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Open Access 2023 | OriginalPaper | Chapter

# 2. Responses of 2DOF Sliding Base Systems Under Harmonic Ground Motions

Author : Hong-Song Hu

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Publisher: Springer Nature Singapore

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## Abstract

Although actual structures would not be subjected to harmonic ground motions, theoretical solutions obtained under harmonic ground motions can often provide useful information for understanding the mechanism of the structures considered. For this reason, this chapter focuses on the responses of 2-degree-of-freedom (2DOF) sliding base (SB) systems subjected to harmonic ground motions. The main contents of the present chapter include the typical response histories of a 2DOF SB system subjected to harmonic ground motions, the influence of the critical parameters on the maximum structural responses, and the derivation and interpretation of the available theoretical solutions.

## 2.1 Equations of Motion

For a 2DOF SB system shown in Fig. 2.1, the dynamic equilibrium equations can be written as
$$\left\{ {\begin{array}{*{20}l} {m\left( {\ddot{u}_{g} + \ddot{u}_{s} + \ddot{u}_{r} } \right) + c\dot{u}_{r} + ku_{r} = 0} \hfill \\ {m_{b} \left( {\ddot{u}_{g} + \ddot{u}_{s} } \right) + m\left( {\ddot{u}_{g} + \ddot{u}_{s} + \ddot{u}_{r} } \right) = f} \hfill \\ \end{array} } \right.$$
(2.1)
in which m, k, and c refer to the top mass, lateral stiffness, and viscous damping coefficient of the superstructure, respectively; mb is the mass of the sliding base; ug(t), us(t), and ur(t) are the ground displacement, sliding displacement, and relative displacement between the top mass and sliding base, respectively; $$\dot{u}_{g}$$, $$\dot{u}_{s}$$, and $$\dot{u}_{r}$$ are the corresponding velocities; $$\ddot{u}_{g}$$, $$\ddot{u}_{s}$$, and $$\ddot{u}_{r}$$ are the corresponding; and f is the friction force between the sliding base and foundation. The first equation of Eq. (2.1) denotes the dynamic equilibrium of the top mass, while the second equation is the dynamic equilibrium of the entire system.
The SB system can display two kinds of phases in its response history: the stick phase and the sliding phase. For the stick phases, the sliding acceleration, $$\ddot{u}_{s}$$, is equal to 0, and the sliding friction force is greater than the friction force, f; therefore, Eq. (2.1) leads to
$$\left\{ {\begin{array}{*{20}l} {m\ddot{u}_{r} + c\dot{u}_{r} + ku_{r} = - m\ddot{u}_{g} } \hfill \\ {\alpha \ddot{u}_{r} + \ddot{u}_{g} = f/\left( {m + m_{b} } \right)} \hfill \\ {\left| f \right| < \left( {m + m_{b} } \right)\mu g} \hfill \\ \end{array} } \right.$$
(2.2)
where
$$\alpha = \frac{m}{{m + m_{b} }}$$
(2.3)
is the mass ratio; $$\mu$$ is the friction coefficient; and g is the gravity acceleration. The first equation of Eq. (2.2) can be written as
$$\ddot{u}_{r} + 2\xi \omega \dot{u}_{r} + \omega^{2} u_{r} = - \ddot{u}_{g}$$
(2.4)
where
$$\omega = \sqrt{\frac{k}{m}} ,\quad \xi = \frac{c}{2\omega m}$$
(2.5)
$$\omega$$ and $$\xi$$ are the natural frequency and damping ratio of the corresponding fixed base structure, respectively. Equation (2.4) is the differential equation governing the relative displacement of a single-degree-of-freedom (SDOF) system under ground acceleration, $$\ddot{u}_{g}$$.
According to the second and third equations of Eq. (2.2), we obtain
$$\left| {\alpha \ddot{u}_{r} + \ddot{u}_{g} } \right| < \mu g$$
(2.6)
This is the precondition for the stick phases. Sliding occurs when it is no longer satisfied, and
$$f = \delta \left( {m + m_{b} } \right)\mu g$$
(2.7)
in which $$\delta$$ represents the direction of the friction force, and $$\left| \delta \right| = 1$$. While transitioning from the stick phase to the sliding phase, the direction of the friction force remains unchanged. Thus, it can be inferred from the second equation of Eq. (2.2) that $$\delta$$ and $$\alpha \ddot{u}_{r} + \ddot{u}_{g}$$ have the same sign at the moment before sliding. By substituting Eq. (2.7) into the second equation of Eq. (2.1), we obtain
$$\ddot{u}_{s} = \delta \mu g - \alpha \ddot{u}_{r} - \ddot{u}_{g}$$
(2.8)
Substituting Eq. (2.8) into the first equation of Eq. (2.1) to obtain
$$\left( {1 - \alpha } \right)m\ddot{u}_{r} + c\dot{u}_{r} + ku_{r} = - \delta \mu mg$$
(2.9)
Dividing Eq. (2.9) by (1 − $$\alpha$$)m to obtain
$$\ddot{u}_{r} + 2\xi_{1} \omega_{1} \dot{u}_{r} + \omega_{1}^{2} u_{r} = - \frac{\delta \mu g}{{1 - \alpha }}$$
(2.10)
where
$$\omega_{1} = \frac{\omega }{{\sqrt {1 - \alpha } }},\quad \xi_{1} = \frac{\xi }{{\sqrt {1 - \alpha } }}$$
(2.11)
Equation (2.10) is the differential equation of a SDOF system with the natural frequency of $$\omega_{1}$$ and damping ratio of $$\xi_{1}$$ under a step force. The static displacement corresponding to the step force is
$$u_{st} = \frac{ - \delta \mu g}{{\omega_{1}^{2} \left( {1 - \alpha } \right)}} = \frac{ - \delta \mu g}{{\omega^{2} }}$$
(2.12)
Thus, in the sliding phases, a new natural frequency, $$\omega_{1}$$, and a new damping ratio, $$\xi_{1}$$, of the relative displacement vibration are determined, both of which are linked to the mass ratio, $$\alpha$$. The solution of Eq. (2.10) is
$$\left( {\begin{array}{*{20}c} {u_{r} \left( t \right)} \\ {\dot{u}_{r} \left( t \right)} \\ \end{array} } \right) = {\mathbf{A}}\left( \tau \right)\left( {\begin{array}{*{20}c} {u_{r} \left( {t_{i} } \right)} \\ {\dot{u}_{r} \left( {t_{i} } \right)} \\ \end{array} } \right) + {\mathbf{b}}\left( \tau \right)$$
(2.13)
in which t is the global time, ti is the moment when sliding starts, $$\tau$$ = t − ti, and
\begin{aligned} {\mathbf{A}}\left( \tau \right) & = e^{{ - \xi_{1} \omega_{1} \tau }} \\ & \quad \left[ {\begin{array}{*{20}c} {\cos \left( {\omega_{1d} \tau } \right) + \xi_{1} /\sqrt {1 - \xi_{1}^{2} } \sin \left( {\omega_{1d} \tau } \right)} & {\frac{{\sin \left( {\omega_{1d} \tau } \right)}}{{\omega_{1} \sqrt {1 - \xi_{1}^{2} } }}} \\ { - \frac{{\omega_{1} \sin \left( {\omega_{1d} \tau } \right)}}{{\sqrt {1 - \xi_{1}^{2} } }}} & {\cos \left( {\omega_{1d} \tau } \right) - \xi_{1} /\sqrt {1 - \xi_{1}^{2} } \sin \left( {\omega_{1d} \tau } \right)} \\ \end{array} } \right] \\ \end{aligned}
(2.14)
$${\mathbf{b}}\left( \tau \right) = u_{st} \left[ {\begin{array}{*{20}c} {1 - e^{{ - \xi_{1} \omega_{1} \tau }} \left( {\xi_{1} /\sqrt {1 - \xi_{1}^{2} } \sin \left( {\omega_{1d} \tau } \right) + \cos \left( {\omega_{1d} \tau } \right)} \right)} \\ {e^{{ - \xi_{1} \omega_{1} \tau }} \frac{{\omega_{1} \sin \left( {\omega_{1d} \tau } \right)}}{{\sqrt {1 - \xi_{1}^{2} } }}} \\ \end{array} } \right]$$
(2.15)
$$\omega_{1d} = \omega_{1} \sqrt {1 - \xi_{1}^{2} }$$
(2.16)
The first term in the right-hand side of Eq. (2.13) corresponds to free vibration caused by the initial condition, while the second term represents the vibration caused by the step force.
At the onset of sliding, the velocity of sliding equals to 0, i.e., $$\dot{u}_{s} \left( {t_{i} } \right) = 0$$. Therefore, integrating Eq. (2.8) leads to
$$\dot{u}_{s} \left( t \right) = \delta \mu g\left( {t - t_{i} } \right) - \alpha \left( {\dot{u}_{r} \left( t \right) - \dot{u}_{r} \left( {t_{i} } \right)} \right) - \left( {\dot{u}_{g} \left( t \right) - \dot{u}_{g} \left( {t_{i} } \right)} \right)$$
(2.17)
this round Sliding stops once $$\dot{u}_{s} \left( t \right)$$ returns to 0 again. The sliding of the structure can either stop or persist, based on whether Eq. (2.6) is satisfied or not. For this check of Eq. (2.6), Eq. (2.4) should be used to determine the relative acceleration $$\ddot{u}_{r}$$, under the assumption that sliding stops. The sliding displacement, $$u_{s} \left( t \right)$$, can be obtained by integrating Eq. (2.17):
\begin{aligned} u_{s} \left( t \right) & = u_{s} \left( {t_{i} } \right) + \frac{1}{2}\delta \mu g\left( {t - t_{i} } \right)^{2} - \alpha \left( {u_{r} \left( t \right) - u_{r} \left( {t_{i} } \right) - \dot{u}_{r} \left( {t_{i} } \right)\left( {t - t_{i} } \right)} \right) \\ & \quad - \left( {u_{g} \left( t \right) - u_{g} \left( {t_{i} } \right) - \dot{u}_{g} \left( {t_{i} } \right)\left( {t - t_{i} } \right)} \right) \\ \end{aligned}
(2.18)
Figure 2.2 summarizes the process of computing the response history.

## 2.2 Typical Response Histories

In the following analyses, the ground acceleration is taken as a sinusoidal wave if not specified:
$$\ddot{u}_{g} = a_{g} \sin \left( {\omega_{g} t} \right)$$
(2.19)
where ag and $$\omega_{g}$$ are the amplitude and frequency of the sinusoidal wave, respectively.
Figure 2.3 shows the responses of a sliding base system with $$\mu$$ = 0.2, $$\alpha$$ = 0.5, T = 2$$\pi /\omega$$ = 0.5 s, $$\xi$$ = 5%, Tg = 2$$\pi /\omega_{g}$$ = 1 s, and ag = 0.4g, in which T represents the natural period of the superstructure, and Tg indicates the period of the ground acceleration. As shown in Fig. 2.3, the responses of the system converge to steady periodic responses with the same period as the ground acceleration after several cycles. The sliding base system pauses briefly before sliding in the opposite direction in this instance. This is referred to as the stick-sliding case. After the amplitude of the ground acceleration exceeds a certain value, the sliding base system will continue to slide incessantly during the steady periodic state, as shown in Fig. 2.4, where the ground acceleration amplitude, ag, is increased to 1.2g. This is referred to as the sliding-sliding case. Sliding will not occur if the ground acceleration is sufficiently low, which is known as stick-stick case. The sliding base structure and the fixed base structure have no difference in this case. The following analyses focus on the steady periodic responses.

## 2.3 Occurrence Conditions of the Three Types of Periodic Responses

### 2.3.1 Boundaries Between the Stick-Stick and Stick-Sliding Cases

For the stick-stick case, the steady response of the relative displacement is
$$u_{r} \left( t \right) = - \frac{{a_{g} }}{{\omega^{2} }}R_{d} \sin \left( {\omega_{g} t - \phi } \right)$$
(2.20)
$$\phi = \tan^{ - 1} \frac{{2\xi \left( {\omega_{g} /\omega } \right)}}{{1 - \left( {\omega_{g} /\omega } \right)^{2} }}$$
(2.21)
where $$\phi$$ is the phase angle, which defines the time by which the response lags behind the input ground motion. Substituting Eqs. (2.19) and (2.20) into the precondition for the stick phases (i.e., Eq. 2.6) gives
$$\left| {\alpha R_{a} \sin \left( {\omega_{g} t - \phi } \right) + \sin \left( {\omega_{g} t} \right)} \right| < \mu g/a_{g}$$
(2.22)
where
$$R_{a} = \left( {\omega_{g} /\omega } \right)^{2} R_{d} = \frac{{\left( {\omega_{g} /\omega } \right)^{2} }}{{\sqrt {\left[ {1 - \left( {\omega_{g} /\omega } \right)^{2} } \right]^{2} + \left[ {2\xi \left( {\omega_{g} /\omega } \right)} \right]^{2} } }}$$
(2.23)
is the acceleration response factor for the fixed base structures. Equation (2.22) can be rewritten as
$$\left| {\left( {\alpha R_{a} \cos \phi + 1} \right)\sin \left( {\omega_{g} t} \right) - \alpha R_{a} \sin \phi \cos \left( {\omega_{g} t} \right)} \right| < \mu g/a_{g}$$
(2.24)
Equation (2.24) is always satisfied for the stick-stick cases, so the maximum value of the left-hand side term should be smaller than μg/ag; thus, we have
$$\mu g/a_{g} > \sqrt {\left( {\alpha R_{a} } \right)^{2} + 2\left( {\alpha R_{a} } \right)\cos \phi + 1}$$
(2.25)
Therefore, the boundaries between the stick-stick and stick-sliding cases are
$$\frac{{a_{g} }}{\mu g} = \frac{1}{{\sqrt {\left( {\alpha R_{a} } \right)^{2} + 2\left( {\alpha R_{a} } \right)\cos \phi + 1} }}$$
(2.26)
The variations of the boundaries between the stick-stick and stick-sliding cases for different $$\omega_{g} /\omega$$ and $$\alpha$$ are shown in Fig. 2.5.

### 2.3.2 Boundaries Between the Stick-Sliding and Sliding-Sliding Cases

As derived in Sect. 2.2, Eq. (2.10) is the governing equation of the relative displacement during the sliding phase. Equation (2.10) is the differential equation of a SDOF fixed base system with the natural frequency of $$\omega_{1}$$ and damping ratio of $$\xi_{1}$$ subjected to a step force. The static displacement corresponding to this step force, ust, is given in Eq. (2.12), which has the same sign as the sliding direction. As shown in Fig. 2.4, for the sliding-sliding case, the sliding base system slides in one direction for half period of the ground motion, 0.5Tg, and then slides in the opposite direction for another 0.5Tg, and so on. Therefore, the equivalent step force also changes direction in 0.5Tg when the sliding direction changes, as shown in Fig. 2.6. Thus, the relative displacements of the two opposite half sliding cycles have the same absolute values but opposite signs.
For a certain half sliding cycle, the solution of Eq. (2.10) (i.e. Eq. 2.13) can be rewritten as
$$\left( {\begin{array}{*{20}c} {u_{r} \left( t \right)} \\ {\dot{u}_{r} \left( t \right)} \\ \end{array} } \right) = {\mathbf{A}}\left( \tau \right)\left( {\begin{array}{*{20}c} {u_{r0} } \\ {\dot{u}_{r0} } \\ \end{array} } \right) + {\mathbf{b}}\left( \tau \right)$$
(2.27)
where ur0 and $$\dot{u}_{r0}$$ are the relative displacement and velocity at the moment when this half sliding cycle starts, and this moment is denoted as ti; $$\tau$$ = t − ti is the local time during this half sliding cycle, so $$0 \le \tau \le \pi /\omega_{g}$$. When $$\tau = \pi /\omega_{g}$$, the opposite half sliding cycle starts, the relative displacement and velocity at this moment are equal to −ur0 and $$- \dot{u}_{r0}$$, respectively. Therefore,
$$- \left( {\begin{array}{*{20}c} {u_{r0} } \\ {\dot{u}_{r0} } \\ \end{array} } \right) = {\mathbf{A}}\left( {\pi /\omega_{g} } \right)\left( {\begin{array}{*{20}c} {u_{r0} } \\ {\dot{u}_{r0} } \\ \end{array} } \right) + {\mathbf{b}}\left( {\pi /\omega_{g} } \right)$$
(2.28)
The solutions of Eq. (2.28) are
\begin{aligned} u_{r0} & = u_{st} \frac{{ - \sinh \theta_{1} + \left( {\xi_{1} /\sqrt {1 - \xi_{1}^{2} } } \right)\sin \theta_{2} }}{{\cosh \theta_{1} + \cos \theta_{2} }} \\ & = \frac{ - \delta \mu g}{{\omega^{2} }}\frac{{ - \sinh \theta_{1} + \left( {\xi_{1} /\sqrt {1 - \xi_{1}^{2} } } \right)\sin \theta_{2} }}{{\cosh \theta_{1} + \cos \theta_{2} }} \\ \end{aligned}
(2.29)
$$\dot{u}_{r0} = - u_{st} \omega_{1} \frac{{\sin \theta_{2} /\sqrt {1 - \xi_{1}^{2} } }}{{\cosh \theta_{1} + \cos \theta_{2} }} = \frac{{\delta \mu g\omega_{1} }}{{\omega^{2} }}\frac{{\sin \theta_{2} /\sqrt {1 - \xi_{1}^{2} } }}{{\cosh \theta_{1} + \cos \theta_{2} }}$$
(2.30)
where
$$\theta_{1} = \xi_{1} \frac{{\omega_{1} }}{{\omega_{g} }}\pi ,\,\theta_{2} = \sqrt {1 - \xi_{1}^{2} } \frac{{\omega_{1} }}{{\omega_{g} }}\pi = \frac{{\omega_{1d} }}{{\omega_{g} }}\pi$$
(2.31)
and $$\delta$$ represents the direction of the friction force, i.e., the opposite direction of sliding.
By substituting Eqs. (2.29) and (2.30) into Eq. (2.27), the relative displacement and velocity at any time of this half sliding cycle are obtained:
\begin{aligned} & u_{r} \left( {t_{i} + \tau } \right) = \frac{{ - \delta \mu g}}{{\omega ^{2} }} \\ & \left( \begin{aligned} & 1 - e^{{ - \xi _{1} \omega _{1} \tau }} \\ & \frac{{\cos \left( {\theta _{2} - \omega _{{1d}} \tau } \right)\sqrt {1 - \xi _{1}^{2} } - \xi _{1} \sin \left( {\theta _{2} - \omega _{{1d}} \tau } \right) + \xi _{1} \sin \left( {\omega _{{1d}} \tau } \right)e^{{\theta _{1} }} + \sqrt {1 - \xi _{1}^{2} } \cos \left( {\omega _{{1d}} \tau } \right)e^{{\theta _{1} }} }}{{\sqrt {1 - \xi _{1}^{2} } \left( {\cosh \theta _{1} + \cos \theta _{2} } \right)}} \\ \end{aligned} \right) \\ \end{aligned}
(2.32)
$$\dot{u}_{r} \left( {t_{i} + \tau } \right) = \frac{{ - \delta \mu g\omega_{1} }}{{\omega^{2} }}e^{{ - \xi_{1} \omega_{1} \tau }} \frac{{\sin \left( {\omega_{1d} \tau } \right)\left( {e^{{\theta_{1} }} + \cos \theta_{2} } \right) - \cos \left( {\omega_{1d} \tau } \right)\sin \theta_{2} }}{{\sqrt {1 - \xi_{1}^{2} } \left( {\cosh \theta_{1} + \cos \theta_{2} } \right)}}$$
(2.33)
When $$\tau = \pi /\omega_{g}$$, the opposite half sliding cycle initiates. From Eq. (2.17), we have
$$\dot{u}_{s} \left( {t_{i} + \pi /\omega_{g} } \right) = \delta \mu g\pi /\omega_{g} - \alpha \left( { - \,2\dot{u}_{r0} } \right) - 2a_{g} /\omega_{g} \cos \left( {\omega_{g} t_{i} } \right) = 0$$
(2.34)
Therefore,
$$\cos \left( {\omega_{g} t_{i} } \right) = \frac{\delta \pi \mu g}{{2a_{g} }} + \frac{{\alpha \omega_{g} \dot{u}_{r0} }}{{a_{g} }} = A_{1} \frac{\delta \mu g}{{a_{g} }}$$
(2.35)
where
$$A_{1} = \frac{{\alpha \omega_{g} \omega_{{1}} }}{{\omega^{2} \sqrt {1 - \xi_{1}^{2} } }}\frac{{\sin \theta_{2} }}{{\cosh \theta_{1} + \cos \theta_{2} }} + \frac{\pi }{2}$$
(2.36)
Since a certain half sliding cycle initiates at t = ti,
\begin{aligned} \left| {\alpha \ddot{u}_{r} + \ddot{u}_{g} } \right|_{{t = t_{i} }} & = \delta \left( {\alpha \left( { - \, 2\xi \omega \dot{u}_{r0} - \omega^{2} u_{r0} - a_{g} \sin \left( {\omega_{g} t_{i} } \right)} \right) + a_{g} \sin \left( {\omega_{g} t_{i} } \right)} \right) \\ & = - A_{2} \mu g + \delta \left( {1 - \alpha } \right)a_{g} \sin \left( {\omega_{g} t_{i} } \right) \ge \mu g \\ \end{aligned}
(2.37)
where
$$A_{2} = \frac{{\alpha \left( {\xi_{1} \sin \theta_{2} /\sqrt {1 - \xi_{1}^{2} } + \sinh \theta_{1} } \right)}}{{\cosh \theta_{1} + \cos \theta_{2} }}$$
(2.38)
From Eqs. (2.35) and (2.37), we have
$$\left( {1 - \alpha } \right)^{2} \left( {a_{g}^{2} - A_{1}^{2} \left( {\mu g} \right)^{2} } \right) \ge \left( {1 + A_{2} } \right)^{2} \left( {\mu g} \right)^{2}$$
(2.39)
Thus, the boundaries between the stick-sliding and sliding-sliding cases are
$$\frac{{a_{g} }}{\mu g} = \sqrt {\left( {1 + A_{2} } \right)^{2} /\left( {1 - \alpha } \right)^{2} + A_{1}^{2} }$$
(2.40)
The variations of the boundaries between the stick-sliding and sliding-sliding cases for different $$\omega_{g} /\omega$$ and $$\alpha$$ are shown in Fig. 2.7.
When $$\omega_{g} /\omega_{1} \to 0$$, $$\theta_{1}$$ and $$\theta_{2} \to + \infty$$, $$A_{1} \to \pi /2$$, and $$A_{2} \to \alpha$$; thus, Eq. (2.40) becomes
$$\frac{{a_{g} }}{\mu g} = \sqrt {\left( {1 + \alpha } \right)^{2} /\left( {1 - \alpha } \right)^{2} + \left( {\pi /2} \right)^{2} }$$
(2.41)
When $$\omega_{g} /\omega_{1} \to + \infty$$, $$\theta_{1}$$ and $$\theta_{2} \to 0$$, $$A_{1} \to \pi /\left( {2\left( {1 - \alpha } \right)} \right)$$, and $$A_{2} \to 0$$; thus, Eq. (2.40) becomes
$$\frac{{a_{g} }}{\mu g} = \frac{{\sqrt {1 + \left( {\pi /2} \right)^{2} } }}{1 - \alpha }$$
(2.42)
As can be seen in Fig. 2.7, there are also some local peak values when $$\theta_{2}$$ = (2n − 1)$$\pi$$. These points are corresponding to the resonance.

## 2.4 Parametric Study for the Maximum Responses

### 2.4.1 Maximum Pseudo Acceleration of the Top Mass

The maximum shear force applied to the superstructure can be represented by the maximum pseudo acceleration of the top mass (Chopra, 2001), A, as follows:
$$A = \omega^{2} \times \left( {\max \left| {u_{r} (t)} \right|} \right) = \frac{k}{m}\left( {\max \left| {u_{r} (t)} \right|} \right)$$
(2.43)
The function of $$a_{g} ,\mu g,\omega_{g} /\omega ,\alpha$$, and $$\xi$$ is related to A in the 2DOF sliding base system depicted in Fig. 2.1. Figure 2.8 illustrates how the normalized maximum pseudo acceleration, A/$$\mu$$g, relates to the frequency ratio, $$\omega_{g} /\omega$$, for varying values of $$\alpha$$ and $$a_{g} /\mu g$$. A sliding base structure, unlike fixed base structures, has multiple resonant frequencies when sliding occurs. $$a_{g} /\mu g$$ and $$\alpha$$ can also affect these resonant frequencies. As shown in Fig. 2.8a, b, as $$a_{g} /\mu g$$ increases, several resonant frequency ratios appear in the region of $$\omega_{g} /\omega$$ < 1, and these ratios shift towards higher values gradually and eventually reach upper limits, which correspond to the sliding-sliding cases. As shown in Fig. 2.8c, d, resonances are more prone to happen with smaller mass ratios, $$\alpha$$.
As shown in Fig. 2.8c, d, the maximum pseudo acceleration typically reduces with an increase in the mass ratio, $$\alpha$$. The reason for this is that the damping ratio corresponding to the sliding phase, $$\xi_{1}$$, is equal to $$\xi /\sqrt {1 - \alpha }$$ as given in Eq. (2.11), which increases as $$\alpha$$ increases. However, in the resonant frequency ranges of a specific sliding base structure, the maximum pseudo acceleration may be greater compared to other structures with lower mass ratios.
Figure 2.9 shows the relationship between $$A/\mu g$$ and $$a_{g} /\mu g$$ for different values of $$\alpha$$ and $$\omega_{g} /\omega$$, where the circle and triangle represent the boundary between the stick-stick and stick-sliding cases, and the boundary between the stick-sliding and sliding-sliding cases, respectively. The $$A/\mu g$$ versus $$a_{g} /\mu g$$ curves are segregated into three areas, each corresponding to cases of stick-stick, stick-sliding, and sliding-sliding. In the stick-stick region, responses remain independent of mass ratio, $$\alpha$$, with the $$A/\mu g$$ versus $$a_{g} /\mu g$$ curve taking the form of a straight line. The slope of this straight line is equal to the displacement response factor, Rd, for a fixed base structure:
$$R_{d} = \frac{1}{{\sqrt {\left[ {1 - \left( {\omega_{g} /\omega } \right)^{2} } \right]^{2} + \left[ {2\xi \left( {\omega_{g} /\omega } \right)} \right]^{2} } }}$$
(2.44)
In the stick-sliding region, due to additional resonances when $$\omega_{g} /\omega$$ < 1, the maximum pseudo acceleration of a sliding base structure may exceed that of the related fixed base structure. This phenomenon can be observed in Fig. 2.9a for $$\omega_{g} /\omega$$ = 0.1 and 0.5. In the sliding-sliding region, the upper limit of the maximum pseudo acceleration is reached, and does not change even when there are larger ground accelerations. This upper limit reflects the effectiveness of sliding base structures for isolating extremely large earthquakes. As shown in Fig. 2.9b, in normal circumstances, as the mass ratio increases, the likelihood of the sliding-sliding case occurring becomes more difficult. This trend can be more clearly observed in Fig. 2.7.

### 2.4.2 Amplitude of the Sliding Displacement

As shown in Fig. 2.4b, the maximum sliding displacement results from the accumulation of transient responses that occur before reaching the steady periodic state. Hence, the maximum sliding displacement is considerably influenced by the initial ground acceleration, e.g., there is a significant disparity in the maximum sliding displacement between the sine and cosine ground accelerations. To represent the extent of sliding when exposed to harmonic ground motions, the amplitude of the sliding displacement, us,ap, is a more appropriate response quantity compared to the maximum sliding displacement. It is defined as the difference between the maximum and minimum sliding displacements during the steady state; therefore, the value of it is exclusively linked to the responses of the steady state.
Figure 2.10 depicts the correlation between the frequency ratio, $$\omega_{g} /\omega$$, and the normalized sliding displacement amplitude, $$u_{s,ap} /\left( {a_{g} /\omega_{g}^{2} } \right)$$, for varying $$\alpha$$ and $$a_{g} /\mu g$$ values. As shown in Fig. 2.10a, b, when $$\omega_{g} /\omega$$ A is less than 1, the normalized sliding displacement amplitude is not greatly affected by the mass, $$\alpha$$, ratio or frequency ratio, $$\omega_{g} /\omega$$. There is a noticeable reduction of the sliding displacement amplitude when $$\omega_{g} /\omega$$ increases, after the frequency ratio, $$\omega_{g} /\omega$$, surpasses 1. This result is in agreement with Fig. 2.5, where the critical $$a_{g} /\mu g$$ for the occurrence of the stick-sliding case increases as $$\omega_{g} /\omega$$ increases once $$\omega_{g} /\omega$$ surpasses 1. As shown in Fig. 2.10b, if the normalized ground acceleration amplitude, $$a_{g} /\mu g$$, is sufficiently large, as $$\omega_{g} /\omega$$ surpasses a certain value, the sliding displacement amplitude will rise alongside an increase in $$\omega_{g} /\omega$$. Figure 2.5 is also in accordance with this, where the critical value of $$a_{g} /\mu g$$ leading to stick-sliding decreases with increasing value of $$\omega_{g} /\omega$$ once the peak point of critical value $$a_{g} /\mu g$$ is reached. Figure 2.10b also indicates that there is a tendency generally for the sliding displacement amplitude to decrease as the mass ratio increases in the area of large frequency ratios.
Figure 2.11 depicts the correlation between the normalized sliding displacement amplitude, $$u_{s,ap} /\left( {a_{g} /\omega_{g}^{2} } \right)$$, and the normalized ground acceleration amplitude, $$a_{g} /\mu g$$, for different $$\omega_{g} /\omega$$ and $$\alpha$$. As shown in Fig. 2.11, after sliding occurs, the normalized sliding displacement amplitude increases as $$a_{g} /\mu g$$ increases, but with a gradually decreasing speed increase. The normalized sliding displacement amplitude has an upper bound value that is 2. The reason for this is that when $$a_{g} /\mu g$$ is very large, the absolute acceleration of the sliding base (which is limited by the friction coefficient) is negligible compared with the ground acceleration, so the sliding base can be considered motionless from the perspective of ground, and us,ap is equal to the vibration amplitude of the ground displacement, $$2a_{g} /\omega_{g}^{2}$$.

## 2.5 Theoretical Solutions for the Responses of the Sliding-Sliding Case

For the sliding-sliding case, the maximum pseudo acceleration reaches the upper limit, and is no longer dependent on the amplitude of the ground acceleration. This upper limit response has important physical meaning, and reflects the effectiveness of the sliding base system for reducing the superstructure response.
In this section, the theoretical solutions for the maximum pseudo acceleration and sliding displacement amplitude for the sliding-sliding case are derived. The derived results are further used to explain the mechanism of the sliding base system. For the stick-sliding case, theoretical solutions cannot be obtained because of the complicated transformation between stick and sliding phases.

### 2.5.1 Solutions for the Maximum Pseudo Acceleration

The maximum relative displacement will be obtained when the relative velocity is equal to 0, i.e. $$\dot{u}_{r} = 0$$; thus, the corresponding local time, $$\tau_{j}$$, can be obtained by taking the right-hand side of Eq. (2.33) equal to zero:
$$\tan \left( {\omega_{1d} \tau_{j} } \right) = \frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}$$
(2.45)
From Eq. (2.45), several solutions for $$\tau_{j}$$ may be obtained, and the number of solutions is dependent on the value of $$\theta_{2}$$. By substituting Eq. (2.45) into Eq. (2.32), the relative displacement at the moment of $$\tau_{j}$$ can be obtained:
$$u_{r} \left( {t_{i} + \tau_{j} } \right) = \frac{ - \delta \mu g}{{\omega^{2} }}\left( {1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \frac{{\cos \left( {\theta_{2} - \omega_{1d} \tau_{j} } \right) + \cos \left( {\omega_{1d} \tau_{j} } \right)e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} \right)$$
(2.46)
Using Eq. (2.46), the normalized maximum pseudo acceleration can be written as
\begin{aligned} A/\mu g & = \max \left| {\frac{{u_{r} \left( {t_{i} + \tau_{j} } \right)\omega^{2} }}{\mu g}} \right| \\ & = \max \left| {1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \frac{{\cos \left( {\theta_{2} - \omega_{1d} \tau_{j} } \right) + \cos \left( {\omega_{1d} \tau_{j} } \right)e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} \right| \\ \end{aligned}
(2.47)
Equation (2.47) is not an explicit equation yet, since there may be more than one solution for $$\tau_{j}$$ and the corresponding pseudo acceleration. The maximum pseudo acceleration is the maximum of these candidates. To obtain the explicit form of Eq. (2.47), different cases of $$\theta_{2}$$ need to be considered. The following presents the complete derivations for the explicit expressions of the normalized maximum pseudo acceleration, $$A/\mu g$$.
$$\sin^{2} \left( {\omega_{1d} \tau_{j} } \right) = \frac{{\tan^{2} \left( {\omega_{1d} \tau_{j} } \right)}}{{\tan^{2} \left( {\omega_{1d} \tau_{j} } \right) + 1}} = \frac{{\sin^{2} \theta_{2} }}{{2e^{{\theta_{1} }} \left( {\cosh \theta_{1} + \cos \theta_{2} } \right)}}$$
(2.48)
If $$\theta_{2} \ne n\pi \,\left( {n = 1,2, \ldots } \right)$$, $$\sin \theta_{2} \ne 0$$, so $$\sin \left( {\omega_{1d} \tau_{j} } \right) \ne 0$$ from Eq. (2.48). Therefore, Eq. (2.47) can be rewritten as Eq. (2.49) by using Eq. (2.45).
$$A/\mu g = \max \left| {1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \frac{{\sin \theta_{2} }}{{\sin \left( {\omega_{1d} \tau_{j} } \right)\left( {\cosh \theta_{1} + \cos \theta_{2} } \right)}}} \right|$$
(2.49)
Case 1. If $$0 < \theta_{2} < \pi$$, $$\sin \theta_{2} > 0$$. From Eq. (2.45), we have
$$\tan \left( {\omega_{1d} \tau_{j} } \right) = \frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }} < \frac{{\sin \theta_{2} }}{{1 + \cos \theta_{2} }} = \tan \left( {{\theta_{2}}}/2 \right)$$
(2.50)
Since $$0 \le \tau_{j} \le \pi /\omega_{g}$$, $$0 \le \omega_{1d} \tau_{j} \le \theta_{2}$$, so there is only one solution for $$\tau_{j}$$, and
$$\omega_{1d} \tau_{j} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) < \theta_{2} /2 < \pi /2$$
(2.51)
From Eq. (2.51), we have $$\sin \left( {\omega_{1d} \tau_{j} } \right) > 0$$, so from Eq. (2.48), we have
$$\frac{{\sin \theta_{2} }}{{\sin \left( {\omega_{1d} \tau_{j} } \right)}} = \sqrt {{2}e^{{\theta_{1} }} \left( {\cosh \theta_{1} + \cos \theta_{2} } \right)}$$
(2.52)
Substituting Eq. (2.52) into Eq. (2.49) gives
$$A/\mu g = \left| {1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} } \right|$$
(2.53)
Case 2. If $$\pi < \theta_{2} < 2\pi$$, $$\sin \theta_{2} < 0$$, From Eq. (2.45), we have
$$\tan \left( {\theta_{2} /2} \right) = \frac{{\sin \theta_{2} }}{{1 + \cos \theta_{2} }} < \tan \left( {\omega_{1d} \tau_{j} } \right) = \frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }} < 0$$
(2.54)
so the smallest solution for $$\tau_{j}$$ is
$$\pi /2 < \theta_{2} /2 < \omega_{1d} \tau_{j} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) + \pi < \pi$$
(2.55)
Since
$$\arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) + 2\pi > \theta_{2} /2 + \pi > \theta_{2}$$
(2.56)
there is no other solutions for $$\tau_{j}$$ because $$0 \le \omega_{1d} \tau_{j} \le \theta_{2}$$. From Eq. (2.55), we have $$\sin \left( {\omega_{1d} \tau_{j} } \right) > 0$$, so from Eq. (2.48), we have
$$\frac{{\sin \theta_{2} }}{{\sin \left( {\omega_{1d} \tau_{j} } \right)}} = - \sqrt {{2}e^{{\theta_{1} }} \left( {\cosh \theta_{1} + \cos \theta_{2} } \right)}$$
(2.57)
Substituting Eq. (2.57) into Eq. (2.49) gives
$$A/\mu g = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.58)
Case 3. If $$\theta_{2} > 2\pi$$ and $$\theta_{2} \ne n\pi \,\left( {n = 1,2, \ldots } \right)$$, there are at least two solutions for $$\tau_{j}$$. If $$\sin \theta_{2} > 0$$, i.e., $$2n\pi < \theta_{2} < \left( {2n + 1} \right)\pi$$, then from Eq. (2.45), we have $$\tan \left( {\omega_{1d} \tau_{j} } \right) > 0$$. Therefore, the first solution for $$\tau_{j}$$ is
$$0 < \omega_{1d} \tau_{j} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) < \pi /2$$
(2.59)
and the corresponding pseudo acceleration is
$$\left| {\frac{{u_{r} \left( {t_{i} + \tau_{j} } \right)\omega^{2} }}{\mu g}} \right| = \left| {1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\phi }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} } \right|$$
(2.60)
where
$$\phi = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right)$$
(2.61)
The second solution for $$\tau_{j}$$ is
$$\pi < \omega_{1d} \tau_{j} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) + \pi < 3\pi /2$$
(2.62)
and the corresponding pseudo acceleration is
$$\left| {\frac{{u_{r} \left( {t_{i} + \tau_{j} } \right)\omega^{2} }}{\mu g}} \right| = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\left( {\phi + \pi } \right)}} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.63)
The pseudo accelerations corresponding to other solutions of $$\tau_{j}$$ are all smaller than the larger one of Eqs. (2.60) and (2.63), since the oscillation of the relative displacement gradually decays because of the damping. If Eq. (2.60) is equal to
$$\left| {\frac{{u_{r} \left( {t_{i} + \tau_{j} } \right)\omega^{2} }}{\mu g}} \right| = 1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\phi }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.64)
it is clear that Eq. (2.63) is larger than Eq. (2.60). If Eq. (2.60) is equal to
$$\left| {\frac{{u_{r} \left( {t_{i} + \tau_{j} } \right)\omega^{2} }}{\mu g}} \right| = e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\phi }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} - 1$$
(2.65)
Equation (2.60) minus Eq. (2.63) is
\begin{aligned} & e^{{ - \frac{{\xi _{1} }}{{\sqrt {1 - \xi _{1}^{2} } }}\phi }} \sqrt {\frac{{{\text{2}}e^{{\theta _{1} }} }}{{\cosh \theta _{1} + \cos \theta _{2} }}} \left( {1 - e^{{ - \frac{{\xi _{1} }}{{\sqrt {1 - \xi _{1}^{2} } }}\pi }} } \right) - 2 \\ & \quad \quad < \sqrt {\frac{{{\text{2}}e^{{\theta _{1} }} }}{{\cosh \theta _{1} + \cos \theta _{2} }}} \left( {1 - e^{{ - \frac{{\theta _{1} }}{{\theta _{2} }}\pi }} } \right) - 2 \\ & \quad \quad < \sqrt {\frac{{{\text{2}}e^{{\theta _{1} }} }}{{\cosh \theta _{1} - 1}}} \left( {1 - e^{{ - \theta _{1} }} } \right) - 2 \\ & \quad \quad = \frac{{e^{{\theta _{1} }} - 1}}{{\cosh \theta _{1} - 1}}\left( {1 - e^{{ - \theta _{1} }} } \right) - 2 = 0 \\ \end{aligned}
(2.66)
Therefore, Eq. (2.63) is always larger than Eq. (2.60), and the maximum pseudo acceleration is obtained at
$$\pi < \omega_{1d} \tau_{j} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) + \pi < 3\pi /2$$
(2.67)
and the maximum pseudo acceleration is
$$A/\mu g = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.68)
Case 4. If $$\sin \theta_{2} < 0$$, i.e., $$\left( {2n + 1} \right)\pi < \theta_{2} < \left( {2n + 2} \right)\pi$$, then from Eq. (2.45), we have $$\tan \left( {\omega_{1d} \tau_{j} } \right) < 0$$. Therefore, the maximum pseudo acceleration will be obtained at the first solution of $$\tau_{j}$$, that is
$$\pi /2 < \omega_{1d} \tau_{j} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) + \pi < \pi$$
(2.69)
and the maximum pseudo acceleration is
$$A/\mu g = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.70)
Case 5. If $$\theta_{2} = n\pi \,\left( {n = 1,2, \ldots } \right)$$, $$\sin \theta_{2} = 0$$, and from Eq. (2.45), we have $$\tan \left( {\omega_{1d} \tau_{j} } \right) = 0$$, so the first solution for $$\tau_{j}$$ is
$$\omega_{{1{\text{d}}}} \tau_{j} = 0$$
(2.71)
Substituting Eq. (2.71) into Eq. (2.46) gives the corresponding pseudo acceleration:
$$\left| {\frac{{u_{r} \left( {t_{i} + \tau_{j} } \right)\omega^{2} }}{\mu g}} \right| = \frac{{\left( { - \,1} \right)^{n} + e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \left( { - \,1} \right)^{n} }} - 1$$
(2.72)
The second solution for $$\tau_{j}$$ is
$$\omega_{1d} \tau_{j} = \pi$$
(2.73)
and Substituting Eq. (2.73) into Eq. (2.46) gives the corresponding pseudo acceleration:
\begin{aligned} \left| {\frac{{u_{r} \left( {t_{i} + \tau_{j} } \right)\omega^{2} }}{\mu g}} \right| & = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\pi }} \frac{{\left( { - \,1} \right)^{n} + e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \left( { - \,1} \right)^{n} }} \\ & = 1 + e^{{ - \theta_{1} /n}} \frac{{\left( { - \,1} \right)^{n} + e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \left( { - \,1} \right)^{n} }} \\ \end{aligned}
(2.74)
Equation (2.72) minus Eq. (2.74) is
\begin{aligned} & \frac{{\left( { - \,1} \right)^{n} + e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \left( { - \,1} \right)^{n} }}\left( {1 - e^{{ - {{\theta_{1} }/n}}} } \right) - 2 \le \frac{{\left( { - \,1} \right)^{n} + e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \left( { - \,1} \right)^{n} }}\left( {1 - e^{{ - \theta_{1} }} } \right) - 2 \\ & \quad = \left\{ {\begin{array}{*{20}l} {\frac{{ - \,2\left( {1 + e^{{\theta_{1} }} } \right)}}{{\cosh \theta_{1} + 1}} < 0} & {\left( {n\,{\text{is}}\,{\text{even}}} \right)} \\ 0 & {\left( {n\,{\text{is}}\,{\text{odd}}} \right)} \\ \end{array} } \right. \\ \end{aligned}
(2.75)
Therefore, Eq. (2.72) is always smaller than Eq. (2.74). For $$\theta_{2} = n\pi \,\left( {n = 1,2, \ldots } \right)$$,
$$\frac{{\left( { - \,1} \right)^{n} + e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \left( { - \,1} \right)^{n} }} = \sqrt {\frac{{2e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \left( { - \,1} \right)^{n} }}} = \sqrt {\frac{{2e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.76)
so the maximum pseudo acceleration can be written as
$$A/\mu g = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\pi }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.77)
The results of Cases 2–5 can be combined in one equation, that is, for $$\theta_{2} \ge \pi$$, the normalized maximum pseudo acceleration is
$$A/\mu g = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{j} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.78)
where
$$\pi /2 < \omega_{1d} \tau_{j} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) + \pi < 3\pi /2$$
(2.79)
In summary, the explicit solutions for the maximum pseudo acceleration are as follows:
For $$0 < \theta_{2} < \pi$$, the normalized maximum pseudo acceleration is
$$A/\mu g = \left| {1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{0} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} } \right|$$
(2.80)
where
$$\omega_{1d} \tau_{0} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) < \pi /2$$
(2.81)
For $$\theta_{2} \ge \pi$$, the normalized maximum pseudo acceleration is
$$A/\mu g = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{0} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}}$$
(2.82)
where
$$\pi /2 < \omega_{1d} \tau_{0} = \arctan \left( {\frac{{\sin \theta_{2} }}{{e^{{\theta_{1} }} + \cos \theta_{2} }}} \right) + \pi < 3\pi /2$$
(2.83)
In the above equations, $$\tau_{0}$$ is the local time corresponding to the maximum pseudo acceleration. $$\theta_{1}$$ and $$\theta_{2}$$ in Eqs. (2.80)–(2.83) are only related to $$\xi_{1}$$ and $$\omega_{1} /\omega_{g}$$ as given in Eq. (2.31); thus, the normalized maximum pseudo acceleration, $$A/\mu g$$, is only dependent on $$\xi_{1}$$ and $$\omega_{1} /\omega_{g}$$ for the sliding-sliding case. $$\xi_{1}$$ and $$\omega_{1}$$ are the natural frequency and damping ratio for the sliding phase, which are related to the mass ratio, $$\alpha$$, as shown in Eq. (2.11).

### 2.5.2 Interpretation of the Solutions for the Maximum Pseudo Acceleration

Figure 2.12 plots the relationship between $$A/\mu g$$ and $$\omega_{g} /\omega$$ using the derived theoretical solutions. As shown in Fig. 2.12a, the results from the theoretical solutions and numerical methods are the same, which verifies the accuracy of the theoretical solutions. As shown in Fig. 2.12b, as the mass ratio, $$\alpha$$, increases, the frequency ratios, $$\omega_{g} /\omega$$, of resonance shift towards larger values and the general responses decrease. This is because $$\xi_{1}$$ and $$\omega_{1}$$ increase as $$\alpha$$ increases for given $$\xi$$ and $$\omega$$.
In the following paragraphs, we will examine various frequency ratio regions, and elucidate the outcomes illustrated in Fig. 2.12b. When $$\omega_{g} /\omega_{1} \to 0$$, $$\theta_{1}$$ and $$\theta_{2} \to + \infty$$, so from Eq. (2.83), we have $$\omega_{1d} \tau_{0} \to \pi$$, and Eq. (2.82) becomes
$$A/\mu g = 1 + e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{0} }} \sqrt {\frac{{{2}e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} \to 1 + 2e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\pi }}$$
(2.84)
If the damping ratio, $$\xi$$ = 0, the right-hand side of Eq. (2.84) is equal to 3, which implies that the maximum pseudo acceleration is equal to 3 times μg. For actual structures with damping, the normalized maximum pseudo acceleration, $$A/\mu g$$, is always between 1 and 3, as demonstrated by Eq. (2.84). This is because when $$\omega_{g} /\omega_{1} \to 0$$, the ground motion period is much larger than the natural period of the sliding system, so there is sufficient time for the oscillation of the relative displacement to decrease and stabilize to the static displacement, $$u_{st} = - \delta \mu g/\omega^{2}$$, prior to the onset of the opposite sliding. Once sliding changes direction, the relative displacement initiates oscillation around the new static equilibrium displacement, −ust. If the damping is sufficiently small, the relative displacement can approach − 3ust during the first oscillation cycle; if the damping is very large, the oscillation of the relative displacement diminishes rapidly and is only capable of attaining the new static equilibrium displacement, −ust. Figure 2.13 shows the steady state response of the normalized pseudo acceleration for $$\alpha$$ = 0.8, $$\xi$$ = 5%, T = 1 s and Tg = 10 s ($$\omega_{g} /\omega_{1}$$ = 0.045).
Equation (2.82) can also be written as
$$A/\mu g = 1 + 2e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{0} }} \sqrt {\frac{1}{{\left( {e^{{ - \theta_{1} }} - 1} \right)^{2} + 2e^{{ - \theta_{1} }} \left( {\cos \theta_{2} + 1} \right)}}}$$
(2.85)
If $$\xi$$ = 0 [which means $$\theta_{1}$$ = 0 from Eq. (2.31)] and $$\theta_{2}$$ = (2n − 1)$$\pi$$ (where n is a positive integer), the normalized maximum pseudo acceleration, $$A/\mu g$$, becomes infinite from Eq. (2.85). If $$\xi \ne 0$$, a local peak value will be obtained for $$A/\mu g$$ when $$\theta_{2}$$ = (2n − 1)$$\pi$$, i.e., $$\omega_{1d}$$ = (2n − 1)$$\omega_{g}$$ from Eq. (2.31), in which $$\omega_{1d}$$ is the natural frequency of the damped vibration during the sliding phase:
$$A/\mu g = 1 + 2e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\pi }} \frac{1}{{1 - e^{{ - \theta_{1} }} }}$$
(2.86)
This result is consistent with the observations from Fig. 2.12b that there are multiple resonant frequencies in the area of $$\omega_{g} /\omega_{1}$$ < 1.
For $$\xi$$ = 0, Fig. 2.14 displays the normalized pseudo acceleration response for $$\omega_{1} = \omega_{g}$$ and $$\omega_{1} = 3\omega_{g}$$. The ground acceleration, $$\ddot{u}_{g}$$, is taken as $$a_{g} \cos \left( {\omega_{g} t} \right)$$, where ag is a large value so that sliding can occur during the whole time of investigation. Since $$\omega_{1d} = \omega_{1} = (2n - 1)\omega_{g}$$ for $$\xi$$ = 0, the relative displacement reaches the peak value of a certain half sliding cycle when the direction of sliding changes, which is the furthest location from the static equilibrium displacement of the next half sliding cycle. From Eq. (2.13), the relative displacement at the end of one half sliding cycle for $$\xi$$ = 0 is
$$u_{r} \left( {t_{i} + T_{g} /2} \right) = \cos \left( {\omega_{1} T_{g} /2} \right)u_{{\text{r}}} \left( {t_{i} } \right) + u_{st} - u_{st} \cos \left( {\omega_{1} T_{g} /2} \right) = - u_{{\text{r}}} \left( {t_{i} } \right) + 2u_{st}$$
(2.87)
So
$$\left| {u_{r} \left( {t_{i} + T_{g} /2} \right)} \right| - \left| {u_{r} \left( {t_{i} } \right)} \right| = 2\mu g/\omega^{2}$$
(2.88)
Thus, during one cycle of the ground motion, Tg, the maximum relative displacement can increase by $$4\mu g/\omega^{2}$$, and can continue increasing until the ground acceleration is not large enough anymore to start sliding. For actual structures, the maximum relative displacement will be reached after several cycles of ground motion because of the damping, as shown in Fig. 2.15 for $$\xi$$ = 5%, $$\alpha$$ = 0.8, T = 1 s and Tg = 0.45 s $$\left( {\omega_{g} /\omega_{1d} = 1} \right)$$. When $$\omega_{g} /\omega_{1} \to 0$$, $$\theta_{1} \to + \infty$$ from Eq. (2.31), so Eq. (2.86) tends toward Eq. (2.84). This means that when $$\omega_{g} /\omega$$ decreases, the resonance will slowly disappear, as shown in Fig. 2.12b.
When $$\omega_{g} /\omega_{1} \to + \infty$$, $$\theta_{1}$$ and $$\theta_{2} \to 0$$ from Eq. (2.31), we have $$\omega_{1d} \tau_{0} \to \theta_{2} /2$$ from Eq. (2.81), so Eq. (2.80) becomes
$$A/\mu g = \left| {1 - e^{{ - \frac{{\xi_{1} }}{{\sqrt {1 - \xi_{1}^{2} } }}\omega_{1d} \tau_{0} }} \sqrt {\frac{{2e^{{\theta_{1} }} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} } \right| \to \left| {\frac{{1 - 2\xi_{1}^{2} }}{8}\left( {\frac{{\omega_{1} }}{{\omega_{g} }}\pi } \right)^{2} } \right| \to 0$$
(2.89)
This is because when $$\omega_{g} /\omega_{1} \to + \infty$$, the frequency of ground acceleration is considerably larger than the natural frequency of the sliding system, so the oscillation of the relative displacement can be hardly stimulated because of the frequently changed sliding direction.

### 2.5.3 Solutions for the Sliding Displacement Amplitude

In Sect. 2.4.2, we have obtained
\begin{aligned} \cos \left( {\omega_{g} t_{i} } \right) & = \frac{\delta \pi \mu g}{{2a_{g} }} + \frac{{\alpha \omega_{g} \dot{u}_{r0} }}{{a_{g} }} \\ & = \frac{\delta \mu g}{{a_{g} }}\left( {\frac{{\alpha \omega_{g} \omega_{{1}} }}{{\omega^{2} \sqrt {1 - \xi_{1}^{2} } }}\frac{{\sin \theta_{2} }}{{\cosh \theta_{1} + \cos \theta_{2} }} + \frac{\pi }{2}} \right) \\ \end{aligned}
(2.90)
By substituting $$t = t_{i} + \pi /\omega_{g}$$ into Eq. (2.18) and using Eq. (2.90), the amplitude of the sliding displacement can be obtained as follows:
\begin{aligned} u_{s,ap} & = \left| {u_{s} \left( {t_{i} + \pi /\omega_{g} } \right) - u_{s} \left( {t_{i} } \right)} \right| \\ & = \left| {\frac{1}{2}\delta \mu g\left( {\pi /\omega_{g} } \right)^{2} + \alpha \left( {2u_{r0} + \dot{u}_{r0} \left( {\pi /\omega_{g} } \right)} \right) - \frac{{2a_{g} }}{{\omega_{g}^{2} }}\sin \left( {\omega_{g} t_{i} } \right) - \frac{{\pi a_{g} }}{{\omega_{g}^{2} }}\cos \left( {\omega_{g} t_{i} } \right)} \right| \\ & = \left| {\frac{{ - \, 2a_{g} }}{{\omega_{g}^{2} }}\sin \left( {\omega_{g} t_{i} } \right) + 2\alpha u_{r0} } \right| \\ & = \left| {\frac{{2a_{g} }}{{\omega_{g}^{2} }}\sin \left( {\omega_{g} t_{i} } \right) + 2\alpha \frac{\delta \mu g}{{\omega^{2} }}\frac{{ - \sinh \theta_{1} + \left( {\xi_{1} /\sqrt {1 - \xi_{1}^{2} } } \right)\sin \theta_{2} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} \right| \\ \end{aligned}
(2.91)
Therefore, the normalized sliding displacement amplitude is
$$u_{s,ap} /\left( {a_{g} /\omega_{g}^{2} } \right) = \left| {2\sin \left( {\omega_{g} t_{i} } \right) + 2\alpha \frac{{\delta \mu g\omega_{g}^{2} }}{{a_{g} \omega^{2} }}\frac{{ - \sinh \theta_{1} + \left( {\xi_{1} /\sqrt {1 - \xi_{1}^{2} } } \right)\sin \theta_{2} }}{{\cosh \theta_{1} + \cos \theta_{2} }}} \right|$$
(2.92)
As revealed by Eq. (2.92), unlike the maximum pseudo acceleration, the sliding displacement amplitude is dependent on the amplitude of the ground acceleration, ag, for the sliding-sliding case. When $$a_{g} /\mu g \to + \infty$$, $$\cos \left( {\omega_{g} t_{i} } \right) \to 0$$ as revealed by Eq. (2.90), so the normalized sliding displacement amplitude, $$u_{s,ap} /\left( {a_{g} /\omega_{g}^{2} } \right)$$, in Eq. (2.92) tends toward to 2, which is consistent with the results presented in Sect. 2.5.2.

## 2.6 Conclusions

This chapter presents the responses of 2DOF sliding base systems subjected to harmonic ground motions. The response history of a SB system can exhibit two types of phases: the stick phase and the sliding phase. During the sliding phase, the vibration of the relative displacement has a higher natural frequency and damping ratio compared with the stick phase, which is related to the mass ratio. The equivalent dynamic force for the vibration of the relative displacement during the sliding phase is a step force, which results a static displacement as given in Eq. (2.12). The responses of the sliding base system under harmonic ground motions converge rapidly to steady periodic responses with the same period as the ground acceleration. The sliding base system displays three types of motions, namely stick-stick, stick-sliding, and sliding-sliding, as the ground acceleration amplitude increases.
A sliding base structure has multiple resonant frequencies when it slides, unlike fixed base structures. For the sliding-sliding case, resonance happens when the period of ground motion is odd times of the natural period of vibration of the relative displacement that occurs during sliding. As the mass ratio increases, the maximum pseudo acceleration generally decreases. However, Resonance can result in obtaining larger maximum pseudo acceleration for larger mass ratios in some particular cases. Moreover, in certain instances, the sliding base structure can have a higher maximum pseudo acceleration than its fixed base counterpart due to extra resonances. In the sliding-sliding case, the upper limit of pseudo acceleration is reached, regardless of the amplitude of the ground acceleration. Equations (2.80)–(2.83) can be used to calculate the maximum pseudo acceleration in the sliding-sliding case.
The amplitude of the sliding displacement is a suitable response quantity to represent the extent of sliding, and it can be normalized by the half of the vibration amplitude of the ground displacement, $$a_{g} /\omega_{g}^{2}$$. The normalized sliding displacement amplitude, $$u_{s,ap} /\left( {a_{g} /\omega_{g}^{2} } \right)$$, is more affected by the mass ratio and frequency ratio, $$\omega_{g} /\omega$$, in region $$\omega_{g} /\omega$$ ≥ 1 compared to region $$\omega_{g} /\omega$$ < 1. The normalized sliding displacement amplitude exhibits a higher value as $$a_{g} /\mu g$$ increases following the sliding occurring, yet its rate of increase gradually reduces, and it eventually reaches an upper bound of 2.
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