## 2 The Transverse Isotropic Model of Layered Rock Mass
### A Semi-Infinite Plane Subjected to Vertically Distributed Stresses  
For the case of the isotropic elastic half-plane subjected to vertical stresses on the point, the coordinate system is established as shown in Figure 2. The XOY plane is assumed to the transverse isotropic plane [44]. If \(F_{1}\) is the applied stress, the basic equation can be obtained as follows:  
Figure 1: Transversely isotropic constitutive model. Figure 2: A transversely isotropic elastic half-plane surface subjected to vertically concentrated stress. 1. Equilibrium differential equation \[\left\{ \begin{array}{l}\frac{\partial\sigma_{x}}{\partial x}+\frac{\partial\tau_ {xz}}{\partial z}+f_{x}=0\\ \frac{\partial\sigma_{y}}{\partial z}+\frac{\partial\tau_{xz}}{\partial x}+f_{ z}=0\end{array} \right.\] (2) where \(f_{x}\) and \(f_{z}\) are the volume forces along the X and Z axes, respectively; \(\sigma_{x}\) and \(\sigma_{z}\) are the normal stresses along the X and Z axes, respectively; and \(\tau_{xz}\) is the shear stress along the X direction. 2. geometric equation \[\varepsilon_{x}=\frac{\partial u}{\partial x},\varepsilon_{y}=\frac{\partial w }{\partial z},\gamma_{yz}=\frac{\partial w}{\partial x}+\frac{\partial u}{ \partial z}\] (3) where \(u\) and \(w\) are the displacements along the X and Z axes, respectively. 3. physical equations \[\left\{ \begin{array}{l}\varepsilon_{x}=\frac{1}{E_{1}}\big{(}1-{\mu_{1}}^{ 2}\big{)}\sigma_{x}-\frac{1}{E_{2}}(1+\mu_{1})\mu_{2}\sigma_{z}\\ \varepsilon_{z}=-\frac{1}{E_{2}}(1+\mu_{1})\mu_{2}\sigma_{x}+\frac{1}{E_{2}} \big{(}1-\frac{E_{1}}{E_{2}}{\mu_{2}}^{2}\big{)}\sigma_{z}\\ \gamma_{xz}=\frac{1}{E_{2}}\tau_{xz}\end{array} \right.\] (4) where \(E_{1}\) and \(\mu_{1}\) are the elastic modulus and Poisson's ratio in the transverse isotropic plane; \(E_{2}\) and \(\mu_{2}\) are the elastic modulus and Poisson's ratio in the vertical isotropic plane (z-axis direction); and \(G_{2}\) is the shear modulus in the vertical isotropic plane. 4. boundary conditions  When \(z=0\), \(\tau_{xz}=0\), \(\sigma_{z}=-p\delta(x)\). where: \(\delta(x)\) is the function of \(Dirac\), \(\delta(x)=\left\{ \begin{array}{l}\infty,x=0\\ 0,x\
eq 0\end{array} \right..\) \[\frac{\partial^{2}\varepsilon_{x}}{\partial z^{2}}+\frac{\partial^{2} \varepsilon_{z}}{\partial x^{2}}=\frac{\partial^{2}\gamma_{xz}}{\partial x \partial z}\] (5)  
Without considering physical strength, the stress component of the transversely isotropic plane problem is also expressed as:  
\[\sigma_{x}=\frac{\partial^{2}\varphi}{\partial z^{2}},\sigma_{z}=\frac{ \partial^{2}\varphi}{\partial x^{2}},\tau_{xz}=-\frac{\partial^{2}\varphi}{ \partial x\partial z} \tag{6}\]  
where \(\varphi=\varphi(x,z)\) is a \(Airy\) function. The stress function \(\varphi=\varphi(x,z)\) automatically satisfied the equilibrium Equation (2) [45]. The physical Equation (4) is substituted into the deformation coordination Equation (5), and then Equation (6) is deduced to obtain the compatible equation, which is expressed by the stress function as:  
\[\frac{1}{E_{1}}\Big{(}1-u_{1}^{2}\Big{)}\frac{\partial^{4}\varphi}{\partial z ^{4}}+\left[\frac{1}{G_{2}}-\frac{2}{E_{2}}(1+u_{1})u_{2}\right]\frac{ \partial^{4}\varphi}{\partial x^{2}\partial z^{2}}+\frac{1}{E_{2}}\Bigg{(}1- \frac{E_{1}u_{2}^{2}}{E_{2}}\Bigg{)}\frac{\partial^{4}\varphi}{\partial x^{4}}=0 \tag{7}\]  
Note that the solution of the transversely isotropic problem is reduced to seek the stress function satisfying the differential Equation (7) and boundary conditions \(\varphi=\varphi(x,z)\) without regard to physical force. It is very convenient for this differential equation to use the Fourier transform to solve Equation (7). By applying the Fourier transform of Equation (7) to X, it can be obtained as follows:  
\[ \begin{array}{l}\frac{1}{E_{1}}\big{(}1-\mu_{1}^{2}\big{)}\frac{d^{4} \varphi(\xi,z)}{dz^{4}}-\xi^{2}\Big{[}\frac{1}{G_{2}}-\frac{2}{E_{2}}(1+\mu_{ 1})\mu_{2}\Big{]}\frac{d^{2}\varphi(\xi,z)}{dz^{2}}\\ +\xi^{4}\frac{1}{E_{2}}\Big{(}1-\frac{E_{1}u_{2}^{2}}{E_{2}}\Big{)}\overline{ \varphi}(\xi,z)=0\end{array}