\left( {1 - \gamma^{2/3} } \right)} {.}$$

(29)

1.4 D. Derivation of Eq. (35)

From Eq. (34), Taylor series expansion is

$$ f\left( x \right) = f\left( {x_{o} } \right) + \sum\limits_{n = 1}^{N} {\frac{{f^{n} \left( {x_{o} } \right)}}{n!}\left( {x - x_{o} } \right)^{n} } $$

(34)

Here,

$$ f_{{{\text{QZS}}}} (x) = 8x - 4\mu \left( {\frac{1}{{\sqrt {\gamma^{2} + x^{2} } }} - 1} \right)x $$

(24)

Finding the derivatives of $f\left( {x_{0} } \right)$

$$ f(x_{0} ) = 8x_{0} - 4\mu \left( {\frac{{x_{0} }}{{\sqrt {\gamma^{2} + x_{0}^{2} } }} - x_{0} } \right) $$

(85)

$$ f^{\prime}\left( {x_{0} } \right) = 8 + 4\mu - \frac{{4\mu \gamma^{2} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{3/2} }} $$

(86)

$$ f^{\prime\prime}\left( {x_{0} } \right) = \frac{{12\mu x_{0} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{3/2} }} - \frac{{12\mu x_{0}^{3} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{5/2} }} $$

(87)

$$ f^{\prime\prime\prime}\left( {x_{0} } \right) = \frac{12\mu }{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{3/2} }} - \frac{{72\mu x_{0}^{2} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{5/2} }} + \frac{{60\mu x_{0}^{4} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{7/2} }} $$

(88)

$$ f^{iv} \left( {x_{0} } \right) = \frac{{600\mu x_{0}^{3} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{7/2} }} - \frac{{420\mu x_{0}^{5} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{9/2} }} - \frac{{180\mu x_{0} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{5/2} }} $$

(89)

$$ f^{v} \left( {x_{0} } \right) = \frac{{2700\mu x_{0}^{2} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{7/2} }} - \frac{180\mu }{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{5/2} }} - \frac{{6300\mu x_{0}^{4} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{9/2} }} + \frac{{3780\mu x_{0}^{6} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{11/2} }} $$

(90)

$$ f^{vi} \left( {x_{0} } \right) = \frac{{79380\mu x_{0}^{5} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{11/2} }} - \frac{{44100\mu x_{0}^{3} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{9/2} }} - \frac{{41580\mu x_{0}^{7} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{13/2} }} + \frac{{6300\mu x_{0} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{7/2} }} $$

(91)

$$ f^{vii} \left( {x_{0} } \right) = \frac{6300\mu }{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{7/2} }} - \frac{{176400\mu x_{0}^{2} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{9/2} }} + \frac{{793800\mu x_{0}^{4} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{11/2} }} - \frac{{1164240\mu x_{0}^{6} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{13/2} }} + \frac{{540540\mu x_{0}^{8} }}{{\left( {\gamma^{2} + x_{0}^{2} } \right)^{15/2} }} $$

(92)

Substituting $x_{0} = 0$ in Eq. (92),

$$ f\left( 0 \right) = 0\quad f^{\prime}\left( 0 \right) = 8 + 4\mu \left( {1 - \frac{1}{\gamma }} \right)\quad f^{\prime\prime}\left( 0 \right) = 0\quad f^{\prime\prime\prime}\left( 0 \right) = \frac{12\mu }{{\gamma^{3} }} $$

(93)

$$ f^{iv} \left( 0 \right) = 0\quad f^{v} \left( 0 \right) = - \frac{180\mu }{{\gamma^{5} }}\quad f^{vi} \left( 0 \right) = 0\quad f^{vii} \left( 0 \right) = - \frac{6300\mu }{{\gamma^{7} }} $$

(94)

By expanding Eq. (24) using Eq. (34), the expression for the force–displacement can be approximated as Eq. (35):

$$ f\left( x \right) = \left[ {8 + 4\mu \left( {\frac{\gamma - 1}{\gamma }} \right)} \right]x + \frac{2\mu }{{\gamma^{3} }}x^{3} - \frac{3\mu }{{2\gamma^{5} }}x^{5} + \frac{5\mu }{{4\gamma^{7} }}x^{7} {.}$$

(35)

1.5 E. Derivation of Eq. (49)

First representing Eq. (47a) and (47b)

$$ - \Omega^{2} A + \left( {\frac{3}{4}\alpha A^{3} } \right) + \left( {\frac{5}{8}\delta A^{5} } \right) = - z_{0} \cos \left( \phi \right) $$

(47a)

$$ - 2\xi A\Omega = - z_{0} \sin \left( \phi \right) $$

(47b)

Squaring and adding Eq. (47a) and (47b), and thereafter eliminating $\phi$,

$$ \Omega^{4} A^{2} + \frac{9}{16}\alpha^{2} A^{6} + \frac{25}{{64}}\delta^{2} A^{10} - \frac{3}{2}\Omega^{2} \alpha A^{4} - \frac{5}{4}\Omega^{2} \delta A^{6} + \frac{15}{{16}}\delta \alpha A^{8} + 4\xi^{2} \Omega^{2} A^{2} = z_{0}^{2} $$

(95)

$$ \Omega^{4} A^{2} + \Omega^{2} \left[ {4\xi^{2} A^{2} - \frac{3}{2}\alpha A^{4} - \frac{5}{4}\delta A^{6} } \right] + \left[ {\frac{9}{16}\alpha^{2} A^{6} + \frac{25}{{64}}\delta^{2} A^{10} \frac{15}{{16}}\delta \alpha A^{8} - z_{0}^{2} } \right] = 0 $$

(96)

Let, $\Omega^{2} = w$

$$ w^{2} A^{2} + w\left[ {4\xi^{2} A^{2} - \frac{3}{2}\alpha A^{4} - \frac{5}{4}\delta A^{6} } \right] + \left[ {\frac{9}{16}\alpha^{2} A^{6} + \frac{25}{{64}}\delta^{2} A^{10} \frac{15}{{16}}\delta \alpha A^{8} - z_{0}^{2} } \right] = 0 $$

(97)

$$ w = - \frac{b}{2a} \pm \frac{{\sqrt {b^{2} - 4ac} }}{2a} $$

(98)

$$ w = \left( { - 2\xi^{2} + \frac{3}{4}\alpha A^{2} + \frac{5}{8}\delta A^{4} } \right) \pm \sqrt {\left( {2\xi^{2} - \frac{3}{4}\alpha A^{2} - \frac{5}{8}\delta A^{4} } \right)^{2} - \left( {\frac{9}{16}\alpha^{2} A^{4} + \frac{25}{{64}}\delta^{2} A^{8} + \frac{15}{{16}}\delta \alpha A^{6} - \frac{{z_{0}^{2} }}{{A^{2} }}} \right)} $$

(99)

Substituting, $w = \Omega^{2}$

$$ \begin{gathered} \Omega = \pm \left\{ {\left( {\frac{3}{4}\alpha A^{2} + \frac{5}{8}\delta A^{4} - 2\xi^{2} } \right) \pm \left[ {\left( { - \frac{3}{4}\alpha A^{2} - \frac{5}{8}\delta A^{4} + 2\xi^{2} } \right)^{2} - \left( {\frac{9}{16}\alpha^{2} A^{4} + \frac{25}{{64}}\delta^{2} A^{8} + \frac{15}{{16}}\delta \alpha A^{6} - \frac{{z_{0}^{2} }}{{A^{2} }}} \right)} \right]^{1/2} } \right\}^{1/2} \hfill \\ \hfill \\ \end{gathered} {.}$$

(49)

1.6 F. Derivation of Eqs. (50), (51), (52), and (53)

At the peak resonance point, both the intersections coincide with each other, i.e., Ω₁ became equal to Ω₂ of Eq. (49).

Considering only the real part, Ω₁ and Ω₂ can be written as

$$ \Omega_{1} = \left\{ {\left( {\frac{3}{4}\alpha A^{2} + \frac{5}{8}\delta A^{4} - 2\xi^{2} } \right) + \left[ {\left( { - \frac{3}{4}\alpha A^{2} - \frac{5}{8}\delta A^{4} + 2\xi^{2} } \right)^{2} - \left( {\frac{9}{16}\alpha^{2} A^{4} + \frac{25}{{64}}\delta^{2} A^{8} + \frac{15}{{16}}\delta \alpha A^{6} - \frac{{z_{0}^{2} }}{{A^{2} }}} \right)} \right]^{1/2} } \right\}^{1/2} $$

(49a)

$$ \Omega_{2} = \left\{ {\left( {\frac{3}{4}\alpha A^{2} + \frac{5}{8}\delta A^{4} - 2\xi^{2} } \right) - \left[ {\left( { - \frac{3}{4}\alpha A^{2} - \frac{5}{8}\delta A^{4} + 2\xi^{2} } \right)^{2} - \left( {\frac{9}{16}\alpha^{2} A^{4} + \frac{25}{{64}}\delta^{2} A^{8} + \frac{15}{{16}}\delta \alpha A^{6} - \frac{{z_{0}^{2} }}{{A^{2} }}} \right)} \right]^{1/2} } \right\}^{1/2} $$

(49b)

Equating Eq. (49a) and (49b),

$$ 2\left[ {\left( { - \frac{3}{4}\alpha A^{2} - \frac{5}{8}\delta A^{4} + 2\xi^{2} } \right)^{2} - \left( {\frac{9}{16}\alpha^{2} A^{4} + \frac{25}{{64}}\delta^{2} A^{8} + \frac{15}{{16}}\delta \alpha A^{6} - \frac{{z_{0}^{2} }}{{A^{2} }}} \right)} \right]^{1/2} = 0 $$

(100)

$$ \frac{9}{16}\alpha^{2} A^{4} + \frac{25}{{64}}\delta^{2} A^{8} + 4\xi^{4} + \frac{15}{{16}}\delta \alpha A^{6} - 3\alpha A^{2} \xi^{2} - \frac{5}{2}\delta A^{4} \xi^{2} - \frac{9}{16}\alpha^{2} A^{4} - \frac{25}{{64}}\delta^{2} A^{8} - \frac{15}{{16}}\delta \alpha A^{6} + \frac{{z_{0}^{2} }}{{A^{2} }} = 0 $$

(101)

$$ 4\xi^{4} - 3\alpha A^{2} \xi^{2} - \frac{5}{2}\delta A^{4} \xi^{2} + \frac{{z_{0}^{2} }}{{A^{2} }} = 0 $$

(102)

$$ 3\alpha A^{4} + \frac{5}{2}\delta A^{6} - 4\xi^{2} A^{2} - \frac{{z_{0}^{2} }}{{\xi^{2} }} = 0 $$

(103)

Denoting A as peak amplitude (A_p)

$$ \frac{5}{2}\delta A_{{\text{p}}}^{6} + 3\alpha A_{{\text{p}}}^{4} - 4\xi^{2} A_{{\text{p}}}^{2} - \frac{{z_{0}^{2} }}{{\xi^{2} }} = 0 $$

(50)

At this peak resonance, peak frequency ($\Omega_{{\text{p}}}$) occurs, using Eq. (49) to derive the relation of $\Omega_{{\text{p}}}$. As it is clear that, at resonance condition, the nested square root of Eq. (49) is zero, so $\Omega_{{\text{p}}}$ can be calculated as

$$ \Omega_{1} = \left\{ {\left( {\frac{3}{4}\alpha A_{{\text{p}}}^{2} + \frac{5}{8}\delta A_{{\text{p}}}^{4} - 2\xi^{2} } \right) + 0} \right\}^{1/2} $$

(104)

$$ \Omega_{2} = \left\{ {\left( {\frac{3}{4}\alpha A_{{\text{p}}}^{2} + \frac{5}{8}\delta A_{{\text{p}}}^{4} - 2\xi^{2} } \right) + 0} \right\}^{1/2} $$

(105)

$$ \Omega_{p} = \sqrt {\frac{3}{4}\alpha A_{{\text{p}}}^{2} + \frac{5}{8}\delta A_{{\text{p}}}^{4} - 2\xi^{2} } $$

(51)

Equating Eq. (51) to zero,

$$ \Omega_{{\text{p}}} = \sqrt {\frac{3}{4}\alpha A_{{\text{p}}}^{2} + \frac{5}{8}\delta A_{{\text{p}}}^{4} - 2\xi^{2} } = 0 $$

(106)

$$ \frac{5}{8}\delta A_{{\text{p}}}^{4} + \frac{3}{4}\alpha A_{{\text{p}}}^{2} - 2\xi^{2} = 0 $$

(107)

$$ \frac{5}{8}\delta \left( {A_{{\text{p}}}^{2} } \right)^{2} + \frac{3}{4}\alpha \left( {A_{{\text{p}}}^{2} } \right) - 2\xi^{2} = 0 $$

(108)

$$ A_{{\text{p}}}^{2} = \frac{{\left( { - \frac{3}{4}\alpha } \right) \pm \sqrt {\left( {\frac{3}{4}\alpha } \right)^{2} + 5\delta \xi^{2} } }}{{\frac{5}{4}\delta }} $$

(109)

$$ A_{{\text{p}}}^{2} = \frac{1}{\delta }\left( {\sqrt {\frac{9}{25}\alpha^{2} + \frac{16}{5}\delta \xi^{2} } - \frac{3}{5}\alpha } \right) $$

(52)

Substituting Eq. (52) into Eq. (50) and ignoring the higher-order amplitude term,

$$ z_{0}^{2} = 4\xi^{4} \frac{1}{\delta }\left( {\sqrt {\frac{9}{25}\alpha^{2} + \frac{16}{5}\delta \xi^{2} } - \frac{3}{5}\alpha } \right) $$

(110)

$$z_{0} = 2\xi^{2} \sqrt{ {{\frac{1}{\delta }}} \left({\sqrt {\frac{9}{25}\alpha^{2} + \frac{16}{5}\delta \xi^{2} } - \frac{3\alpha }{{5\delta }}}\right)}{.}$$

(53)

1.7 G. Derivation of Eqs. (55) and (56)

Considering Eq. (54),

$$ T = \frac{{\sqrt {\left( {\Omega^{4} A^{2} } \right) - (2\Omega^{2} Az_{0} \cos \phi ) + z_{0}^{2} } }}{{z_{0} }} $$

(54)

Finding out the value of $\cos \left( \phi \right)$ form Eq. (47a), and substituting in Eq. (54),

$$ \cos \left( \phi \right) = - \frac{{ - \Omega^{2} A + \left( {\frac{3}{4}\alpha A^{3} } \right) + \left( {\frac{5}{8}\delta A^{5} } \right)}}{{z_{0} }} $$

(111)

$$ T = \frac{{\sqrt {\left( {\Omega^{4} A^{2} } \right) - (2\Omega^{2} Az_{0} \cos \phi ) + z_{0}^{2} } }}{{z_{0} }} $$

(112)

$$ T = \frac{{\sqrt {\left( {\Omega^{4} A^{2} } \right) + \left( { - 2\Omega^{4} A^{2} + \left( {\frac{3}{2}\Omega^{2} \alpha A^{4} } \right) + \left( {\frac{5}{4}\Omega^{2} \delta A^{6} } \right)} \right) + z_{0}^{2} } }}{{z_{0} }} $$

(113)

$$ T = \sqrt { - \frac{{\Omega^{4} A^{2} }}{{z_{0}^{2} }} + \frac{{\frac{3}{2}\Omega^{2} \alpha A^{4} }}{{z_{0}^{2} }} + \frac{{\frac{5}{4}\Omega^{2} \delta A^{6} }}{{z_{0}^{2} }} + 1} $$

(114)

$$ T = \sqrt {\left( {\frac{\Omega A}{{z_{0} }}} \right)^{2} \left( {\frac{5}{4}\delta A^{4} + \frac{3}{2}\alpha A^{2} - \Omega^{2} } \right) + 1} $$

(55)

By substituting the value of $\Omega$ as $\Omega_{{\text{p}}}$ from Eq. (50) and $A$ as $A_{{\text{p}}}$, the value of peak transmissibility $T_{{\text{p}}}$ can be obtained:

$$ \Omega = \Omega_{{\text{p}}} = \sqrt {\frac{3}{4}\alpha A_{{\text{p}}}^{2} + \frac{5}{8}\delta A_{{\text{p}}}^{4} - 2\xi^{2} } $$

(115)

$$ A^{2} = A_{{\text{p}}}^{2} = \frac{1}{\delta }\left( {\sqrt {\frac{9}{25}\alpha^{2} + \frac{16}{5}\delta \xi^{2} } - \frac{3}{5}\alpha } \right) $$

(116)

$$ T_{{\text{p}}} = \sqrt {\frac{{A_{{\text{p}}}^{2} }}{{z_{0}^{2} }}\left[ {\left( {\frac{5}{8}\delta A_{{\text{p}}}^{2} + \frac{3}{4}\alpha } \right)^{2} A_{{\text{p}}}^{4} - 4\xi^{4} } \right] + 1}{.} $$

(56)

1.8 H. Derivation of Eqs. (59), (61), and (62)

Substituting Eq. (58) into Eq. (43),

$$ y( \tau ) = A\cos \left( {\Omega \tau + \phi } \right) + \varepsilon ( \tau ) $$

(58)

$$ y^{\prime\prime} + 2\xi y^{\prime} + \alpha y^{3} + \delta y^{5} = - z_{0} \cos \left( {\Omega \tau } \right) $$

(43)

Finding out the value of variables,

$$ \Omega \tau + \phi = \theta $$

(117)

$$ - z_{0} \cos \left( {\Omega \tau } \right) = - z_{0} \cos \theta \cos \phi - z_{0} \sin \theta \sin \phi $$

(118)

$$ y( \tau ) = A\cos \left( {\Omega \tau + \phi } \right) + \varepsilon ( \tau ) $$

(119)

$$ y^{\prime}( \tau ) = - A\Omega \sin \theta + \varepsilon^{\prime}( \tau ) $$

(120)

$$ y^{\prime\prime} = - A\Omega^{2} \cos \theta + \varepsilon^{\prime\prime}( \tau ) $$

(121)

Ignoring the term of order higher than $O\left( {\varepsilon^{2} } \right)$, and also assuming a harmonic response,

$$ y^{3} = \left( {A^{3} \cos^{3} \theta + 3\varepsilon A^{2} \cos^{2} \theta } \right) $$

(122)

$$ \cos^{3} \theta = \frac{3}{4}\cos \theta $$

(123)

$$ \cos^{2} \theta = \frac{\cos 2\theta + 1}{2} $$

(124)

$$ y^{3} = \left( {A^{3} \frac{3}{4}\cos \theta + 3\varepsilon A^{2} \frac{\cos 2\theta + 1}{2}} \right) $$

(125)

Substituting all these values into Eq. (43), and performing the first-order perturbation analysis by ignoring the term of order higher than $O(\epsilon^2 )$, and also assuming a harmonic response, the equation of motion can be expressed in the form of

$$ \varepsilon^{\prime\prime} + 2\mu \varepsilon^{\prime} + \left( {p - 2q\cos 2\theta } \right)\varepsilon = 0 $$

(59)

Here,

$$ p = \frac{{3\alpha A^{2} }}{{2}{\Omega^{2} }},\quad q = - \frac{{3\alpha A^{2} }}{{4\Omega^{2} }},\quad \mu = \frac{\xi }{\Omega } $$

(60)

The solution of Eq. (59) is unstable if it lies outside the parabola:

$$ p = 1 \pm \sqrt {q^{2} - 4\mu^{2} } $$

(61)

By substituting Eq. (60) into Eq. (61),

$$ \frac{{3\alpha A^{2} }}{{2\Omega^{2} }} = 1 \pm \sqrt {\left( { - \frac{{3\alpha A^{2} }}{{4\Omega^{2} }}} \right)^{2} - 4\left( {\frac{\xi }{\Omega }} \right)^{2} } $$

(126)

$$ 3\alpha A^{2} = 2\Omega^{2} \pm \sqrt {\left( {\frac{{9\alpha^{2} A^{4} }}{4}} \right) - 16\xi^{2} \Omega^{2} } $$

(127)

$$ 3\alpha A^{2} - 2\Omega^{2} = \pm \sqrt {\left( {\frac{{9\alpha^{2} A^{4} }}{4}} \right) - 16\xi^{2} \Omega^{2} } $$

(128)

$$ 9\alpha^{2} A^{4} + 4\Omega^{4} - 12\alpha A^{2} \Omega^{2} = \frac{{9\alpha^{2} A^{4} }}{4} - 16\xi^{2} \Omega^{2} $$

(129)

$$ 4\Omega^{4} + \frac{{27\alpha^{2} A^{4} }}{4} + \Omega^{2} \left( {16\xi^{2} - 12\alpha A^{2} } \right) = 0 $$

(130)

$$ \Omega^{4} + \Omega^{2} \left( {4\xi^{2} - 3\alpha A^{2} } \right) + \frac{{27\alpha^{2} A^{4} }}{16} = 0 $$

(131)

$$ \Omega_{{{\text{us}}1}} = \sqrt {\frac{3}{2}\alpha A^{2} - 2\xi^{2} + \frac{1}{2}\sqrt {\left( {4\xi^{2} - 3\alpha A^{2} } \right)^{2} - \frac{27}{4}\alpha^{2} A^{4} } } $$

(62a)

$$ \Omega_{{{\text{us}}2}} = \sqrt {\frac{3}{2}\alpha A^{2} - 2\xi^{2} - \frac{1}{2}\sqrt {\left( {4\xi^{2} - 3\alpha A^{2} } \right)^{2} - \frac{27}{4}\alpha^{2} A^{4} } } . $$

(62b)