# Shaft Wall Damage to High-Depth Inclined Ore Passes under Impact Wear Behavior
## 2 Shaft Wall Damage Model for High-Depth Inclined Ore Passes
### Theoretical Solution  
#### 2.3.1 Ore Indentation Depth \(\delta\)  
Based on the contact mechanics theory, the deformation process resulting from the ore colliding with the shaft wall can be roughly divided into two stages: 1 the elastic deformation of the shaft wall occurs at the initial stage of contact between the ore and the shaft wall; 2 when the deformation of the shaft wall exceeds the elastic limit and reaches the initial yield state of the shaft wall rock mass, plastic deformation of the shaft wall occurs. (1) Elastic deformation stage  
Due to the low velocity when the ore collides with the shaft wall, the contact between the ore and the shaft wall can be regarded as a quasi-static contact [25]. Based on the quasi-static contact mechanics theory [26], the normal impact force \(F_{\mathrm{n}}\) during elastic deformation can be expressed as follows:  
\[F_{\mathrm{n}}=\frac{4}{3}E^{*}R^{\frac{1}{2}}\delta^{\frac{3}{2}} \tag{1}\]  
where \(R\) is the radius of the spherical ore; \(P\) is defined as the ore block size, with \(P=2R\); and \(E^{*}\) is the equivalent elastic modulus, satisfying  
\[\frac{1}{E^{*}}=\frac{1-\
u_{1}}{E_{1}}+\frac{1-\
u_{2}}{E_{2}} \tag{2}\]  
where \(E_{1}\), \(\
u_{1}\), \(E_{2}\), and \(\
u_{2}\) are the elastic modulus and the Poisson ratio of the ore and the shaft wall, respectively. Based on [PERSON]'s second law of motion, the condition for elastic deformation is satisfied that  
\[-m\frac{d^{2}\delta}{dt^{2}}=\frac{4}{3}E^{*}R^{\frac{1}{2}}\delta^{\frac{3}{2}} \tag{3}\]  
where \(m\) is the mass of the spherical ore, with \(m=4\pi\rho R^{3}/3\); \(\rho\) is the density of the ore; and \(d^{2}\delta/dt^{2}\) is the ore's normal acceleration at collision. Combined with the initial collision conditions, Equation (4) is obtained via integrating Equation (3) as follows:  
\[\frac{m}{2}[{v_{n}}^{2}-(\frac{d\delta}{dt})^{2}]=\frac{8}{15}E^{*}R^{\frac{1 }{2}}\delta^{\frac{5}{2}} \tag{4}\]  
where \(v_{n}\) is the ore's normal velocity before collision, and \(d\delta/dt\) is the ore's normal velocity at collision. From the general collision process, the ore's normal impact velocity decreases continuously from the time when the ore starts to contact the shaft wall to the time when the ore indentation depth reaches the maximum. When \(d\delta/dt=0\), the maximum ore indentation depth \(\delta_{\mathrm{m}}\) in the elastic stage can be solved as follows:  
\[\delta_{m}=(\frac{15m{v_{n}}^{2}}{16E^{*}R^{\frac{1}{2}}}) \tag{5}\]  
(2) Plastic deformation stage  
Based on the elastic-plastic contact mechanics theory [27], when the shaft wall begins to deform plastically, it is satisfied that  
\[\sigma_{n}=\eta\sigma_{q} \tag{6}\]where \(\sigma_{\rm n}\) is the normal contact stress between the ore and the shaft wall; \(\sigma_{\rm q}\) is the initial yield strength of the rock mass of the shaft wall; and \(\eta\) is a dimensionless factor, with \(\eta=1\) when the rock mass of the shaft wall is initially yielding. From Equation (6), the normal impact force \(F_{\rm n}\) during the plastic deformation of the shaft wall can be written as follows:  
\[F_{n}^{*}=\pi a^{*2}\sigma_{n} \tag{7}\]  
where \(a^{*}\) is the contact radius between the ore and the shaft wall during plastic deformation, and \(a^{*}=a_{\rm q}\) when the rock mass of the shaft wall is initially yielding. The relation between the contact radius \(a\) and the ore indentation depth \(\delta\) is expressed as follows:  
\[a^{2}=R\delta \tag{8}\]  
Combining Equations (1), (7) and (8), the ore indentation depth \(\delta_{\rm q}\) at the initial yield can be solved as follows:  
\[\delta_{q}=\frac{9\pi^{2}\sigma_{q}{}^{2}R}{16E^{*2}} \tag{9}\]  
According to the law of conservation of energy, the ore's normal impact kinetic energy is transformed into the elastic deformation energy of the shaft wall and the energy consumed in the plastic deformation.