# Evaluating Pile-Supported Embankment Considering the Soil Anisotropy Effect
## 2 Description of the Model
### Anisotropic Plasticity Theory  
ABAQUS uses the plastic potential function expression proposed by [PERSON] and [PERSON] [47]. It is hyperbolic on the meridian plane and elliptical on the \(\pi\) plane. As shown in Figure 2, the plastic potential surface is completely smooth, avoiding the singularity of the numerical integration at the corner of the Mohr-Coulomb yield function (Equation (2)). The Mohr-Coulomb correlation yield function is proposed in Appendix A to derive the vector \(a\).  
\[G=\sqrt{\left(\varepsilon c|_{0}\tan\Psi\right)^{2}+\left(R_{mw}q\right)^{2}}-p \tan\Psi=0 \tag{2}\]  
In Equation (2), \(\varepsilon\) represents the meridian eccentricity, which governs the resemblance between the hyperbolic shape of the plastic potential function on the meridian plane and its asymptote. The dilatancy angle, denoted by \(\psi\), and the initial cohesion \(c\,|\,0\) are also defined. The value of \(c\,|\,0\) signifies the cohesion prior to any plastic deformation, and in the absence of hardening or softening, it remains constant. The triple symmetric elliptic form of the plastic potential function on the partial plane is denoted by \(R_{mw}(\theta,e)\) and can be expressed using Equations (3) and (4).  
\[R_{mw}(\theta,e)=\frac{4(1-e^{2})\cos^{2}\theta+\left(2e-1\right)^{2}}{2(1-e^{ 2})\cos\theta+\left(2e-1\right)\sqrt{4(1-e^{2})\cos^{2}\theta+5e^{2}-4e}}R_{mc} (\frac{\pi}{3},\varphi) \tag{3}\]  
\[R_{mc}(\frac{\pi}{3},\varphi)=\frac{3-\sin\varphi}{6\cos\varphi} \tag{4}\]  
Figure 1: Iteration diagram of the backward Euler algorithm.  
Figure 2: Mohr-Coulomb plastic potential surface. (**a**) Plastic potential surface on the meridian plane; (**b**) The plastic potential surface on the deviatoric plane.  
In Equation (4), \(\varphi\) represents the friction angle of the soil, while \(e\) refers to the eccentricity on the \(\pi\) plane. The eccentricity \(e\) governs the shape of the plastic potential surface within the range of \(\theta\) = 0 to \(\pi/3\) on the \(\pi\) plane, as depicted in (Equation (5)).  
\[e=\frac{3-\sin\varphi}{3+\sin\varphi} \tag{5}\]  
Let the vector \(b\) denote the derivative of the plastic potential function \(G\) to the stress (Equation (6)).  
\[b=\frac{\partial G}{\partial\sigma}=C_{1}\frac{\partial p}{\partial\sigma}+C_ {2}\frac{\partial q}{\partial\sigma}+C_{3}\frac{\partial\theta}{\partial\sigma} \tag{6}\]  
where  
\[C_{1}=\frac{\partial G}{\partial p},\;C_{2}=\frac{\partial G}{\partial q},\;C _{3}=\frac{\partial G}{\partial\theta}\]  
Specifically expressed as  
\[C_{1}=-\tan\Psi \tag{7}\]  
\[C_{2}=\frac{q.R_{muv}^{2}}{\sqrt{\left(\varepsilon c|_{0}\tan\Psi\right)^{2}+ \left(R_{muv}q\right)^{2}}} \tag{8}\]  
\[C_{3}=\frac{q^{2}.R_{muv}}{\sqrt{\left(\varepsilon c|_{0}\tan\Psi\right)^{2}+ \left(R_{muv}q\right)^{2}}}.\frac{\partial R_{muv}}{\partial\theta} \tag{9}\]  
Since the expression of \(R_{muv}(\theta,e)\) is more complex, for the convenience of derivation, a total of five temporary variables are recorded:  
\[TEMP1=4(1-e^{2})\cos^{2}\theta+\left(2e-1\right)^{2} \tag{10}\]  
\[TEMP2=2(1-e^{2})\cos\theta \tag{11}\]  
\[TEMP3=4(1-e^{2})\cos^{2}\theta+5e^{2}-4e \tag{12}\]  
\[TEMP4=2(1-e^{2})\cos\theta+(2e-1)\sqrt{4(1-e^{2})\cos^{2}\theta+5e^{2}-4e} \tag{13}\]  
\[TEMP5=\frac{\partial TEMP4}{\partial\theta}=-TEMP2\left[\tan\theta+\frac{2(2 e-1)}{\sqrt{TEMP3}}\cdot\sin\theta\right] \tag{14}\]  
The following is available:  
\[\frac{\partial R_{muv}}{\partial\theta}=-R_{mc}(\frac{\pi}{3},\varphi)\cdot \frac{4(1-e^{2})\sin 2\theta\cdot TEMP4+TEMP1\cdot TEMP5}{\left(TEMP4\right)^{2}} \tag{15}\]  
The partial derivatives of \(p\), \(q\), and \(\theta\) to stress are consistent with the expressions in the yield function.  
The expression of vector \(b\) is  
\[b=\frac{\partial G}{\partial\sigma}=\frac{1}{3}\tan\psi\left[\begin{array}{ c}1\\ 1\\ 1\\ 0\\ 0\end{array}\right]+\frac{3R_{muv}\left(R_{muv}+\frac{2R_{muv}}{\delta\theta} \frac{1}{3}\right)}  
It can be simplified as  
\[b=\frac{\tan\psi}{3}[1\ 1\ 1\ 0\ 0\ 0]^{T}+E_{2}I-E_{3}J \tag{17}\]