# Dynamic Behavior of an Embedded Foundation under Horizontal Vibration in a Poroelastic Half-Space
## 2 Problem Definition and Governing Equations
### General Solutions of Horizontal Vibration of Embedded Foundation
#### 2.3.1 Reaction Force at the Side of the Embedded Foundation
\right] \tag{16}\]  
Equations (15) and (16) can be compiled as  
\[\Big{(}x^{2}-c_{1}x+c_{2}\Big{)}\varphi_{\rm s,f}=0 \tag{17}\]  
\[\Big{(}x-{d_{1}}^{2}\Big{)}\psi_{s,f}=0 \tag{18}\]  
where \(x=\
abla^{2}\), \(c_{1}=\frac{2\alpha M\rho_{f}\omega^{2}-\rho\omega^{2}M-\big{(}\lambda+2G_{\rm s }+a^{2}M\big{)}\big{(}a\omega^{2}-bi\omega\big{)}}{(\lambda+2G_{\rm s})M}\), \(c_{2}=\frac{\rho\omega^{2}\big{(}a\omega^{2}-bi\omega\big{)}-\big{(}\rho_{f} \omega^{2}\big{)}^{2}}{(\lambda+2G_{\rm s})M}\), \(d_{3}{}^{2}=\frac{-\rho\omega^{2}\big{(}a\omega^{2}-bi\omega\big{)}+\big{(}\rho_{f} \omega^{2}\big{)}^{2}}{G_{\rm s}(a\omega^{2}-bi\omega)}\) Equations (17) and (18) can be written as  
\[\Big{(}x-{d_{1}}^{2}\Big{)}\varphi_{1}=0\ \Big{(}x-{d_{2}}^{2}\Big{)}\varphi_{2}=0 \tag{19}\]  
with \(x_{1,2}={d_{1,2}}^{2}=\frac{c_{1}\pm\sqrt{c_{1}^{2}-4c_{2}}}{2}\), \(\varphi_{\rm s}=\varphi_{1}+\varphi_{2}\), \(\varphi_{1}\) is assumed as \(\varphi_{1}=H(r)I(\theta)e^{i\omega t}\) in order to separate variables. The following equation can be obtained:  
\[I(\theta)\frac{d^{2}H(r)}{dr^{2}}+I(\theta)\frac{dH(r)}{rdr}+H(r)\frac{d^{2}I( \theta)}{r^{2}d\theta^{2}}-{d_{1}}^{2}H(r)I(\theta)=0 \tag{20}\]  
The solution of Equation (20) is:  
\[H(r)=S_{a}K_{n1}(d_{1}r)+S_{b}I_{n1}(d_{1}r)\ \ I(\theta)=S_{c}\sin n_{1}\theta+S_{d} \cos n_{1}\theta \tag{21}\]  
The formulation of \(\varphi_{1}\) and \(\varphi_{2}\) can be obtained as follows,  
\[\varphi_{1}=H(r)I(\theta)e^{i\omega t}=[S_{a}K_{n1}(d_{1}r)+S_{b}I_{n1}(d_{1}r)] (S_{c}\sin n_{1}\theta+S_{d}\cos n_{1}\theta)e^{i\omega t} \tag{22}\]  
\[\varphi_{2}=H(r)I(\theta)e^{i\omega t}=[S_{aa}K_{n2}(d_{2}r)+S_{bb}I_{n2}(d_{2}r)] (S_{cc}\sin n_{2}\theta+S_{dd}\cos n_{2}\theta)e^{i\omega t} \tag{23}\]where \(I_{\rm{n1}}\), \(I_{\rm{n2}}\), \(K_{\rm{n1}}\), \(K_{\rm{n2}}\) denote modified Bessel function of the first and second kind. With the property of Bessel function and the odevity of \(e\), \(S_{\rm{b}}=S_{\rm{bb}}=0\), \(S_{\rm{c}}=S_{\rm{cc}}=0\),  
\[\varphi_{s}=[S_{1}K_{1}(d_{1}r)\cos\theta+S_{11}K_{1}(d_{2}r)]e^{i\omega t} \quad]\varphi_{f}=[S_{2}K_{1}(d_{1}r)\cos\theta+S_{22}K_{1}(d_{2}r)]e^{i\omega t} \tag{24}\]  
\[\psi_{s}=S_{3}K_{1}(d_{3}r)\sin\theta e^{i\omega t}\quad\psi_{f}=S_{4}K_{1}(d_ {3}r)\sin\theta e^{i\omega t} \tag{25}\]  
By substituting the expression of Equations (24) and (25) to Equations (13) and (14),  
\[S_{2}=\chi_{1}S_{1},\ S_{22}=\chi_{2}S_{11},\ S_{4}=\chi_{3}S_{3},\]  
where  
\[\Big{(}x-{d_{1}}^{2}\Big{)}\psi_{s,f}=0 \tag{18}\]0  
By substituting the expression of \(\varphi_{s},\ \varphi_{f},\ \psi_{s},\ \psi_{f}\) to the potential function and using the boundary condition of \(u_{r}=u_{h}e^{i\omega t}\cos\theta\), \(u_{\theta}=-u_{h}e^{i\omega t}\sin\theta\), \(w_{\rm{r}}\) (\(r_{0}\),\(\theta\),\(z\)) = 0, the following equations can be obtained,  
\[\Big{(}x-{d_{1}}^{2}\Big{)}\psi_{s,f}=0 \tag{18}\]1  
\[\Big{(}x-{d_{1}}^{2}\Big{)}\psi_{s,f}=0 \tag{18}\]2  
\[\Big{(}x-{d_{1}}^{2}\Big{)}\psi_{s,f}=0 \tag{18}\]3  
The roots of \(S_{1}\), \(S_{11}\), \(S_{3}\) can be obtained by solving a \(3\times 3\) matrix. The force resultant around the foundation side is \(q_{\rm{h}}\). The horizontal resultant of reaction stress can be obtained by direct integration of the horizontal stress. With the negative pore water pressure considered, \(\sigma_{r}=\sigma^{\prime}-p_{f}\). \[\Big{(}x-{d_{1}}^{2}\Big{)}\psi_{s,f}=0 \tag{18}\]4  
The expression of \(q_{\rm{h}}\) can be obtained as follows by a series of substitution of normal stress, shear stress, and pore water pressure. \[q_{h}=-\pi re^{i\omega t}\left[ \begin{array}{c}(\lambda+2G_{s}+\alpha M+ \chi_{1}M)S_{1}{d_{1}}^{2}K_{1}(d_{1}r)+\\ (\lambda+2G_{s}+\alpha M+\chi_{2}M)S_{2}{d_{2}}^{2}K_{1}(d_{2}r)+G_{s}S_{3}{d_ {3}}^{2}K_{1}(d_{3}r)\end{array} \right] \tag{30}\]  
The reaction force along the foundation side is:  
\[F_{s}=q_{h}\times L=G_{s}u_{h}L(f_{ss}+if_{sd}) \tag{31}\]  
where \(f_{ss}={\rm{Re}}[\frac{q_{h}}{G_{s}u_{h}}]\), \(f_{sd}={\rm{Im}}[\frac{q_{h}}{G_{s}u_{h}}]\).