## 2 Two Degrees of Freedom (2-DOF) Model
### Damping  
The 2-DOF model considers the damping as the second effect of SSI. Assuming that the damping is low enough to maintain the orthogonality properties of the mode shapes, it is possible to write:  
\[\ddot{u_{1}}+2\xi_{1}\omega_{1}\dot{u_{1}}+\omega_{1}^{2}x_{1}=-\alpha_{1} \ddot{u_{b}} \tag{21}\]  
\[\ddot{u_{2}}+2\xi_{2}\omega_{2}\dot{u_{2}}+\omega_{2}^{2}x_{2}=-\alpha_{2} \ddot{u_{b}} \tag{22}\]  
where:  
\(\xi_{1}\) and \(\xi_{2}\) are the damping ratios, respectively: \(\xi_{1}=\frac{C_{1}}{2\omega_{1}M_{1}}\) and \(\xi_{2}=\frac{C_{2}}{2\omega_{2}M_{2}}\)  
\(\alpha_{1}\) and \(\alpha_{2}\) are the modal participation factors for the two mode shapes of the system and \(M_{1}\) and \(M_{2}\) are the masses in the first and second mode:  
\(M_{1}=\Big{(}A^{(1)}\Big{)}^{T}\textbf{M}\cdot\Big{(}A^{(1)}\Big{)}\) and \(M_{2}=\Big{(}A^{(2)}\Big{)}^{T}\textbf{M}\cdot\Big{(}A^{(2)}\Big{)}\)  
Finally, \(A^{(1)}\) and \(A^{(2)}\) are the mode shapes, that can be calculated by introducing (17) and (18) inside:  
\[\Big{(}\textbf{K}-\omega_{1}^{2}\textbf{M}\Big{)}\left(\textbf{A}\right)=\left( 0\right) \tag{23}\]  
Solving by setting \(A_{1}^{(i)}=1\):The first mode shape \((A^{(1)})\) is given (by setting \(A^{(1)}_{1}=1\)) :  
\[(A^{(1)})=\left(\begin{array}{c}A^{(1)}_{1}\\ A^{(1)}_{2}\end{array}\right)=\left(\begin{array}{c}1\\ \frac{2\alpha-2-\beta}{\beta(2+\beta)}\end{array}\right) \tag{24}\]  
and the second mode shape \((A^{(2)})\) is given (by setting \(A^{(2)}_{1}=1\)) :  
\[(A^{(2)})=\left(\begin{array}{c}A^{(2)}_{1}\\ A^{(2)}_{2}\end{array}\right)=\left(\begin{array}{c}1\\ -\frac{1}{2\alpha+\beta}\end{array}\right) \tag{25}\]  
Therefore:  
\[\xi_{1}=\frac{\Big{(}A^{(1)}\Big{)}^{T}\textbf{C}\cdot\Big{(}A^{(1)}\Big{)}}{2 \omega_{1}\big{(}A^{(1)}\big{)}^{T}\textbf{M}\cdot\big{(}A^{(1)}\big{)}}=\frac {\xi_{g}}{\sqrt{\frac{\beta+2}{2\alpha}}\bigg{(}1+\frac{\left(2\alpha-2-\beta \right)^{2}}{\beta(\beta+2)^{2}}}+\frac{\xi_{g}\frac{\left(2\alpha-2-\beta \right)^{2}}{\beta^{2}(\beta+2)^{2}}}{\sqrt{\frac{\beta+2}{2}}\bigg{(}\frac{1} {\beta}+\frac{\left(2\alpha-2-\beta\right)^{2}}{\beta^{2}(\beta+2)^{2}}\bigg{)}} \tag{26}\]  
\[\xi_{2}=\frac{\Big{(}A^{(2)}\Big{)}^{T}\textbf{C}\cdot\Big{(}A^{(2)}\Big{)}}{2 \omega_{2}\big{(}A^{(2)}\big{)}^{T}\textbf{M}\cdot\big{(}A^{(2)}\big{)}}=\frac {\xi_{g}}{\sqrt{\frac{2\alpha+\beta}{2\alpha}}\bigg{(}1+\frac{\beta}{\left(2 \alpha+\beta\right)^{2}}}+\frac{\xi_{g}\frac{1}{\left(2\alpha+\beta\right)^{2 }}}{\sqrt{\frac{2\alpha+\beta}{2}}\bigg{(}\frac{1}{\beta}+\frac{1}{\left(2 \alpha+\beta\right)^{2}}\bigg{)}} \tag{27}\]  
Equation (27) may be rewritten as:  
\[\xi_{2}=\frac{\xi_{g}\sqrt{\frac{2\alpha}{2\alpha+\beta}}}{\bigg{(}1+\frac{ \beta}{\left(2\alpha+\beta\right)^{2}}\bigg{)}}+\frac{\xi_{g}\frac{1}{\left(2 \alpha+\beta\right)^{2}}\sqrt{\frac{2}{2\alpha+\beta}}}{\bigg{(}\frac{1}{ \beta}+\frac{1}{\left(2\alpha+\beta\right)^{2}}\bigg{)}} \tag{28}\]  
Equation (28) shows that the damping of the system consists of two contributions: the first is proportional to the soil damping (\(\xi_{g}\)) and the second term depends on the contribution of the structural damping (\(\xi_{s}\)). In particular, Equation (28) indicates also that the damping of the equivalent system is larger than the damping ratio of the structure itself, confirming previous findings, such as [1].