## 2 Methods
### Spring Coefficient  
The spring coefficient (k) of the reference sample A is calculated by dividing the force F by the length change \(\Delta\)L, as shown in Equation (5), where the force and the length change are computed at a strain of 1%. When calculating the external force, the axial differential stress \((\sigma_{1}-\sigma_{3})\) in the formula is obtained from the static triaxial undrained CU test results of [PERSON] et al. [34]. It is found that the axial differential stress 78 kPa can be obtained at the 1% strain. \[\text{k}=\frac{\text{F}}{\Delta\text{L}} \tag{5}\]  
In the formula, the force \(\text{F}=(\sigma_{1}-\sigma_{3})\times\text{A}\), the axial difference stress is calculated at a strain of 1%. The obtained axial differential stress \((\sigma_{1}-\sigma_{3})\) is 78 kPa, and we can convert it to 0.78 kgf/cm\({}^{2}\). Based on specimen area equaling 39.59 cm\({}^{2}\), we can find F = 30.88 kg. Meantime, \(\Delta\)L = \(\varepsilon\times\text{L}\) = 0.01 \(\times\) 14.2 cm = 0.14 cm. The spring coefficient k can be obtained by dividing the force F obtained above by the length change \(\Delta\)L. Using Equation (5), the spring coefficient (k) is calculated as 220 kgf/cm, the spring coefficient of the first cycle. Many papers [35; 36; 37] mentioned that G or E value decreases with increasing strain. However, few papers mentioned the relationship between k and strain. The model in this article requires the relationship between k value and strain; the strain increases due to the variation of k value as the cyclic number increases. We used the relationship between G and the cyclic number as the relationship between the spring coefficient k and the cyclic number. After the first cycle's spring coefficient is calculated, the shear modulus (G) of the first five cycles of the specimen at A-2 is calculated, as shown in Table 2. The shear modulus (G) curve equation, which is G = 5884.9e\({}^{-0.1\text{t}}\) kPa, is solved in Table 2 with the help of a command in Excel. This study uses the shear modulus (G) curve equation of the first five cycles at point A as the basis for the spring coefficient (k) curve equation. The spring coefficient (k) of the first cycle is brought in from the above calculation k = 220 kgf/cm; a spring coefficient (k) curve equation similar to the shear modulus (G) curve equation can be obtained, as shown in Figure 2. Since the number of liquefaction times of the sample at point A is 15, the number of cycles on the horizontal axis is changed to 15. Point A's spring coefficient can be obtained from this method as k = 244e\({}^{-0.1\text{t}}\) kgf/cm. The spring coefficient is a variable that changes with time and will decrease as time increases. Table 2 uses the above spring coefficient equation to calculate the values for each of the 15 cycles. \begin{table}
\begin{tabular}{c c c c c c} \hline  & & **A-2** & & \\ \hline Cyclic no. & 1 & 2 & 3 & 4 & 5 \\ \hline shear modulus G (kPa) & 5047.94 & 4163.75 & 3478.77 & 2977.94 & 2561.36 \\ \hline \end{tabular}
\end{table} 
Table 2: Determine shear modulus using cyclic triaxial test in A-2. Damping Coefficient  
[PERSON] et al. and [PERSON] [17; 19] both mentioned the damping ratio (D) analysis method, such as Equation (6). \[\mathrm{D}=\frac{1}{4\pi}\frac{\mathrm{A_{L}}}{\mathrm{A_{T}}} \tag{6}\]  
In which \(\mathrm{A_{L}}\) is the area of the hysteresis loop, and \(\mathrm{A_{T}}\) is the area of a shaded right triangle as Figure 3. \(\mathrm{A_{L}}\) can be calculated by \((\varepsilon_{1},\sigma_{\mathrm{d1}})\), \((\varepsilon_{2},\sigma_{\mathrm{d2}})\), \((\varepsilon_{3},\sigma_{\mathrm{d3}})\)... \((\varepsilon_{\mathrm{n}},\sigma_{\mathrm{dn}})\) into Equation (7)  
\[\mathrm{A_{L}}=\frac{1}{2}\left| \begin{matrix}\varepsilon_{1}&\varepsilon_{2}& \varepsilon_{3}&\ldots&\varepsilon_{\mathrm{n}}&\varepsilon_{1}\\ \sigma_{\mathrm{d1}}&\sigma_{\mathrm{d2}}&\sigma_{\mathrm{d3}}&\ldots&\sigma _{\mathrm{dn}}&\sigma_{\mathrm{d1}}\end{matrix}