## 3 The Role of Internal Resistance in Distribution of Horizontal Stress
\begin{array}{l}\left(\frac{\partial\sigma_{x}}{\partial x}+\frac{ \partial\tau_{yx}}{\partial y}+\frac{\partial\tau_{zx}}{\partial z}-R_{x} \right)+f_{x}=\rho\ddot{u}_{x}\\ \left(\frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\sigma_{y}}{\partial y }+\frac{\partial\tau_{zy}}{\partial z}-R_{y}\right)+f_{y}=\rho\ddot{u}_{y}\\ \left(\frac{\partial\tau_{xz}}{\partial x}+\frac{\partial\tau_{yz}}{\partial y }+\frac{\partial\sigma_{z}}{\partial z}-R_{z}\right)+f_{z}=\rho\ddot{u}_{z} \end{array} \right\}, \tag{2}\]  
where \(R_{i}\) is the internal resistance, \(f_{i}\) the body force and \(\ddot{u}_{s,i}\) the imposed seismic force per unit mass in the \(i\)-th direction (\(i\in\{x,y,z\}\)), whilst \(\rho\) is the density of the material. Intuitively, the internal resistance is given as  
\[R_{i}=\frac{n}{z}\tau_{R,m}, \tag{3}\]  
where \(n\) is the number of planes of the material element resisting to deformation (and thus, to the transmission of stress) and \(\tau_{R,m}\) is the mobilized shear strength of soil. The cubic material element of Figure 1 with planes perpendicular to the coordinate axes of Cartesian coordinate system is assumed. In Mohr-Coulomb terms the shear strength at depth \(z\) is  
\[\tau_{R,m}=c_{m}+\gamma z\tan\varphi_{m}. \tag{4}\]  
If no shear is assumed to take place in the vertical direction, \(R_{z}\) in Equation (2) is set to zero. This is a rational assumption especially when loads are distributed uniformly in a semi-infinite mass (as in the case of the self-weight of a material with horizontal surface and/or seismic loading), causing no relative movement of soil particles in the vertical direction. Regarding the acceleration term, the pseudo-static concept of analysis is adopted and the imposed seismic force per unit mass will be [30]  
\[\ddot{u}_{s,i}=k_{i}g, \tag{5}\]where \(k_{i}\) is the seismic coefficient in the \(x\), \(y\), or \(z\) direction (usually, \(k_{x}=k_{y}=k_{h}\) and \(k_{z}=k_{v}\)) and \(g\) is the acceleration of gravity. The seismic coefficients may take negative values indicating the change in direction of the seismic force during an earthquake. In geotechnical engineering gravity is usually the only body force, thus, when body forces are considered in the analysis, \(f_{z}\) equals \(\rho g\), whilst \(f_{x}=f_{y}=0\). Based on the above and replacing \(\rho g\) with \(\gamma\) and, also, considering the downward direction of forces as positive, Equation (2) becomes:  
\[\left. \begin{array}{l}\frac{\partial x_{x}}{\partial x}+\frac{ \partial\tau_{yx}}{\partial y}+\frac{\partial\tau_{zx}}{\partial z}-\frac{n}{ z}(c_{m}+(1-k_{v}))\gamma z\tan\varphi_{m})=-k_{h}\gamma\\ \frac{\partial\tau_{xy}}{\partial x}+\frac{\partial x_{y}}{\partial y}+\frac{ \partial\tau_{xy}}{\partial z}-\frac{n}{z}(c_{m}+(1-k_{v}))\gamma z\tan \varphi_{m})=-k_{h}\gamma\\ \frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\tau_{yz}}{\partial y}+ \frac{\partial\tau_{yz}}{\partial z}-\gamma=-k_{v}\gamma\end{array} \right\}, \tag{6}\]  
whilst, in plane strain conditions it simplifies to:  
\[\frac{\partial\sigma_{x}}{\partial x}+\frac{\partial\tau_{xy}}{ \partial z}-\frac{2}{z}(c_{m}+(1-k_{v}))\gamma z\tan\varphi_{m})=-k_{h}\gamma \tag{7a}\] \[\frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\sigma_{z}}{ \partial z}=(1-k_{v})\gamma \tag{7b}\]