## 2 Description of MABPA
### Prediction of the Peak Response using the Momentary Energy Input
#### 2.2.2 Formulation of the Effective Period and the Hysteretic Dissipated Energy in a Half Cycle  
To predict the peak responses of the first two modes, bilinear idealization of the equivalent acceleration-equivalent displacement relationship was made as follows. In the following discussion, the formulations are made for the first modal response. Figure 2 shows the bilinear idealization procedure used in this study.  
Then, the effective period (\(T_{\text{1}eff}\)) and the hysteretic dissipated energy in the half cycle per unit mass (\(\Delta E_{\mu\text{1}U}*/M_{\text{1}U}*\)) were formulated. Figure 3 shows the modeling of the structural response in a half cycle used to predict the peak response. In Figure 3, the parameter \(\eta\) is the displacement ratio in the positive and negative directions and was assumed to be in the range from 0 to 1.  
Figure 3: Modeling of the structural response in a half cycle used to predict the peak response: (**a**) The effective slope (\(\kappa_{\text{1}eff}\)) and (**b**) the hysteretic dissipated energy in a half cycle per unit mass (\(\Delta E_{\mu\text{1}U}/M_{\text{1}U}\)).  
Figure 2: Bilinear idealization of the equivalent acceleration-equivalent displacement relationship.  
In this study, the effective period of the first mode corresponding to \(D_{1U}{}^{*}\) was calculated as:  
\[T_{1 eff}(D_{1U}{}^{*})=2\pi/\sqrt{\kappa_{1 eff}(D_{1U}{}^{*})}. \tag{6}\]  
In addition, the equivalent velocity of the hysteretic dissipated energy in a half cycle corresponding to \(D_{1U}{}^{*}\) was calculated as:  
\[V_{\Delta E_{1U}{}^{*}}(D_{1U}{}^{*})=\sqrt{\frac{2\Delta E_{1U}{}^{*}}{M_{1U} {}^{*}}(D_{1U}{}^{*})}. \tag{7}\]  
The effective slope corresponding to \(D_{1U}{}^{*}\) was calculated as:  
\[\kappa_{1 eff}(D_{1U}{}^{*}) =\int\limits_{0}^{1}\kappa_{1 eff}(D_{1U}{}^{*},\eta)d\eta \tag{8}\] \[=\left\{\begin{array}{ll}\kappa_{1}&:D_{1U}{}^{*}\leq D_{1U}{}^ {*}y\\ (2\ln 2)\frac{A_{1U}{}^{*}}{D_{1U}{}^{*}}+\kappa_{2}-\left\{\ln\!\left(1+\frac{D _{1U}{}^{*}y}{D_{1U}{}^{*}}\right)\right\}\!\left(\frac{A_{1U}{}^{*}y}{D_{1U} {}^{*}}+\kappa_{1}-\kappa_{2}\right)+\frac{D_{1U}{}^{*}y}{D_{1U}{}^{*}}(\kappa _{1}-\kappa_{2})&:D_{1U}{}^{*}>D_{1U}{}^{*}y\end{array}\right.\]  
The hysteretic dissipated energy in a half cycle per unit mass corresponding to \(D_{1U}{}^{*}\) was calculated as:  
\[\frac{\Delta E_{1U}{}^{*}}{M_{1U}{}^{*}}(D_{1U}{}^{*}) =\int\limits_{0}^{1}\frac{\Delta E_{1U}{}^{*}}{M_{1U}{}^{*}}(D_{1 U}{}^{*},\eta)d\eta \tag{9}\] \[=\left\{\begin{array}{ll}\frac{1}{3}\kappa_{1}D_{1U}{}^{*}{}^{2} \\ \frac{1}{3}\kappa_{2}D_{1U}{}^{*}{}^{2}+\frac{3}{2}A_{1U}{}^{*}{}_{0}D_{1U}{}^ {*}\!\left\{\left(1-\frac{D_{1U}{}^{*}y}{D_{1U}{}^{*}}\right)^{2}+\frac{2}{3} \frac{D_{1U}{}^{*}y}{D_{1U}{}^{*}}\!\left(1-\frac{2}{3}\frac{D_{1U}{}^{*}y}{ D_{1U}{}^{*}}\right)\right\}&:D_{1U}{}^{*}>D_{1U}{}^{*}y\end{array}\right.\]  
The predicted peak response \(D_{1U}{}^{*}{}_{\text{max}}\) was obtained from the following equation:  
\[V_{\Delta E\mu 1}u^{*}\Big{\{}T_{1 eff}(D_{1U}{}^{*}{}_{\text{max}})\Big{\}}=V_{ \Delta E}\Big{\{}T_{1 eff}(D_{1U}{}^{*}{}_{\text{max}})\Big{\}}. \tag{10}\]  
Note that no viscous damping was considered because the isolation layer was assumed to have no viscous damping.