3.1 Equations of Motion

Figure 3.1a shows a schematic diagram of an N-story SB structure, the sliding base of which rests on the foundation. By lumping the structure mass at the corresponding floor level, the multistory SB structure can be analyzed using the simplified model shown in Fig. 3.1b. The lateral displacements of the ith floor in the x and y directions with respect to the sliding base are denoted as urxi and uryi, respectively. By further assuming that the structure is symmetrical about the x and y axes, the equations of dynamic equilibrium of the structure subjected to three-component excitations are as follows:

$$ \left\{ {\begin{array}{*{20}l} { {\mathbf{m}}\left[ {\left( {\ddot{u}_{gx} + \ddot{u}_{sx} } \right){\mathbf{1}} + {\ddot{\mathbf{u}}}_{rx} } \right] + {\mathbf{c}}_{x} {\dot{\mathbf{u}}}_{rx} + {\mathbf{k}}_{x} {\mathbf{u}}_{rx} = {\mathbf{0}}} \hfill \\ { m_{b} \left( {\ddot{u}_{gx} + \ddot{u}_{sx} } \right) + {\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\left[ {\left( {\ddot{u}_{gx} + \ddot{u}_{sx} } \right){\mathbf{1}} + {\ddot{\mathbf{u}}}_{rx} } \right] = f_{x} } \hfill \\ { {\mathbf{m}}\left[ {\left( {\ddot{u}_{gy} + \ddot{u}_{sy} } \right){\mathbf{1}} + {\ddot{\mathbf{u}}}_{ry} } \right] + {\mathbf{c}}_{y} {\dot{\mathbf{u}}}_{ry} + {\mathbf{k}}_{y} {\mathbf{u}}_{ry} = {\mathbf{0}}} \hfill \\ { m_{b} \left( {\ddot{u}_{gy} + \ddot{u}_{sy} } \right) + {\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\left[ {\left( {\ddot{u}_{gy} + \ddot{u}_{sy} } \right){\mathbf{1}} + {\ddot{\mathbf{u}}}_{ry} } \right] = f_{y} } \hfill \\ \end{array} } \right. $$
(3.1)
Fig. 3.1
A schematic diagram and a simplified model. a. It indicates the roof diaphragm, floor diaphragm, vertical structural element, sliding surface and base, and foundation. b. It indicates base mass m b, N th floor mass m N, k x N of k y N, and u r x N of u r y N.

N-story SB structure: a schematic diagram; and b simplified model

where m, cx (cy), and kx (ky) are the mass, damping, and stiffness matrices, respectively, when the structure base is fixed; mb is the mass of the sliding base; 0 and 1 are the vectors whose elements are all zero and unity, respectively. \(\ddot{u}_{gx} \left( t \right)\) and \(\ddot{u}_{gy} \left( t \right)\) are the x and y components, respectively, of the ground acceleration; \(\ddot{u}_{sx} \left( t \right)\) and \(\ddot{u}_{sy} \left( t \right)\) are the sliding accelerations with respect to the ground in the x and y directions, respectively; \({\mathbf{u}}_{rx} = \left[ {u_{rx1} ,u_{rx2} , \ldots ,u_{rxN} } \right]\) and \({\mathbf{u}}_{ry} = \left[ {u_{ry1} ,u_{ry2} , \ldots ,u_{ryN} } \right]\) are the floor displacement vectors in the x and y directions, respectively; \({\dot{\mathbf{u}}}_{rx}\) and \({\dot{\mathbf{u}}}_{ry}\), and \({\ddot{\mathbf{u}}}_{rx}\) and \({\ddot{\mathbf{u}}}_{ry}\) are the corresponding velocity and acceleration vectors, respectively; and fx(t) and fy(t) are the x and y components, respectively, of the friction force at the sliding interface. In Eq. (3.1), the first and second (third and fourth) equations represent the dynamic equilibrium of each floor mass and the whole structure, respectively, in the x-direction (y-direction).

For the model shown in Fig. 3.1b, the mass matrix m is

$$\mathbf{m} = \left[ {\begin{array}{*{20}c} {m_{1} } & 0 & 0 & \cdots & 0 \\ {} & {m_{2} } & 0 & \cdots & 0 \\ {} & {} & {m_{3} } & \cdots & 0 \\ {} & {{\text{sym}}.} & {} & \ddots & \vdots \\ {} & {} & {} & {} & {m_{N} } \\ \end{array}} \right]$$
(3.2)

where mi (i = 1, 2, 3, …, N) is the mass of the ith floor. The stiffness matrices kx and ky are

$${\mathbf{k}_{x}} = \left[ {\begin{array}{*{20}c} {k_{{x1}} + k_{{x2}} } & { - k_{{x2}} } & 0 & \cdots & 0 & 0 \\ {} & {k_{{x2}} + k_{{x3}} } & { - k_{{x3}} } & \cdots & 0 & 0 \\ {} & {} & {k_{{x3}} + k_{{x4}} } & \cdots & 0 & 0 \\ {} & {} & {} & \ddots & \vdots & \vdots \\ {} & {} & {} & {} & { - k_{{x(N - 1)}} } & 0 \\ {} & {{\text{sym}}.} & {} & {} & {k_{{x(N - 1)}} + k_{{xN}} } & { - k_{{xN}} } \\ {} & {} & {} & {} & {} & {k_{{xN}} } \\ \end{array}} \right]$$
(3.3)

and

$${\mathbf{k}_{y}} = \left[ {\begin{array}{*{20}c} {k_{{y1}} + k_{{y2}} } & { - k_{{y2}} } & 0 & \cdots & 0 & 0 \\ {} & {k_{{y2}} + k_{{y3}} } & { - k_{{y3}} } & \cdots & 0 & 0 \\ {} & {} & {k_{{y3}} + k_{{y4}} } & \cdots & 0 & 0 \\ {} & {} & {} & \ddots & \vdots & \vdots \\ {} & {} & {} & {} & { - k_{{y(N - 1)}} } & 0 \\ {} & {{\text{sym}}.} & {} & {} & {k_{{y(N - 1)}} + k_{{yN}} } & { - k_{{yN}} } \\ {} & {} & {} & {} & {} & {k_{{yN}} } \\ \end{array}} \right]$$
(3.4)

where kxi and kyi are the lateral stiffness of the ith story in the x and y directions, respectively. The damping of the superstructure can be assumed to be of the Rayleigh type; thus, when the damping ratios of two specified modes of the superstructure are given, the damping matrix cx and cy can then be determined.

The response history of an SB structure can exhibit two types of phases: stick phase and sliding phase. For the stick phases, during which sliding does not occur, the sliding acceleration is equal to 0, and the friction force, f, is smaller than the static friction force; thus, we have

$$ \left\{ \begin{aligned} & \ddot{u}_{sx} = \ddot{u}_{sy} = 0 \\ & f = \sqrt {f_{x}^{2} + f_{y}^{2} } < \left( {m + m_{b} } \right)\left( {g + \ddot{u}_{gz} } \right)\mu_{s} \\ \end{aligned} \right. $$
(3.5)

where \(\ddot{u}_{gz} \left( t \right)\) is the z (vertical) component of the ground acceleration, g is the gravity acceleration, and \(\mu_{s}\) is the static friction coefficient. Combining Eqs. (3.1) and (3.5) leads to

$$ \left\{ \begin{aligned} & {\mathbf{m}}\left[ {\ddot{u}_{gx} {\mathbf{1}} + {\ddot{\mathbf{u}}}_{rx} } \right] + {\mathbf{c}}_{x} {\dot{\mathbf{u}}}_{rx} + {\mathbf{k}}_{x} {\mathbf{u}}_{rx} = {\mathbf{0}} \\ & {\mathbf{m}}\left[ {\ddot{u}_{gy} {\mathbf{1}} + {\ddot{\mathbf{u}}}_{ry} } \right] + {\mathbf{c}}_{y} {\dot{\mathbf{u}}}_{ry} + {\mathbf{k}}_{y} {\mathbf{u}}_{ry} = {\mathbf{0}} \\ & \sqrt {\left( {\frac{{{\mathbf{1}}^{{\mathbf{T}}} {\mathbf{m}}\ddot{\mathbf{u}}_{rx} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} + \ddot{u}_{gx} } \right)^{2} + \left( {\frac{{{\mathbf{1}}^{{\mathbf{T}}} {\mathbf{m}}\ddot{\mathbf{u}}_{rx} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} + \ddot{u}_{gy} } \right)^{2} } < \left( {g + \ddot{u}_{gz} } \right)\mu_{s} \\ \end{aligned} \right. $$
(3.6)

The first and second equations of Eq. (3.6) govern the response of an SB structure during the stick phases. The third equation of Eq. (3.6) is the precondition for the stick phases; when it is no longer satisfied, a sliding phase starts.

During the sliding phases, the direction of the friction force is opposite to the direction of the sliding velocity; thus, we have

$$ \left\{ \begin{aligned} & f_{x} = \frac{{ - \dot{u}_{sx} }}{{\sqrt {\dot{u}_{sx}^{2} + \dot{u}_{sy}^{2} } }}\left( {\sum\limits_{i = 1}^{N} {m_{i} } + m_{b} } \right)\left( {g + \ddot{u}_{gz} } \right)\mu \\ & f_{y} = \frac{{ - \dot{u}_{sy} }}{{\sqrt {\dot{u}_{sx}^{2} + \dot{u}_{sy}^{2} } }}\left( {\sum\limits_{i = 1}^{N} {m_{i} } + m_{b} } \right)\left( {g + \ddot{u}_{gz} } \right)\mu \\ \end{aligned} \right. $$
(3.7)

where \(\mu\) is the dynamic friction coefficient. Combining Eqs. (3.1) and (3.7) yields

$$ \left\{ \begin{aligned} & \ddot{u}_{sx} {\mathbf{m1}} + {\mathbf{m}}\ddot{\mathbf{u}}_{rx} + {\varvec{c}}_{x} {\dot{\mathbf{u}}}_{rx} + {\mathbf{k}}_{x} {\mathbf{u}}_{rx} = - \ddot{u}_{gx} {\mathbf{m1}} \\ & \ddot{u}_{sx} + \frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\ddot{\mathbf{u}}_{rx} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} + \frac{{\dot{u}_{sx} }}{{\sqrt {\dot{u}_{sx}^{2} + \dot{u}_{sy}^{2} } }}\left( {g + \ddot{u}_{gz} } \right)\mu = - \ddot{u}_{gx} \\ & \ddot{u}_{sy} {\mathbf{m1}} + {\mathbf{m}}\ddot{\mathbf{u}}_{ry} + {\varvec{c}}_{y} {\dot{\mathbf{u}}}_{ry} + {\mathbf{k}}_{y} {\mathbf{u}}_{ry} = - \ddot{u}_{gy} {\mathbf{m1}} \\ & \ddot{u}_{sy} + \frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\ddot{\mathbf{u}}_{ry} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} + \frac{{\dot{u}_{sy} }}{{\sqrt {\dot{u}_{sx}^{2} + \dot{u}_{sy}^{2} } }}\left( {g + \ddot{u}_{gz} } \right)\mu = - \ddot{u}_{gy} \\ \end{aligned} \right. $$
(3.8)

Equation (3.8) governs the response of an SB structure during the sliding phases. When the sliding velocity during a sliding phase becomes 0, this round of sliding ends. Afterwards, the structure may continue to slide or enter a stick phase depending on whether the third equation of Eq. (3.6) is satisfied. Based on the governing equations for the stick and sliding phases and the transition conditions between different phases, as presented above, a program was developed for computing the responses of SB structures subjected to three-dimensional excitations. The numerical methods for solving Eq. (3.8) will be presented in the next section.

3.2 Numerical Computation Methods

The first and second equations of Eq. (3.6), which govern the response of an SB structure during the stick phases, are the same as the governing equations of a fixed base structure. The numerical methods for solving these equations can be found in Chopra (2001).

The system of differential equations presented in Eq. (3.8) can be solved by the time-stepping method. The response quantities \(\left( {\ddot{u}_{sx} } \right)_{i + 1}\), \(\left( {\ddot{u}_{sy} } \right)_{i + 1}\), \(\left( {{\mathbf{u}}_{rx} } \right)_{i + 1}\), \(\left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i + 1}\), \(\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i + 1}\), \(\left( {{\mathbf{u}}_{ry} } \right)_{i + 1}\), \(\left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i + 1}\) and \(\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i + 1}\) at time i + 1 satisfy Eq. (3.8) at time i + 1, i.e.,

$$ \left\{ \begin{aligned} & \left( {\ddot{u}_{sx} } \right)_{i + 1} {\mathbf{m1}} + {\mathbf{m}}\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i + 1} + {\varvec{c}}_{x} \left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i + 1} + {\mathbf{k}}_{x} \left( {{\mathbf{u}}_{rx} } \right)_{i + 1} = - \left( {\ddot{u}_{gx} } \right)_{i + 1} {\mathbf{m1}} \\ & \left( {\ddot{u}_{sx} } \right)_{i + 1} + \frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i + 1} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} + \frac{{\left( {\dot{u}_{sx} } \right)_{i + 1} }}{{\sqrt {\left[ {\left( {\dot{u}_{sx} } \right)_{i + 1} } \right]^{2} + \left[ {\left( {\dot{u}_{sy} } \right)_{i + 1} } \right]^{2} } }} \\ & \quad \quad \quad \quad \quad \quad \quad \left[ {g + \left( {\ddot{u}_{gz} } \right)_{i + 1} } \right]\mu = - \left( {\ddot{u}_{gx} } \right)_{i + 1} \\ & \left( {\ddot{u}_{sy} } \right)_{i + 1} {\mathbf{m1}} + {\mathbf{m}}\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i + 1} + {\mathbf{c}}_{y} \left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i + 1} + {\mathbf{k}}_{y} \left( {{\mathbf{u}}_{ry} } \right)_{i + 1} = - \left( {\ddot{u}_{gy} } \right)_{i + 1} {\mathbf{m1}} \\ & \left( {\ddot{u}_{sy} } \right)_{i + 1} + \frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i + 1} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} + \frac{{\left( {\dot{u}_{sy} } \right)_{i + 1} }}{{\sqrt {\left[ {\left( {\dot{u}_{sx} } \right)_{i + 1} } \right]^{2} + \left[ {\left( {\dot{u}_{sy} } \right)_{i + 1} } \right]^{2} } }} \\ & \quad \quad \quad \quad \quad \quad \quad \left( {g + \ddot{u}_{gz} } \right)\mu = - \left( {\ddot{u}_{gy} } \right)_{i + 1} \\ \end{aligned} \right. $$
(3.9)

Using Newmark’s equations (Newmark, 1959) for the relationships between the response quantities at time i + 1 and the corresponding quantities at time i, we have

$$ \left\{ \begin{aligned} & \left( {\dot{u}_{sx} } \right)_{i + 1} = \left( {\dot{u}_{sx} } \right)_{i} + \frac{1}{2}\Delta t\left( {\ddot{u}_{sx} } \right)_{i} + \frac{1}{2}\Delta t\left( {\ddot{u}_{sx} } \right)_{i + 1} \\ & \left( {{\mathbf{u}}_{rx} } \right)_{i + 1} = \left( {{\mathbf{u}}_{rx} } \right)_{i} + \Delta t\left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i} + \left[ {\left( {0.5 - \beta } \right)\left( {\Delta t} \right)} \right]\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i} + \left[ {\beta \left( {\Delta t} \right)^{2} } \right]\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i + 1} \\ & \left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i + 1} = \left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i} + \frac{1}{2}\Delta t\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i} + \frac{1}{2}\Delta t\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i + 1} \\ & \left( {\dot{u}_{sy} } \right)_{i + 1} = \left( {\dot{u}_{sy} } \right)_{i} + \frac{1}{2}\Delta t\left( {\ddot{u}_{sy} } \right)_{i} + \frac{1}{2}\Delta t\left( {\ddot{u}_{sy} } \right)_{i + 1} \\ & \left( {{\mathbf{u}}_{ry} } \right)_{i + 1} = \left( {{\mathbf{u}}_{ry} } \right)_{i} + \Delta t\left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i} + \left[ {\left( {0.5 - \beta } \right)\left( {\Delta t} \right)} \right]\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i} + \left[ {\beta \left( {\Delta t} \right)^{2} } \right]\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i + 1} \\ & \left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i + 1} = \left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i} + \frac{1}{2}\Delta t\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i} + \frac{1}{2}\Delta t\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i + 1} \\ \end{aligned} \right. $$
(3.10)

where \(\Delta t\) is the time step, and \(\beta\), which ranges from 1/6 to 1/4, is a parameter that controls the variation in the acceleration over a time step. Equation (3.10) can be converted to

$$ \left\{ \begin{aligned} & \left( {\ddot{u}_{sx} } \right)_{i + 1} = \frac{2}{\Delta t}\left[ {\left( {\dot{u}_{sx} } \right)_{i + 1} - \left( {\dot{u}_{sx} } \right)_{i} } \right] - \left( {\ddot{u}_{sx} } \right)_{i} \\ & \left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i + 1} = \frac{1}{{\beta \left( {\Delta t} \right)^{2} }}\left[ {\left( {{\mathbf{u}}_{rx} } \right)_{i + 1} - \left( {{\mathbf{u}}_{rx} } \right)_{i} } \right] - \frac{1}{\beta \Delta t}\left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i} - \left( {\frac{1}{2\beta } - 1} \right)\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i} \\ & \left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i + 1} = \frac{1}{2\beta \Delta t}\left[ {\left( {{\mathbf{u}}_{rx} } \right)_{i + 1} - \left( {{\mathbf{u}}_{rx} } \right)_{i} } \right] + \left( {1 - \frac{1}{2\beta }} \right)\left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i} + \left( {\Delta t - \frac{\Delta t}{{4\beta }}} \right)\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i} \\ & \left( {\ddot{u}_{sy} } \right)_{i + 1} = \frac{2}{\Delta t}\left[ {\left( {\dot{u}_{sy} } \right)_{i + 1} - \left( {\dot{u}_{sy} } \right)_{i} } \right] - \left( {\ddot{u}_{sy} } \right)_{i} \\ & \left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i + 1} = \frac{1}{{\beta \left( {\Delta t} \right)^{2} }}\left[ {\left( {{\mathbf{u}}_{ry} } \right)_{i + 1} - \left( {{\mathbf{u}}_{ry} } \right)_{i} } \right] - \frac{1}{\beta \Delta t}\left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i} - \left( {\frac{1}{2\beta } - 1} \right)\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i} \\ & \left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i + 1} = \frac{1}{2\beta \Delta t}\left[ {\left( {{\mathbf{u}}_{ry} } \right)_{i + 1} - \left( {{\mathbf{u}}_{ry} } \right)_{i} } \right] + \left( {1 - \frac{1}{2\beta }} \right)\left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i} + \left( {\Delta t - \frac{\Delta t}{{4\beta }}} \right)\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i} \\ \end{aligned} \right. $$
(3.11)

Substituting Eq. (3.11) into Eq. (3.9) gives

$$ \left\{ \begin{aligned} & \frac{2}{\Delta t}\left( {\dot{u}_{sx} } \right)_{i + 1} {\mathbf{m1}} + \left[ {\frac{1}{{\beta \left( {\Delta t} \right)^{2} }}{\mathbf{m}} + \frac{1}{2\beta \Delta t}{\mathbf{c}}_{x} + {\mathbf{k}}_{x} } \right]\left( {{\mathbf{u}}_{rx} } \right)_{i + 1} = {\mathbf{p}}_{1} \\ & \frac{2}{\Delta t}\left( {\dot{u}_{sx} } \right)_{i + 1} + \frac{{\left( {\dot{u}_{sx} } \right)_{i + 1} }}{{\sqrt {\left[ {\left( {\dot{u}_{sx} } \right)_{i + 1} } \right]^{2} + \left[ {\left( {\dot{u}_{sy} } \right)_{i + 1} } \right]^{2} } }}\left[ {g + \left( {\ddot{u}_{gz} } \right)_{i + 1} } \right]\mu \\ & \quad \quad \quad \quad \quad + \frac{1}{{\beta \left( {\Delta t} \right)^{2} }}\frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\left( {{\mathbf{u}}_{rx} } \right)_{i + 1} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} = p_{2} \\ & \frac{2}{\Delta t}\left( {\dot{u}_{sy} } \right)_{i + 1} {\mathbf{m1}} + \left[ {\frac{1}{{\beta \left( {\Delta t} \right)^{2} }}{\mathbf{m}} + \frac{1}{2\beta \Delta t}{\mathbf{c}}_{y} + {\mathbf{k}}_{y} } \right]\left( {{\mathbf{u}}_{ry} } \right)_{i + 1} = {\mathbf{p}}_{3} \\ & \frac{2}{\Delta t}\left( {\dot{u}_{sy} } \right)_{i + 1} + \frac{{\left( {\dot{u}_{sy} } \right)_{i + 1} }}{{\sqrt {\left[ {\left( {\dot{u}_{sx} } \right)_{i + 1} } \right]^{2} + \left[ {\left( {\dot{u}_{sy} } \right)_{i + 1} } \right]^{2} } }}\left[ {g + \left( {\ddot{u}_{gz} } \right)_{i + 1} } \right]\mu \\ & \quad \quad \quad \quad \quad + \frac{1}{{\beta \left( {\Delta t} \right)^{2} }}\frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}\left( {{\mathbf{u}}_{ry} } \right)_{i + 1} }}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }} = p_{4} \\ \end{aligned} \right. $$
(3.12)

where

$$ \left\{ \begin{aligned} & {\mathbf{p}}_{1} = \left[ { - \left( {\ddot{u}_{gx} } \right)_{i + 1} + \frac{2}{\Delta t}\left( {\dot{u}_{sx} } \right)_{i} + \left( {\ddot{u}_{sx} } \right)_{i} } \right]{\mathbf{m1}} + \left[ {\frac{1}{{\beta \left( {\Delta t} \right)^{2} }}{\mathbf{m}} + \frac{1}{2\beta \Delta t}{\mathbf{c}}_{x} } \right]\left( {{\mathbf{u}}_{rx} } \right)_{i} \\ & \quad \quad + \left[ {\frac{1}{\beta \Delta t}{\mathbf{m}} - \left( {1 - \frac{1}{2\beta }} \right){\mathbf{c}}_{x} } \right]\left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i} + \left[ {\left(\frac{1}{2\beta } - 1 \right){\mathbf{m}} - \left( {\Delta t - \frac{\Delta t}{{4\beta }}} \right){\mathbf{c}}_{x} } \right]\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i} \\ & p_{2} = - \left( {\ddot{u}_{gx} } \right)_{i + 1} + \frac{2}{\Delta t}\left( {\dot{u}_{sx} } \right)_{i} + \left( {\ddot{u}_{sx} } \right)_{i} \\ & \quad \quad + \frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}}}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }}\left[ {\frac{1}{{\beta \left( {\Delta t} \right)^{2} }}\left( {{\mathbf{u}}_{rx} } \right)_{i} + \frac{1}{\beta \Delta t}\left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i} + \left( {\frac{1}{2\beta } - 1} \right)\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i} } \right] \\ & {\mathbf{p}}_{3} = \left[ { - \left( {\ddot{u}_{gy} } \right)_{i + 1} + \frac{2}{\Delta t}\left( {\dot{u}_{sy} } \right)_{i} + \left( {\ddot{u}_{sy} } \right)_{i} } \right]{\mathbf{m1}} + \left[ {\frac{1}{{\beta \left( {\Delta t} \right)^{2} }}{\mathbf{m}} + \frac{1}{2\beta \Delta t}{\mathbf{c}}_{y} } \right]\left( {{\mathbf{u}}_{ry} } \right)_{i} \\ & \quad \quad + \left[ {\frac{1}{\beta \Delta t}{\mathbf{m}} - \left( {1 - \frac{1}{2\beta }} \right){\mathbf{c}}_{y} } \right]\left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i} + \left[ {\left(\frac{1}{2\beta } - 1 \right){\mathbf{m}} - \left( {\Delta t - \frac{\Delta t}{{4\beta }}} \right){\mathbf{c}}_{y} } \right]\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i} \\ & p_{4} = - \left( {\ddot{u}_{gy} } \right)_{i + 1} + \frac{2}{\Delta t}\left( {\dot{u}_{sy} } \right)_{i} + \left( {\ddot{u}_{sy} } \right)_{i} \\ & \quad \quad + \frac{{{\mathbf{1}}^{{\text{T}}} {\mathbf{m}}}}{{\sum\nolimits_{i = 1}^{N} {m_{i} } + m_{b} }}\left[ {\frac{1}{{\beta \left( {\Delta t} \right)^{2} }}\left( {{\mathbf{u}}_{ry} } \right)_{i} + \frac{1}{\beta \Delta t}\left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i} + \left( {\frac{1}{2\beta } - 1} \right)\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i} } \right] \\ \end{aligned} \right. $$
(3.13)

When determining the response quantities at time i + 1, the response quantities at time i are already known; thus, Eq. (3.12) is a system of nonlinear equations with four unknowns, namely, \(\left( {\dot{u}_{sx} } \right)_{i + 1}\), \(\left( {{\mathbf{u}}_{rx} } \right)_{i + 1}\), \(\left( {\dot{u}_{sy} } \right)_{i + 1}\) and \(\left( {{\mathbf{u}}_{ry} } \right)_{i + 1}\). These nonlinear equations can be solved using the Newton–Raphson iteration technique (Chopra, 2001). After determining the values of \(\left( {\dot{u}_{sx} } \right)_{i + 1}\), \(\left( {{\mathbf{u}}_{rx} } \right)_{i + 1}\), \(\left( {\dot{u}_{sy} } \right)_{i + 1}\) and \(\left( {{\mathbf{u}}_{ry} } \right)_{i + 1}\), other response quantities, namely, \(\left( {u_{sx} } \right)_{i + 1}\), \(\left( {\ddot{u}_{sx} } \right)_{i + 1}\), \(\left( {{\dot{\mathbf{u}}}_{rx} } \right)_{i + 1}\), \(\left( {{\ddot{\mathbf{u}}}_{rx} } \right)_{i + 1}\), \(\left( {u_{sy} } \right)_{i + 1}\), \(\left( {\ddot{u}_{sy} } \right)_{i + 1}\), \(\left( {{\dot{\mathbf{u}}}_{ry} } \right)_{i + 1}\) and \(\left( {{\ddot{\mathbf{u}}}_{ry} } \right)_{i + 1}\) at time i + 1 can also be determined using Newmark’s equations.

3.3 Response Histories Under Earthquake Excitation

Figure 3.2 shows the x-component of the ground accelerations recorded at the Mammoth Lakes station during the 1980 Mammoth Lakes earthquake. The peak ground acceleration (PGA) in the x-direction is 3.8 m/s2. Figure 3.3 shows the absolute acceleration history of the top mass of a single-story fixed base structure subjected to this ground motion. The natural periods of the superstructure along the two principal axes are both 0.3 s, and the damping ratios are taken as 5%. The peak absolute acceleration of the top mass is 7.66 m/s2, which is approximately 2 times the corresponding PGA. Figure 3.4 shows the responses of SB structures with different friction coefficients subjected to this ground motion. The superstructure is the same as that of the aforementioned fixed base structure. The mass of the sliding base is 3/7 of the top mass. As shown in Fig. 3.4, the peak absolute acceleration of the top mass decreases quickly as the friction coefficient decreases; when \(\mu = \mu_{s} = 0.05\), the peak absolute acceleration of the top mass is 1.16 m/s2, which is only 15% of the peak absolute acceleration when the base is fixed. The response history of the sliding displacement varies significantly when different friction coefficients are used. As the friction coefficient decreases, sliding occurs more frequently. Furthermore, the maximum sliding displacements may be obtained in the opposite directions for different friction coefficients.

Fig. 3.2
A graph of u power double dots subscript g x in meters per second square versus time in seconds. The plotted trend fluctuates about 0 with the highest spike at (1.3, 3.8).

x-component of the ground accelerations recorded at the Mammoth Lakes station during the 1980 Mammoth Lakes earthquake

Fig. 3.3
A graph of u power double dots subscript g x + u power double dots subscript r x 1 in meters per second square versus time in seconds. The trend fluctuates about 0 with the deepest trough at (2.2, minus 7.66). The oscillations decay after 2 seconds.

Absolute acceleration history of the top mass of a single-story fixed base structure subjected to the Mammoth Lakes record

Fig. 3.4
4 graphs. a and b. Two graphs of u power double dots subscript g x + u power double dots subscript s x + u power double dots subscript r x 1 in meters per second square and u subscript s x in meters versus time in seconds. The sliding starts, sliding stops, and peak values are indicated on the curves.figure 4

Response histories of SB structures with different friction coefficients subjected to the Mammoth Lakes record