## 3 Case Study
At the same time, two tracking parameters must be used in order to identify the current and previous quadrant position of (\(m_{i}\), \(\theta_{i}\)) pair. Figure 2: The backbone curve. This is necessary because different rules apply for the same position depending on the last reloading point. For example, a point in the first quadrant can be on the reloading branch originating in the first or second quadrant. Moreover, taking into account that the reloading branch has variable slopes, the last plastic rotation associated zero bending moment must also be stored. Table 2 presents the memorized variables and their role in the algorithm. \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**If Previous** & **If Current** & **Rule** & **Branch** \\
**Step Value** & **Trial Value** & **Rule** & **Slope** \(\mathbf{k^{br}}\) \\ \hline \multirow{4}{*}{\(m_{k}\in[OA]\)} & \(|m_{k+1}^{trial}|<|m_{y}|\) & Rule 1 & \(k^{br}=0\) \\ \cline{2-5}  & \(|m_{k+1}^{trial}|\geq|m_{y}|\) & Rule 2 & \(k^{br}=k^{k}\) \\ \cline{2-5}  & \(|m_{k}^{trial}|<|m_{s}|\) & Rule 2 & \(k^{br}=k^{k}\) \\ \cline{2-5}  & \(|m_{k}^{trial}|\geq|m_{s}|\) & Rule 5 & \(k^{br}=k^{k}\) \\ \cline{2-5}  & \(|m_{k}\in[EF]\) & \(|m_{k+1}^{trial}|\geq|m_{k}|\) & Rule 5 & \(k^{br}=k^{k}\) \\ \cline{2-5}  & \(|m_{k}\in[FG]\) & \(|m_{k+1}^{trial}|<|m_{r}|\) & Rule 3 & \(k^{br}=k_{ul}\) \\ \cline{2-5}  & \(|m_{k+1}^{trial}|\geq|m_{r}|\) & Rule 5 & \(k^{br}=k^{k}\) \\ \cline{2-5}  & \(m_{k}\in[GH]\) & \(|m_{k+1}^{trial}|<|m_{max}^{trial}|\) & Rule 4 & \(k^{br}=k^{l}\) \\ \cline{2-5}  & \(|m_{k+1}^{trial}|\geq|m_{max}^{trial}|\) & Rule 2 & \(k^{br}=k^{k}\) \\ \cline{2-5}  & \(m_{k}\in[HI]\) & \(|m_{k+1}^{trial}|<|m_{trl}|\) & Rule 3 & \(k^{br}=k_{ul}\) \\ \cline{2-5}  & \(|m_{k+1}^{trial}|\geq|m_{HI}|\) & Rule 4 & \(k^{br}=k^{l}\) \\ \cline{2-5}  & \(m_{k}\in[II]\) & \(|m_{k+1}^{trial}|<|m_{max}^{trial}|\) & Rule 4 & \(k^{br}=k^{l}\) \\ \cline{2-5}  & \(|m_{k+1}^{trial}|\geq|m_{max}^{trial}|\) & Rule 5 & \(k^{br}=k^{k}\) \\ \hline \multirow{2}{*}{Unloading case} & \(m_{k}\in[OA]\) & \((\forall)\)\(m_{k+1}^{trial}\) & Rule 1 & \(k^{br}=0\) \\ \cline{2-5}  & \(m_{k}\
otin[OA]\) & \((\forall)\)\(m_{k+1}^{trial}\) & Rule 3 & \(k^{br}=k_{ul}\) \\ \hline \multirow{2}{*}{Reloading case} & \((\forall)\)\(m_{k}^{trial}\) & \(sign(m_{k+1}^{trial})\
eq sign(m_{k})\) & Rule 4 & \(k^{br}=k^{l}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Hysteresis rules of the peak-oriented model. Figure 3: The general hysteresis mode.1  
In order to conduct the nonlinear dynamic analysis, a spatial moment-resisting frame structural model was chosen as illustrated by Figure 5: a 10-story building with 2 spans (\(L=6\) m), 3 days (\(B=5\) m), and the floor height is \(H=4\) m. The cross-section of the columns is 1.1 m \(\times\) 1.1 m, and for the beams 0.6 m \(\times\) 0.3 m, while the Young modulus \(E=30{,}000{,}000\) kN/m\({}^{2}\) and density \(\rho=2.5035\) tons/m\({}^{3}\). At the same time, a live load of 11 kN/m is considered acting on all beams. Furthermore, assuming a proportional Rayleigh type damping model, the coefficients \(\alpha\) and \(\beta\) are 0.2339 and 0.0107, respectively, and correspond to a 5% damping ratio on the first two eigenmodes. The structural elements are modeled using beam finite elements as presented in [10]. The structural model is constructed by considering 6 degrees of freedom (DOFs) per joint, 2 plastic hinges for each of the 170 beam elements, and 4 plastic hinges--interacting 2 by 2--for each of the 120 column elements. The total number of plastic hinges gives the number of plastic rotation \(\theta^{\prime\prime}\) to be computed in each time step, namely 820 elements. Additionally, in each time step the state vector \(\mathbf{z}\) given in Equation (7) is evaluated. If the full model is considered, there are 720 displacement elements and additional 720 velocity elements included in the state vector.