## 3 Verification against Numerical Simulations  
To verify the proposed analytical solutions under distributed loading in any direction, a model is established with FLAC. The dimensions are 100 m, and the material parameters are shown in Table 1. The value of the distributed loading is \(7\sqrt{2}\ \times\ 10\)MPa, and it is applied on the top of transversely isotropic floor rock at an angle of 45\({}^{\circ}\) to the horizontal plane to analyze the effect of the loading direction on the stress distribution of the model. The 11 measuring points and their positions in the numerical model are shown in Figure 7. The initial time step is 0.5 us, the maximum time is 1 us, and the relative and absolute  
Figure 5: Transversely isotropic elastic half-plane surface subjected to horizontally distributed stress. Figure 6: Transversely isotropic elastic half-plane surface subjected to distributed stress in any direction. tolerances during the calculation are \(1e-5\). It has been noted that the time step was reasonable enough to guarantee the convergence and resolution accurately. Prior to stress redistribution under different coefficients, a logical way to validate an analytical solution is to compare them with numerical solutions. Figure 8 shows several results of comparing the distribution of the stress fields based on the numerical simulation, accounting for the various values of the elastic moduli. To illustrate the factor (E) that affects the stress distribution, take \(\mu_{1}=\mu_{2}=0.3\), as \(G_{2}\) is related to \(\mu_{1}\), \(\mu_{2}\), \(E_{1}\), and \(E_{2}\); given that \(E_{1}=3\) GPa set \(E_{1}/E_{2}\) as 1.0, 2.0, and 3.0, respectively, and get the same depth (z = 1 m) distribution of \(\sigma_{z}\), \(\sigma_{\times}\), and \(\tau_{\times}\), as shown in Figure 8. It indicates that the results, in terms of distributed stresses based on the proposed analytical solution for the transversely isotropic floor rock mass when compared with the numerical simulation, found that numerical solutions align favorably with the analytical solution, even though the amplitude of stress is not captured. From Figure 8a, it is observed that if the analytical solution is applied to the transversely isotropic floor rock mass, horizontal stress increases when x \(\leq\) 47, then it undergoes the region in which there is an obvious decline until the peak stress is reached, corresponding to an overall decrease due to stress applied. For the case of x \(>\) 53, it shows a tendency to increase. In addition, vertical stress shows a sudden decrease and then rebounds, and it can be seen that the changes of shear stress are the exact opposite of vertical stress in Figure 8. Overall, the stress of an analytical solution is slightly greater than that of numerical simulations when x \(\leq\) 47. On the contrary, the value of an analytical solution tends to be less than that of numerical simulations when x \(>\) 53. The reason for this phenomenon may be attributed to the influences of the model mesh. The results confirm the validity of the proposed solution, which can be utilized to estimate the stress distribution under distributed loading in an arbitrary direction. The numerical model utilized in the discussion in this section might have been the same as that on the segment but for the elastic modulus and Poisson's ratio. Take \(E_{1}=E_{2}=3\) GPa, as \(G_{2}\) is related to \(\mu_{1}\), \(\mu_{2}\), \(E_{1}\), and \(E_{2}\), given \(\mu_{1}\)= 0.3 set \(\mu_{1}/\mu_{2}\) as 1.0, 2.0 and 3.0, respectively, and obtain the same depth (z = 1 m) distribution of \(\sigma_{z}\), \(\sigma_{\times}\), and \(\tau_{\times}\), as shown in Figure 9. It is denoted that there is very little difference between them, which indicates the proposed solution is feasible to obtain the stress concentration zone induced by stress redistribution with certain accuracy.