## 2 Theoretical Research  
### The Rockfall Impact Force on Soil  
For buried pipelines under impact loads from rockfalls, the mathematical formula for calculating the impact force on the soil can be derived. When the rockfall impacts the soil (pebble or clay) with a certain speed, it may sink to a certain depth and the velocity rapidly decreases. During this process, because the stiffness of the soil is smaller than that of the rockfall, the rockfall generally does not rebound slightly and, furthermore, the impact process is very short, which causes a larger impact force. The physical model of a buried long pipeline under a rockfall impact load can be simplified as shown in Figure 1, and the curve of the impact force (\(F\)) versus time (\(t\)) during this impact process is shown in Figure 2 based on the Hertz collision theory and relevant references [31], in which \(F_{\text{max}}\) is the maximum impact force and \(t_{1}\) is time when \(F\) reaches to \(F_{\text{max}}\).  
Figure 1: The physical model of a buried long pipeline under an impact load.  
Figure 2: Impact force (\(F\)) versus time (\(t\)) curve.  
Three main calculation methods of \(F_{\rm max}\) are the Hertz collision theory, the energy theory, and the inelastic collision method. The velocity of rockfall may not be perpendicular to the soil, but this experiment mainly focuses on the case where the rockfall impacts the soil vertically based on the Hertz collision theory. It is essential for the Hertz collision theory to make the following assumptions: (1) the rockfall is a rigid body where no deformation occurs during the collision; (2) the rockfall is not separated from the soil during the collision, and the friction between them is not considered; (3) based on a linear elastic half-space theory, the maximum displacement of the soil is in the elastic stage; (4) only the first impulse action is taken into consideration. Based on the above assumptions, the motion equation of \(F_{\rm max}\) on soil can be written as follows:  
\[F_{\rm max}=K^{\frac{2}{5}}\Bigg{[}\frac{5}{4}\
u_{0}^{2}\frac{m_{1}m_{2}}{m_{ 1}+m_{2}}\Bigg{]}^{\frac{3}{5}}, \tag{1}\]  
where \(m_{1}\) and \(m_{2}\) are the masses of the two elastic spheres (rockfall and soil), \(\
u_{0}\) is the instantaneous velocity before the collision, and \(K\) is a stiffness coefficient determined by the following equations:  
\[K=\frac{4}{3\pi}\sqrt{\frac{r_{1}r_{2}}{r_{1}+r_{2}}}\frac{1}{C_{1}+C_{2}}, \tag{2}\]  
\[C_{1}=\frac{1-\mu_{1}^{2}}{\pi E_{1}}, \tag{3}\]  
\[C_{2}=\frac{1-\mu_{2}^{2}}{\pi E_{2}}, \tag{4}\]  
where \(r_{1}\) and \(r_{2}\) are the radii of the two elastic spheres (rockfall and soil), \(E_{1}\) and \(E_{2}\) are the elastic moduli of the two elastic spheres, and \(\mu_{1}\) and \(\mu_{2}\) are the Poisson ratios of the two elastic spheres. When the rockfall impacts the soil, it can be considered that \(E_{1}\) approaches infinity since the stiffness of the rockfall is much larger than that of the soil. The values of \(m_{2}\) and \(r_{2}\) approach infinity since the soil is regarded as a semi-linear elastomer. Therefore, the following equations can be derived:  
\[C_{1}=0, \tag{5}\]  
\[K=\frac{4}{3}\frac{E_{2}}{1-\mu_{2}^{2}}\sqrt{r_{1}}. \tag{6}\]  
Substituting Equation (6) into Equation (1) gives \(F_{\rm max}\) as  
\[F_{\rm max}=1.167\Bigg{[}\frac{E_{2}}{1-\mu_{2}^{2}}\Bigg{]}^{\frac{2}{5}}m_{ 1}^{\frac{2}{3}}\rho_{1}^{-\frac{1}{15}}\
u_{0}^{\frac{6}{5}}, \tag{7}\]  
where \(m_{1}=\frac{4}{3}\pi\
u_{1}^{3}\rho_{1}\), and \(\rho_{1}\) is the density of the rockfall. Assuming that the pressure on the contact surface between the rockfall and the soil is uniformly distributed, then the impact pressure (\(p_{\rm max}\)) can be expressed as  
\[p_{\rm max}=\frac{F_{\rm max}}{A}, \tag{8}\]  
where \(A\) represents the contact area between the rockfall and the soil.