# Solution for Active and Passive Earth Pressure on Rigid Retaining Walls with Narrow Backfill
## 2 Theoretical Analysis  
### Active Limit Earth Pressure  
Retaining walls adjacent to rigid geological/structural boundaries retain a finite soil with constrained lateral dimensions. In accordance with the results of previous experimental studies [32; 37; 38; 39], this study assumes that upon the destruction of the soil behind the wall, the slip surface manifests as a folded surface, with the slip soil forming a trapezoidal wedge. The force analysis of the slip wedge in the active limit state is shown in Figure 1.  
The reaction force (\(E_{x1}\)) exerted by the retaining wall forms an angle with the horizontal direction equal to the wall-soil friction angle (\(\delta\)). The reaction force exerted on the slip wedge by the soil beneath the slip surface is denoted as \(R\) and forms an angle with respect to the normal to the slip surface, which is the internal friction angle (\(\varphi\)). \(W\) represents the weight of the slip wedge. For clay, the cohesion (\(c\)) is uniformly distributed along the slip surface, generating a resultant force denoted as \(C\) that forms an angle with the horizontal direction equal to the rupture angle \(\theta\), with an effective range of (\(50^{\circ},80^{\circ}\)) in the active state. In cohesionless soil, there is no bonding between soil particles, the soil and rock (or structure) remain in contact under the active state. [PERSON] et al. [38] postulated that the earth pressure on both sides above the slip surface inflection point was equal, i.e., the reaction coefficient \(a_{s}=1\). This conclusion is supported by the model test results of [PERSON] et al. [40], and the force analysis for this case is presented in Figure 1a. When the backfill consists of clay, partial separation occurs between the soil and rock (or structure) under the active limit state due to cohesive bonds between soil particles. Consequently, this paper assumes that the earth pressure on the rock (or structure) above the slip surface inflection point is half of that on the left retaining wall, i.e., \(a_{s}=0.5\), with the corresponding stress analysis presented in Figure 1b. To simplify the calculation, the earth pressure is assumed to be triangularly distributed. The counterforce of the rock on the slip wedge is denoted as \(S\). The expressions of \(W\), \(C\), and \(S\) of the slip wedge are as follows:  
\[S=\frac{a_{s}E_{x1}(H-L\tan\theta)^{2}}{H^{2}}\quad W=\gamma(HL-\frac{1}{2}L^ {2}\tan\theta)\quad C=cL/\cos\theta \tag{1}\]  
From the comparison of Figure 1a,b, it can be seen that the force on the slip wedge of cohesionless soil is a special case of the force on the clay slip wedge. Specifically, this occurs when the cohesion \(c=0\) and the reaction coefficient \(a_{s}=1\). Therefore, according to the static equilibrium conditions of the clay slip wedge in the horizontal and vertical directions under the active state, Equation (2) can be obtained. \[ \begin{split} R\sin(\theta-\varphi)+S\cos\delta=C\cos\theta+E_{x1 }\cos\delta\\ R\cos(\theta-\varphi)+S\sin\delta+C\sin\theta+E_{x1}\sin\delta=W \end{split}  \tag{2}\]  
Solving both equations in Equation (2) simultaneously yields Equation (3). \[E_{x1}=\frac{WH^{2}\tan(\theta-\varphi)-CH^{2}\cos\theta-CH^{2}\sin\theta\tan (\theta-\varphi)}{H^{2}\cos\delta-a_{s}(H-L\tan\theta)^{2}\cos\delta+a_{s}(H- L\tan\theta)^{2}\sin\delta\tan(\theta-\varphi)+H^{2}\sin\delta\tan(\theta- \varphi)} \tag{3}\]  
From Equation (3), it can be seen that \(E_{x1}\) is a univariate function of the rupture angle (\(\theta\)). By setting \(\psi=\tan\theta\), Equation (3) can be simplified to Equation (4). \[E_{x1}=\frac{A_{1}\psi^{2}+B_{1}\psi+C_{1}}{D_{1}\psi^{3}+E_{1}\psi^{2}+F_{1} \psi+G_{1}} \tag{4}\]  
Figure 1: Force analysis of the slip wedge in the active state: (**a**) cohesionless soil; (**b**) clay. In which:  
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