# Lateral Loading of a Rock-Socketed Pile Using the Strain Wedge Model Based on Hoek-Brown Criterion
## 2 Parameters Study
### Modification of Hoek-Brown Based on Mohr-Coulomb  
The concept of the instantaneous angle of friction \(\rho\) is expressed using the following formula [18]:  
\[\sin\rho=\frac{dq}{dp} \tag{8}\]  
Combing the Mohr-Coulomb failure envelope \(\tau=\tau(\sigma)\) and the instantaneous angle \(\rho\), shear and normal stress can be expressed as:  
\[\tau=q\cos\rho \tag{9}\]  
\[\sigma=p-q\sin\rho \tag{10}\]  
Using Equations (7) and (10), the following expressions are obtained:  
\[(q*)^{2}+2q*=2\times(p*+\zeta) \tag{11}\]  
Differential on both sides:  
\[2(q*+1)dq*=2\times dp* \tag{12}\]  
Then:  
\[\frac{dq}{dp}=\frac{dq*}{dp*}=\frac{1}{q*+1}=\frac{\beta}{q+\beta} \tag{13}\]  
Therefore:  
\[\frac{q}{\beta}=\frac{1-\sin\rho}{\sin\rho},\frac{p}{\beta}+\zeta=\frac{\cot^ {2}\rho}{2} \tag{14}\]  
Replacing Equation (14) in Equations (9) and (10) the following expression is obtained:  
\[\frac{\tau}{\beta}=\frac{1-\sin\rho}{1-\cos 2\rho}\sin 2\rho,\;\frac{\sigma}{ \beta}+\zeta=\frac{1-\sin\rho}{1-\cos 2\rho}(\cos 2\rho+\sin\rho) \tag{15}\]  
\[\tau^{*}=\frac{1-\sin\rho}{1-\cos 2\rho}\sin 2\rho,\;\sigma_{o}^{*}=\sigma^{*} +\zeta=\frac{1-\sin\rho}{1-\cos 2\rho}(\cos 2\rho+\sin\rho) \tag{16}\]  
\[2\sin^{3}\rho-(2\sigma_{o}^{*}+3)\sin^{2}\rho+1=0 \tag{17}\]  
\[\sin\rho=\left(1+\frac{2}{3}\sigma_{o}^{*}\right)\left[\sin(\lambda-30)+\frac {1}{2}\right] \tag{18}\]  
where \(\lambda=\frac{2}{3}\sin^{-1}\left[\left(1+\frac{2}{3}\sigma_{o}^{*}\right)^{ -3/2}\right]\).  
It shows that the instantaneous friction angle of the rock is a single-valued function of the confining pressure. According to the Hoek-Brown failure criterion, the smaller the confining pressure, the lower the shear strength of the rock mass. The relationship between shear strength and principal stress is nonlinear under low confining pressure [17]. However, Equation (18) is too complex and implicit to apply; therefore, it needs to be simplified.  
It is possible to obtain the following expression from Equation (8)  
\[\sin\rho=\frac{\beta}{q+\beta} \tag{19}\]  
Replacing \(q=\frac{\sigma_{1}-\sigma_{3}}{2}\) and \(\beta=\frac{m_{\rm i}\sigma_{\rm i}}{8}\) in Equation (19) can obtain the following expression:  
\[\sin\rho=\frac{1}{\sqrt{2\left(\frac{\sigma_{3}}{\beta}+\zeta\right)}+1} \tag{20}\]  
\[\rho=\sin^{-1}\frac{1}{\sqrt{2\left(\frac{\sigma_{3}}{\beta}+\zeta\right)}+1}= \sin^{-1}\frac{1}{\sqrt{2\sigma_{o}^{\rm s}}+1} \tag{21}\]  
The expression is simple in form, and the relationship between the instantaneous angle of friction and the confining pressure can be clearly seen.  
when \(\sigma_{3}=0\):  
\[\rho=\sin^{-1}\frac{1}{\sqrt{2\zeta}+1}=\sin^{-1}\frac{1}{\sqrt{\frac{16}{(m_ {\rm i})^{2}}}+1}=\sin^{-1}\frac{1}{\frac{4}{m_{\rm i}}+1}\]  
For hard rock, \(m_{\rm i}\approx 20\)[14],  
\[\rho=\sin^{-1}\frac{1}{\frac{4}{m_{\rm i}}+1}\approx\sin^{-1}\frac{1}{\frac{4 }{20}+1}=\sin^{-1}\frac{1}{1.2}\approx 56.44^{\circ}\]  
The angle between the fracture surface and the principal stress is \(\Theta=45^{\circ}-\rho/2\)\(=16.78^{\circ}\).  
It can be seen from Equation (21) that the friction of rock is changed with confining pressure and material constants of intact rock \(m_{\rm i}\).  
Using Mohr-Coulomb \(\tau=c+\sigma\tan\rho\) and combining Equations (9) and (10) can easily obtain the following expression:  
\[\begin{array}{l}c=\tau-\sigma\tan\rho=q\cos\rho-p\tan\rho+q\sin\rho\tan\rho \\ =q\frac{1}{\cos\rho}-p\tan\rho\end{array} \tag{22}\]  
\[\frac{c}{\beta}=\frac{q}{\beta}\frac{1}{\cos\rho}-\frac{p}{\beta}\tan\rho= \frac{(\sin\rho-1)^{2}}{\sin 2\rho}+\zeta\tan\rho \tag{23}\]  
Therefore:  
\[c=