## 3 Uniaxial Compression Test and Particle Flow Numerical Simulation
### Construction of the Particle Flow Numerical Model
For the compaction phase, an ideal plastic model was adopted; thus, we ignored the internal stress variation within the coal specimen due to deformation, as shown by the yellow solid line in Figure 6. According to the linear elasticity, small deformation assumption, the axial strain \(\varepsilon_{\mathrm{z}}\) of the specimen is defined as \(\varepsilon_{\mathrm{z}}=\delta/l_{0}\), where \(\delta\) is the axial displacement at the elastic stage and \(l_{0}\) is the initial height of the specimen. The radial stress \(\sigma_{r}\) and circumferential stress \(\sigma_{\theta}\) of the coal specimen can be expressed as follows:  
\[\sigma_{r}=\frac{E}{1+\mu}\left[\frac{\mu}{1-2\mu}\left(\varepsilon_{r}+ \varepsilon_{\theta}+\frac{\delta}{l_{0}}\right)+\varepsilon_{r}\right], \tag{1}\]  
Figure 7: Schematic of the deformation and stress principles of the polyurea-coated coal specimen. (**a**) Deformation of coal specimen before and after loading; (**b**) Load distribution on coal specimen; (**c**) Stress boundary conditions of coal specimen; (**d**) Force analysis of infinitesimal element of polyurea thin shell. \[\sigma_{\theta}=\frac{E}{1+\mu}\bigg{[}\frac{\mu}{1-2\mu}\bigg{(}\varepsilon_{r}+ \varepsilon_{\theta}+\frac{\delta}{l_{0}}\bigg{)}+\varepsilon_{\theta}\bigg{]}, \tag{2}\]  
where \(E\), \(\mu\), \(\varepsilon_{r}\), and \(\varepsilon_{\theta}\) are the elastic modulus, Poisson's ratio, radial strain, and circumferential strain of the specimen, respectively. Because the circumferential stress of the specimen is zero under axial loading, the following equation can be obtained from Formula (2):  
\[\varepsilon_{\theta}=\frac{\mu}{\mu-1}\bigg{(}\varepsilon_{r}+\frac{\delta}{ l_{0}}\bigg{)}, \tag{3}\]  
Considering that the radial stress of the specimen is \(P_{2}\), Equation (1) can be rewritten as follows:  
\[\varepsilon_{r}=\frac{(1+\mu)P_{2}l_{0}-\mu\delta E}{El_{0}}, \tag{4}\]  
In the thin shell model, the torque and bending moments within the coating are neglected, as well as variations in the stress and displacement along the thickness direction of the coating. The force equilibrium equation and the relationship between force and displacement can be expressed as follows:  
Figure 5:0  
Figure 5:1  
Figure 5:2  
Figure 5:3  
where \(u\), \(v\), and \(w\) are the axial, tangential, and radial displacements of the coating; \(F_{\text{p1}}\) and \(F_{\text{p2}}\) are the tangential and normal (tensile/compressive) forces within the coating; \(F_{\text{T}}\) is the shear force within the coating; \(z\) and \(\beta\) are the axial and tangential coordinates; and \(\mu_{\text{p}}\), \(E_{\text{p}}\), \(m\), and \(r_{0}\) are the Poisson's ratio, elastic modulus, thickness, and initial radius of the coal specimen, respectively. The coating deformation is consistent with that of the specimen's outermost layer; hence, the displacement can be expressed as follows:  
Figure 5:4  
The relationship between the axial stress and axial displacement of the specimen can be expressed as follows:  
Figure 5:5  
The theoretical results are indicated by the red solid line in Figure 5:6 After the elastic stage, the sudden drop in stress is due to the brittle failure of the coal specimen. The longitudinal cracks in Figure 2b indicate that the coal specimen fails when the radial strain exceeds its limit. According to the maximum strain criterion, the maximum strain at failure of the specimen is \(\varepsilon_{r\text{c}}=\mu E^{-1}\sigma_{\text{c}}\). Considering the relationship between radial strain and stress, \(\varepsilon_{r}=E^{-1}(P_{2}-\mu\sigma_{z})\), and Equation (15), we can deduce that when the radial strain reaches the maximum strain that the specimen can withstand the axial strain of the polyurea-coated specimen is \(\varepsilon_{1}\). Considering the material parameters obtained from the experiment and the strain during the compaction stage, we calculated the axial strain at the failure of the specimen to be 1.57%.