## 3 Numerical Simulation on Deformation of Embankment Combined with Subgrade
### Simplified Calculation of the Plastic Drainage Plate  
Since the subgrade is considered as a plane strain problem, so the sand is transformed into the sand wall; the sand drain space and permeability coefficient of sand are in equivalent adjustment. After sand-drained ground is adjusted equally to sand wall subgrade, the degrees of consolidation in same depth or average degree of consolidation in any depth should be as equal as possible at any time. Plastic drainage plate uses the equivalent sand-wall method and equals 3D sand-drained ground to 2D sand wall subgrade. As vertical consolidation effects little on the consolidation degree of the sand wall, so the vertical permeability of the sand wall is equal with vertical permeability of undisturbed soil. To reflect the real situation of foundation reinforcement where possible, the principle of equal is used to calculate the half width of equivalent sand wall \(B_{w}\), and then calculate the half width of sand wall B and horizontal penetration parameter of sand wall \(k_{hp}\). \[\pi r_{w}^{2}=2B_{w}\times\frac{\sqrt{3}}{2}d\ (\text{Triangular arrangement});\ \pi r_{w}^{2}=2B_{w}\times d\ (\text{Square arrangement}), \tag{5}\]  
where \(r_{w}\) is the sand well radius (plastic drainage plate equals sand well radius), and \(d\) is the sand drain space (plastic drainage plate equals impacted diameter). When the plane strain problem of the vertical wall ground is equal with the ax symmetric problem of the sand-drained ground, the principle that the average consolidation degree is equal at the depth of z is used, and then it shows:  
\[\frac{k_{hp}}{k_{ha}}=\frac{8(B-b_{w})^{2}}{3d_{e}^{2}F_{a}}, \tag{6}\]To make sure that the plane strain problem of vertical wall ground is equal with the ax symmetric problem of sand-drained ground, then \(B_{rp}=B_{nt}\) is needed, so:  
\[B=n^{2}B_{w}, \tag{7}\]  
\[k_{hp}=\frac{8(B-B_{w})^{2}}{3d_{e}^{2}F_{a}}k_{h}, \tag{8}\]  
\[F_{a}=\left(ln\frac{n}{s}+\frac{k_{h}}{k_{s}}lns-0.75\right)\frac{n^{2}}{n^{2}- 1}+\frac{s^{2}}{n^{2}-1}\left(1-\frac{k_{h}}{k_{s}}\right)\left(1-\frac{s^{2}} {4n^{2}}\right)+\frac{k_{h}}{k_{s}}\frac{1}{n^{2}-1}\left(1-\frac{1}{4n^{2}} \right), \tag{9}\]  
where \(n\) is the hole diameter ratio; \(S\) is the smear factor; \(k_{h}\) is the horizontal soil permeability; \(k_{s}\) is the horizontal permeability coefficient of the smear zone soil; \(r_{w}\) is the radius of the vertical drain; \(r_{s}\) is the smear zone radius; \(r_{e}\) is the valid radius of the effected zone; and \(k_{w}\) is the permeability coefficient of the vertical drain. The diameter and spacing of the sand well mainly depends on the characteristics of soft soil consolidation and the time requirement. Narrowing the spacing has a better effect than the lagging diameter, in terms of it being good for the sand drain to be both thin and dense in principle. However, the diameter of the sand drain has a certain relationship with the construction method. The hole diameter of the plastic drainage plate is width \(\times\) thickness (\(b\times\delta\)); the translation diameter is:  
\[d_{w}=\alpha\frac{2(b+\delta)}{\pi}, \tag{10}\]  
where \(\alpha\) is the reduction factor of the drain's permeable ability under the pressure of the surrounding soil, and would generally be \(0.75;b\) and \(\delta\) are the plastic drainage plate's width and thickness; and \(d_{w}\) is the equivalent impact diameter. The plastic drainage plate can be used as a vertical drain, and it can be arranged according to a square, plum shape, or equilateral triangle configuration, among others. An arrangement of squares and triangles are most commonly used.