# Calculation of Nonlimit Active Earth Pressure against Rigid Retaining Wall Rotating about Base
## 4 Derivation of the Analytical Solution for Active Earth Pressure
### Equilibrium Control Equation for Thin Layer Cells  
The equilibrium control equation can be obtained from the static equilibrium condition of the thin-layer unit, which can be expressed as  
\[F_{x}^{DD^{\prime}}-F_{x}^{DE}+F_{x}^{D^{\prime}E^{\prime}}+F_{x}^{EE^{\prime }}=0 \tag{45}\]  
\[F_{y}^{DD^{\prime}}-F_{y}^{DE}+F_{y}^{D^{\prime}E^{\prime}}+F_{y}^{EE^{\prime }}-dA=0 \tag{46}\]  
The forces applied to the unit are then substituted into Equations (45) and (46) to simplify them and obtain the controlling equation for the circularly curved thin layer unit as follows:  
\[\begin{array}{l}a_{1}\sigma_{1m}dy-\frac{a_{2}a_{3}}{2}\sigma_{1}^{a}dy+ \frac{a_{2}a_{3}}{2}(H-y-\frac{dy}{2})d\sigma_{1}^{a}-\frac{\gamma a_{2}^{2} a_{4}}{2}(H-y)\\ -\frac{\gamma a_{2}^{2}\gamma dy}{2}(H-y-\frac{dy}{2})=0\end{array} \tag{47}\]where  
\[a_{1}=(\tan\varphi-\tan\alpha_{s})\tan\delta-(1+\tan\varphi\tan\alpha_{s}) \tag{48}\]  
\[a_{2}=\frac{\cos\alpha_{s}}{\cos(\theta_{1}-\alpha_{s})\cdot\sin(\frac{\theta_{ 0}-\theta_{s}}{2})} \tag{49}\]  
\[a_{3}=b_{2}(\tan\varphi-\tan\alpha_{s})-b_{1}(1+\tan\varphi\tan\alpha_{s}) \tag{50}\]  
\[a_{4}=\frac{b_{1}^{2}}{2}(1+\tan\varphi\tan\alpha_{s})+b_{3}(\tan\varphi-\tan \alpha_{s}) \tag{51}\]  
\[a_{5}=a_{2}\sin\theta_{a}(\theta_{b}-\theta_{a})(\tan\varphi-\tan\alpha_{s}) \tag{52}\]  
Equation (47) consists of two unknowns, \(\sigma_{\rm 1m}\) and \(\sigma_{1}^{a}\), which are determined using Equation (6).  
\[\sigma_{\rm 1m}=a_{6}\sigma_{1}^{a} \tag{53}\]  
where  
\[a_{6}=\cos^{2}\theta_{a}+\frac{\sin^{2}\theta_{a}}{N} \tag{54}\]  
By substituting Equations (53) and (54) into Equation (47), a differential equation for the major principal stress \(\sigma_{1}^{a}\) can be obtained.  
\[\frac{\mathrm{d}\sigma_{1}^{a}}{dy}-\frac{f_{1}}{H-y}\sigma_{1}^{a}-\frac{ \gamma}{f_{2}}=0 \tag{55}\]  
where  
\[f_{1}=1-\frac{2a_{1}a_{6}}{a_{2}a_{3}} \tag{56}\]  
\[f_{2}=\frac{a_{2}a_{3}}{a_{2}^{2}a_{4}+a_{5}} \tag{57}\]  
Solving differentiated Equation (55) yields  
\[\sigma_{1}^{a}=C(y-H)^{-f_{1}}+\frac{\gamma(y-H)}{(1+f_{1})f_{2}} \tag{58}\]  
\[\sigma_{1m}=a_{6}C(y-H)^{-f_{1}}+\frac{a_{6}\gamma(y-H)}{(1+f_{1})f_{2}} \tag{59}\]  
where \(C\) is a constant to be determined. According to the boundary condition, \(\sigma_{\rm 1m}=0\) at \(y=0\), \(C\) can be solved.  
\[C=-\frac{\gamma(-H)^{1+f_{1}}}{(1+f_{1})f_{2}} \tag{60}\]  
The active earth pressure strength \(\sigma_{\rm 1m}\) can be calculated by substituting Equation (60) into (59), which yields  
\[\sigma_{\rm 1m}=K\gamma\left[H(1-\frac{y}{H})^{-f_{1}}+(y-H)\right] \tag{61}\]  
where  
\[K=\frac{a_{6}}{(1+f_{1})f_{2}} \tag{62}\]  
Equation (61) is the formula for the active earth pressure derived in this study.  
The combined active earth pressure force \(E_{a}\) on the retaining wall can be described as  
\[E_{a}=\int_{0}^{H}\sigma_{\rm 1m}dy \tag{63}\]  
The active earth pressure bending moment \(M\) of the retaining wall can be expressed as \[M=\int_{0}^{H}\sigma_{1m}(H-y)dy \tag{64}\]  
The point of application of the resultant force on the retaining wall is \(h_{p}\).  
\[h_{p}=\frac{M}{E_{a}}=\frac{2(1-f_{1})}{3(2-f_{1})}H \tag{65}\]