## 6 The Derived Coefficients of Earth Pressures  
The coefficient of earth pressure at rest will be obtained after selecting the proper value for \(m\) and defining the mobilized cohesion, \(c_{m}\). For selecting the proper \(m\), the reasoning is as follows. [PERSON]'s [11] assumed arbitrarily the following parabolic distribution  
\[\tau_{xz}=\tau_{xz,OB}\left(\frac{x}{x_{OB}}\right)^{2}, \tag{13}\]  
resulting in  
\[K_{o}=\frac{1+\frac{2}{3}\sin\varphi^{\prime}}{1+\sin\varphi^{\prime}}(1-\sin \varphi^{\prime}). \tag{14}\]  
Apparently, during his \"heroic\" effort in 1944 with no electronic means, [PERSON] must have tried the simpler and more rational linear distribution  
\[\tau_{xz}=\tau_{xz,OB}\left(\frac{x}{x_{OB}}\right). \tag{15}\]  
Besides, the distribution of shear stresses in Zone I follows a linear pattern (see Equation (11) in Appendix A); this distribution, however, gives singularity leading to \(K_{o}=1\) (independently of \(\varphi^{\prime}\) value), which is not the point for sand heaps. On the other hand, from the _purely theoretical point of view_, the very simple \(1-\sin\varphi^{\prime}\) formula, works ideally for the two extreme values of \(\varphi^{\prime}\), where for \(\varphi^{\prime}\)= 0\({}^{\rm o}\) it gives \(K_{o}=1\) referring to hydrostatic conditions and for \(\varphi^{\prime}\)= 90\({}^{\rm o}\) it gives \(K_{o}=0\) referring to a frictional material that can stand vertically without support, thus, exerting no lateral pressure. These extreme cases are enough evidence that the correct expression for the coefficient of earth pressure at rest is the \(K_{o}=1-\sin\varphi^{\prime}\). Thus, for \(m=1\) (corresponding to linear distribution of shear stresses in Zone II) Equation (8) gives the proposed generalized coefficient of earth pressure at rest  
\[K_{oe}^{c-\varphi}=(1-\sin\varphi^{\prime})\bigg{(}1+\frac{k_{h}}{1-k_{\rm p}} \tan\varphi^{\prime}\bigg{)}-\frac{1}{1-k_{\rm p}}\frac{2c_{m}}{\gamma z}\tan \left(45^{\circ}-\frac{\varphi^{\prime}}{2}\right), \tag{16}\]  
with  
\[\sigma_{\circ}=K_{oe}^{c-\varphi}(1-k_{\rm p})\gamma z. \tag{17}\]  
For \(k_{h}=k_{\rm p}=0\) (static conditions), the coefficient of earth pressure at rest simplifies to  
\[K_{oe}^{c-\varphi}=(1-\sin\varphi^{\prime})-\frac{2c_{m}}{\gamma z}\tan\left( 45^{\circ}-\frac{\varphi^{\prime}}{2}\right). \tag{18}\]  
For \(m\to\infty\), it is \(\lim\limits_{m\to\infty}\frac{m-1}{m+1}=1\) and \(\lim\limits_{m\to\infty}\frac{m+\sin\varphi^{\prime}}{m+1}=1\), thus, the generalized coefficient of active earth pressure is obtained:  
\[K_{ae}^{c-\varphi}=\frac{1-\sin\varphi^{\prime}}{1+\sin\varphi^{\prime}} \bigg{(}1+2\frac{k_{h}}{1-k_{\rm p}}\tan\varphi^{\prime}\bigg{)}-\frac{1}{1- k_{\rm p}}\frac{2c_{m}}{\gamma z}\tan\left(45^{\circ}-\frac{\varphi^{\prime}}{2} \right), \tag{19}\]  
with  
\[\sigma_{\alpha}=K_{ax}^{c-\varphi}(1-k_{\rm p})\gamma z. \tag{20}\]  
Additionally, if \(k_{h}=k_{\rm p}=0\), it is \(c_{m}=c^{\prime}\) (see Section 8.2). Thus,  
\[K_{ae}^{c-\varphi}=\frac{1-\sin\varphi^{\prime}}{1+\sin\varphi^{\prime}}- \frac{2c^{\prime}}{\gamma z}\tan\left(45^{\circ}-\frac{\varphi^{\prime}}{2} \right), \tag{21}\]  
with  
\[\sigma_{\alpha}=K_{ae}^{c-\varphi}\gamma z=\frac{1-\sin\varphi^{\prime}}{1+ \sin\varphi^{\prime}}\gamma z-2c^{\prime}\tan\left(45^{\circ}-\frac{\varphi^{ \prime}}{2}\right), \tag{22}\]  
that is, [PERSON]'s [4] solution for the active state. Apparently, this is a compelling evidence for the validity of Equation (8) (and of course, Equation (7a)). \(K_{pe}^{c-\varphi}\) can be obtained indirectly by inverting the signs of shear strength parameters in Equation (19) and that, due to the \"symmetry\" between the active and the passive earth pressure coefficient in the case of smooth, not battered walls retaining soil with zero backslope angle (that is, [PERSON]'s conditions where \(\theta=45^{\circ}+\varphi^{\prime}/2\) or \(45^{\circ}-\varphi^{\prime}/2\) for the active and the passive state, respectively).