# Three-Dimensional Theoretical Model for Soil Arching with Inclined Slip Surfaces
## 3 Derivation of 3D Theoretical Model for Soil Arching with Inclined Slip Surfaces  
This study expands [PERSON] and [PERSON] [PERSON] and [PERSON] (2014) 2D theoretical model for soil arching with inclined slip surfaces to 3D conditions. Figure 3 shows the assumed inclined slip surfaces along with soil differential element subjected to stresses. It is also assumed that the soil is homogeneous, isotropic, and semi-infinite. The 3D model for soil arching with inclined slip surfaces is shown in Figure 4. The slip surfaces incline outside along excavation loosening zone. Assuming that the failure of  
Figure 3: Inclined slip surfaces along with soil differential element subjected to stresses in 3D condition. slip surface follows the Mohr-Coulomb failure criterion, a vertical equilibrium equation of the soil differential element can be obtained according to the vertical force balance. \[ \begin{array}{l}\gamma\left[BL+\frac{2(B+L)(H-z)}{\tan a}+\frac{4(H-z)^{2}}{\tan a }\right]\mathrm{d}z=\frac{2(B+L)(H-z)}{\tan a}\mathrm{d}\sigma_{\mathrm{v}}\\ +\frac{4(H-z)^{2}}{\tan^{2}\mathrm{d}\sigma_{\mathrm{v}}}\mathrm{-}\frac{2(B+L )\sigma_{\mathrm{v}}}{\tan a}\mathrm{d}z-\frac{8(H-z)\sigma_{\mathrm{v}}}{ \tan^{2}\mathrm{d}z}\mathrm{d}z+BL\mathrm{d}\sigma_{\mathrm{v}}\\ +\frac{2\sigma_{\mathrm{mw}}}{\tan a}\left(B+\frac{2(H-z)}{\tan a}-\frac{\mathrm{ d}z}{\tan a}\right)\mathrm{d}z+2c\left(B+\frac{2(H-z)}{\tan a}-\frac{\mathrm{d}z}{\tan a }\right)\mathrm{d}z\\ +\frac{2\sigma_{\mathrm{nl}}}{\tan a}\left(L+\frac{2(H-z)}{\tan a}-\frac{ \mathrm{d}z}{\tan a}\right)\mathrm{d}z+2c\left(L+\frac{2(H-z)}{\tan a}-\frac{ \mathrm{d}z}{\tan a}\right)\mathrm{d}z\\ +2\sigma_{\mathrm{nw}}\tan\varphi\left(B+\frac{2(H-z)}{\tan a}-\frac{\mathrm{ d}z}{\tan a}\right)\mathrm{d}z\\ +2\sigma_{\mathrm{nl}}\tan\varphi\left(L+\frac{2(H-z)}{\tan a}-\frac{\mathrm{ d}z}{\tan a}\right)\mathrm{d}z\end{array}  \tag{10}\]  
where \(\gamma\) is the unit weight of soil, \(c\) is the cohesiveness of soil, \(\varphi\) is the internal friction angle of soil, \(B\) is the width of the loose zone induced by excavation, \(H\) is the depth of loosening soil, \(L\) is the length of the loose zone induced by excavation, \(\alpha\) is the angle between slip surface and horizontal direction, \(z\) is the depth of the soil differential element, \(\sigma_{\mathrm{v}}\) is vertical stress, \(\sigma_{\mathrm{nw}}\) is normal stress perpendicular to the width direction, \(\sigma_{\mathrm{nl}}\) is the normal stress perpendicular to the length direction. According to the equilibrium conditions at both ends of the differential element, the following can be obtained:  
\[\sigma_{\mathrm{nw}}=\sigma_{\mathrm{v}}\left(\cos^{2}\!\alpha+K_{\mathrm{w}} \!\sin^{2}\!\alpha\right) \tag{11}\]  
\[\sigma_{\mathrm{nl}}=\sigma_{\mathrm{v}}\left(\cos^{2}\!\alpha+K_{\mathrm{l}} \!\sin^{2}\!\alpha\right) \tag{12}\]  
where \(K_{\mathrm{w}}\) is the coefficient of earth pressure perpendicular to the width direction, \(K_{\mathrm{l}}\) is the coefficient of earth pressure perpendicular to the length direction. Substitute Equations (11) and (12) into Equation (10). By simplifying and ignoring the second-order infinitesimal, the ordinary differential equation of vertical force on soil differential element can be obtained. Figure 4: Forces acting on the soil differential element in 3D condition. \[\gamma\bigg{[}BL+\frac{2(B+L)(H-z)}{\tan\alpha}+\frac{4(H-z)^{2}}{\tan \alpha}\bigg{]}-2c\Big{(}B+L+\frac{4(H-z)}{\tan\alpha}\Big{)} \tag{13}\] \[-\sigma_{\rm v}\left\{ \begin{array}{l}\frac{2(\cos^{2}\!\alpha+K_ {\rm v}\sin^{2}\!\alpha)}{\tan\alpha}\Big{(}B+\frac{2(H-z)}{\tan\alpha}\Big{)} \\ +\frac{2\big{(}\cos^{2}\!\alpha+K_{\rm v}\sin^{2}\!\alpha)}{\tan\alpha}\Big{(} L+\frac{2(H-z)}{\tan\alpha}\Big{)}\\ +2\big{(}\cos^{2}\!\alpha+K_{\rm v}\sin^{2}\!\alpha)\tan\varphi\Big{(}B+\frac {2(H-z)}{\tan\alpha}\Big{)}\\ +2\big{(}\cos^{2}\!\alpha+K_{\rm v}\sin^{2}\!\alpha\big{)}\tan\varphi\Big{(} L+\frac{2(H-z)}{\tan\alpha}\Big{)}\\ -\frac{2(B+L)}{\tan\alpha}-\frac{8(H-z)}{\tan^{2}\alpha}\end{array}