## 2 Theoretical Solution to the Ultimate Bearing Capacity of a Completely Rough Layered Foundation
### The Bearing Capacity of the Foundation Caused by the Weight of the Sliding Zone Soil  
At this point, soil cohesion and overloading on both sides of the foundation are not considered, i.e., \(c=0\), \(\varphi_{0}=\tau_{0}=0\). To obtain the most unfavorable passive soil pressure, the starting center point of the logarithmic spiral curve needs to be determined through trial calculation. Due to the difficulty of completing the calculation, a unified assumption of a logarithmic spiral curve is adopted to simplify the calculation. It is assumed that the initial center point of the logarithmic spiral curve is at point \(A\) in Figure 2. Taking \(ACD^{\prime}E\) on the left side of Figure 2 as the research object, the moment of each force on point \(A\) is taken, and the passive earth pressure on the plane \(AC\) is obtained based on the moment balance. (1) Moment of soil gravity on point \(A\) in the passive zone \(AD^{\prime}E\) is:  
\[M(G_{1})=\frac{\gamma_{1}r_{0}^{3}\sin\eta\cos\varphi_{1}\cos(\beta-\eta)}{3 \cos(\varphi_{1}+\eta)}\Big{[}(1-n)e^{\theta\tan\varphi_{1}}+ne^{\theta\tan \varphi_{2}}\Big{]}^{3} \tag{21}\]  
(2) Moment of soil gravity on point \(A\) in the logarithmic spiral curve area \(ACD^{\prime}\) is:  
\[ \begin{array}{ll}M(G_{2})&=(1-n)\bigg{\{}\frac{\gamma_{1}r_{0}^{3}}{3\big{(}1 +9\tan^{2}\varphi_{1}\big{)}}\big{\{}4\sin\varphi_{1}-e^{3\theta\tan\varphi_{ 1}}[3\tan\varphi_{1}\cos(\varphi_{1}+\theta)+\sin(\varphi_{1}+\theta)]\big{\}} \bigg{\}}\\ &+\frac{\gamma_{2}r_{0}^{2}n}{3\big{(}1+9\tan^{2}\varphi_{2}\big{)}}\big{\{}4 \sin\varphi_{2}-e^{3\theta\tan\varphi_{2}}[3\tan\varphi_{2}\cos(\varphi_{2}+ \theta)+\sin(\varphi_{2}+\theta)]\big{\}}\end{array}  \tag{22}\]  
(3) The angle between the radial reaction force \(F\) on the failure surface of the logarithmic spiral curve and the normal line of the action point is \(\varphi_{1}\). According to the properties of the logarithmic spiral curve, the action line of this force passes through the initial center point \(A\) of the curve, so its moment at point \(A\) is zero. (4) Moment of the resultant force \(R\) on the plane \(D^{\prime}E\) on point \(A\) is:  
\[M(R)=-\frac{\gamma_{1}r_{0}^{3}\sin^{3}\eta\cos\varphi_{1}}{6\cos^{3}(\varphi_ {1}+\eta)}\Big{[}(1-n)e^{\theta\tan\varphi_{1}}+ne^{\theta\tan\varphi_{2}} \Big{]}^{3}\sin\theta\tan\theta \tag{23}\]  
According to the moment balance of each force on point \(A\) in the sliding zone \(ACD^{\prime}\) of the logarithmic spiral curve, the passive earth pressure on the \(AC\) surface can be obtained as:  
\[ \begin{array}{ll}P_{p}&=\frac{\gamma_{1}r_{0}^{3}\sin\eta\cos(\beta-\eta)}{ 2\cos(\varphi_{1}+\eta)}\big{[}(1-n)e^{\theta\tan\varphi_{1}}+ne^{\theta\tan \varphi_{2}}\big{]}^{3}\\ &+(1-n)\bigg{\{}\frac{\gamma_{1}r_{0}^{2}}{2\big{(}1+9\tan^{2}\varphi_{1}\big{)} \cos\varphi_{1}}\big{\{}4\sin\varphi_{1}-e^{3\theta\tan\varphi_{1}}[3\tan \varphi_{1}\cos(\varphi_{1}+\theta)+\sin(\varphi_{1}+\theta)]\big{\}}\bigg{\}} \\ &+n\frac{\gamma_{2}r_{0}^{2}}{2\big{(}1+9\tan^{2}\varphi_{2}\big{)}\cos\varphi _{1}}\big{\{}4\sin\varphi_{2}-e^{3\theta\tan\varphi_{2}}[3\tan\varphi_{2} \cos(\varphi_{2}+\theta)+\sin(\varphi_{2}+\theta)]\big{\}}\\ &-\frac{\gamma_{1}r_{0}^{2}\sin^{3}\eta}{4\cos^{3}(\varphi_{1}+\eta)}\big{[}( 1-n)e^{\theta\tan\varphi_{1}}+ne^{\theta\tan\varphi_{2}}\big{]}^{3}\sin\theta \tan\theta\end{array}  \tag{24}\]  
The angle between the line \(AC\) and the horizontal plane is \(\varphi_{1}\). Based on the balance relationship of vertical forces on the triangular wedge \(AB^{\prime}C\) in Figure 5, the ultimate bearing capacity of the foundation \(q_{u}^{{}^{\prime\prime}}\) is calculated, and Equation (24) is substituted and organized as follows:  
\[{q^{{}^{\prime\prime}}}_{u}=\frac{\gamma_{1}B}{2}N_{\gamma}+\frac{\gamma_{2}B }{2}{N^{\prime}}_{\gamma} \tag{25}\]\[N_{\gamma}=\frac{\tan\varphi_{1}}{2\cos\varphi_{1}}\left\{