# Semianalytical Solutions to the Axisymmetric Free Strain Consolidation Problem of Unsaturated Soils under the Homogeneous and Mixed Boundary Conditions

Authors: Yuanchun Huang; Lei Wang; Sidong Shen
Journal: Geofluids
 (2022)

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## Abstract

Since the wide applicability of axisymmetric consolidation model, the axisymmetric consolidation problem of unsaturated soils has gradually become a hot issue. However, the mixed boundary conditions of axisymmetric free strain consolidation of unsaturated soils have not been fully considered. The boundary condition is the main factor affecting the consolidation process with the determined consolidation model and related parameters. In this paper, the governing equations of water and air in the polar coordinate were given according to two stress state variables theory of unsaturated soils, Darcy’s law, and Fick’s law under the axisymmetric condition. Based on the free strain hypothesis, a series of semianalytical solutions of the axisymmetric consolidation for unsaturated soil grounds under four different boundary conditions are proposed by using the Laplace transform and the finite Hankel transform. After the correctness of the proposed governing equations and semianalytical solutions is verified by degeneration verification, the parametric consolidation characteristics of unsaturated soils under different parameters through numerical examples and parametric studies are analyzed. Combining the four boundary conditions, the entire consolidation process under homogeneous boundary will be overestimated compared to that under the mixed boundary. In other words, the time required to complete consolidation will be extended accordingly if the boundary condition is not homogeneous. It is clear that the study on the axisymmetric consolidation of unsaturated soils considering multiple boundary conditions is conducive to a more comprehensive understanding of unsaturated soils and provides strong theoretical support for engineering practice.


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## Body

## 1. Introduction

In practical engineering projects, a vertical drain can effectively accelerate the dissipation of water and air phases, so it is widely considered to be an efficient foundation treatment method. At present, the research on the axisymmetric consolidation of saturated soils has reached its maturity, while the development of research on unsaturated soils has not been as fast as expected. In fact, when the consolidation model and related parameters are fixed, the consolidation behavior is mainly affected by the boundary conditions. Due to the presence of water and air phases in the pore, the axisymmetric consolidation of unsaturated soils is more complicated than that of saturated soils, and most practical projects are in unsaturated soils. The free strain consolidation assumes that the deformation under external load is free and the deformation at each point within the soil stratum is different [1]. Thus, the free strain consolidation is closer to the actual situation. Moreover, the state of the sand cushion at the top surface of the foundation treatment and the underlying layer at the bottom directly affect the dissipation process of the excess pore pressures.Based on the one-dimensional (1D) consolidation of saturated soils proposed by Terzaghi [2], Barron [3], and Hansbo [4] had extended the consolidation analysis of sand drain wells to that under the axisymmetric condition. Through the comparison with ideal wells, various factors affecting consolidation in nonideal wells, such as smear effect, well resistance, and vertical flow, were considered reasonably in the consolidation of saturated soils with free strain and equal strain assumptions. Taking the effect of well resistance into consideration, Yoshikuni and Nakanodo [5] also presented an analytical solution to the consolidation of saturated soil foundation under the free strain condition. Due to the complex and numerous parameters of the sand drain foundation, many analyses were based only on saturated soils. But there is a trend that more and more scholars have realized the importance of the consolidation of unsaturated soils. With the continuous and in-depth study of unsaturated soils, some basic theories have been built and widely utilized [6–11]. Based on the consolidation theories under the 1D, 2D plane strain, and axisymmetric condition for unsaturated soils presented by Fredlund and his coworkers [12, 13], Conte [14] proposed coupled solutions for both plane strain and axisymmetric consolidation problems by using finite element analysis. Qin et al. [15] presented a semianalytical solution of the unsaturated consolidation with free strain for vertical drain foundation by using the decoupling method. Zhou and Xu [16, 17] soon after used the differential quadrature method to get solutions on axisymmetric consolidation of unsaturated soils. Following that, Zhou et al. [18] obtained the analytical solutions of the axisymmetric consolidation considering two different loading conditions. Furthermore, Ho et al. [19] deserved an analytical solution to axisymmetric consolidation under equal strain condition by adapting separation of variable and Laplace transformation. In the next few years, other factors affecting axisymmetric consolidation of unsaturated soils such as time-dependent loadings and smear effects were also studied by Ho and Fatahi [20, 21].In addition to some of the influence factors studied in the above papers, there are also boundary conditions that will affect the consolidation. On the basis of the 1D consolidation theory of unsaturated soils obtained by Fredlund and Hassan (1979), Qin et al. [22, 23] derived analytical solutions of 1D consolidation under instantaneous and exponential loadings with the unilateral drainage boundary. Shan et al. [24] deduced theoretical analytical solutions for three different loading conditions under various boundary conditions. Zhou and Zhao [25] gave numerical solutions for 1D consolidation of unsaturated soils considering time-dependent loading and proved that initial and boundary conditions have a significant influence on the consolidation of unsaturated soils. Wang et al. [26] presented a series of semianalytical solutions of 1D consolidation in unsaturated soils considering the first, second, and third kinds of boundaries. Huang et al. [27, 28] gave the analytical solutions of two different boundary conditions for the 2D plane strain consolidation. Wang et al. [29] presented a semianalytical solution for 2D dimensional plane strain consolidation for unsaturated soil with lateral semipermeable drainage boundary under time-dependent loading. Huang and Li [28] made an analysis on the 2D consolidation process of unsaturated soils under vertical semipermeable boundary conditions. Wang et al. [30] introduced the axisymmetric theory consolidation of unsaturated soils into the composite ground and obtained the consolidation governing equations and semianalytical solutions for the composite ground with permeable columns under the first type of boundary conditions. The similar works had been completed in the thermal problem for the nuclear waste repository [15, 31–34].Because of the different dissipation properties of the air and water phases, the boundary is the dominant factor for the consolidation, the assumption of the mixed boundary is more practical for the air and water flowing through the same boundary layer. However, there is no series of mathematical solutions, which are applicable to these kinds of boundaries for the axisymmetric consolidation under the free strain condition. Aiming to investigate the axisymmetric consolidation of unsaturated soils with homogeneous and mixed boundaries, this paper deduces the new semianalytical solutions in terms of the excess pore-air and pore-water pressures and the soil layer settlement by utilizing the Laplace transform and finite Hankel transform. Crump’s method [34] is applied to perform the inversion of the Laplace transform to obtain the final solutions in time domain. Then, the verification of proposed governing equations and semianalytical solutions is carried out by comparing results of the existing equations and solutions in a special case with those in the literature. Through the parametric study, such as different ratios of permeability coefficients and different depths, the influences on the excess pore-air and pore-water pressures and settlement are determined.


## 2. Mathematical Model

### 2.1. Basic Assumptions and Model Introduction

Based on the axisymmetric consolidation theory for unsaturated soils proposed by Dakshanamurthy et al. [13], the basic assumptions are as follows:
(1)
The flow of water and air phases are independent and continuous(2)
The interface between soil particles and water maintains equilibrium under external forces(3)
Soil particles and water phases are incompressible, and the total volume change of unsaturated soil during consolidation is equal to the sum of water and air volumes(4)
The coefficients of volumetric change and permeability related to the water and air phase of the unsaturated soil are constant(5)
The soil element is isotropic(6)
The deformation of sand well foundation occurs only in the vertical direction(7)
The derivation is based on the free strain assumption(8)
The smear effect and well resistance are not considered(9)
The saturation keep constant during the consolidation processIn practice, the permeability coefficients for two phases and saturation change all the time during consolidation. With a purpose to simplify calculations, for the sake of focusing on the effect of boundary conditions on axisymmetric consolidation for unsaturated soil, the permeability coefficients are assumed to be constant. In addition, the smear effect is also ignored in this paper. The elastic deformation of unsaturated soil is mainly concerned in the consolidation theory proposed by Fredlund and Morgenstern [1] in this paper. Although plastic deformation does exist in actual engineering, it is ignored in this paper for the time being.The consolidation model is shown in Figure1; a vertical drain with the radius rw is set in an unsaturated soil layer where the thickness is H. A soil element is adopted to illustrate the radial and vertical flow of air and water phases. re is the outer radius of the unsaturated soil column. kar,kwr,kaz, and kwz are the permeability coefficients of air and water phases flowing in radial and vertical directions, respectively. To a certain extent, the model is to stimulate the drainage and exhaust of unsaturated soils under axisymmetric conditions. The selection of the coordinate axis position is shown in the figure below. In addition, in order to focus on the influence of the top and bottom boundary conditions, the smear effect and well resistance in the radial direction are ignored.Figure 1
A simplified model of axisymmetric consolidation in unsaturated soils.


### 2.2. Governing Equations

According to the assumption of the incompressibility of the solid phase and liquid phase, the total volume change of an unsaturated soil element is equal to the sum of the volume changes in the air and water phase in the element.(1)εv=ΔVvV0=ΔVwV0+ΔVaV0,where ΔVv is change of pore volume in soil and ΔVw and ΔVa are the volume changes of water phase and air phase in soil, respectively. V0 is the initial total volume of an unsaturated soil element.The constitutive relations of unsaturated soils under an axisymmetric condition can be depicted as follows(2)ΔVvV0=m1ksΔσz+σr+σθ3−ua+m2sΔua−uw,(3)ΔVwV0=m1kwΔσz+σr+σθ3−ua+m2wΔua−uw,(4)ΔVaV0=m1kaΔσz+σr+σθ3−ua+m2aΔua−uw,where m1ks, m1kw, and m1ka (kPa−1) are the coefficients of volume change for the soil element, water, and air phase corresponding to the change of net normal stress (Δσz+σr+σθ/3−ua), respectively. m1ks=m1kw+m1ka. m2s, m2w, and m2a (kPa−1) are the coefficient of volume change for the soil element, water, and air phase corresponding to the change of matric suction (Δua−uw), respectively. m2s=m2w+m2a. ua and uw (kPa) are, respectively, excess pore-air pressure and excess pore-water pressure. σz, σr, and σθ are total stress in z-direction, r-direction, and θ-direction, respectively.The first-order partial derivative of Equation (2) on time t can be given as
(5)∂εv∂t=∂∆Vv/V0∂t=m1ks∂σz+σr+σθ/3−ua∂t+m2s∂ua−uw∂t.On the basis of the assumption of the continuity of water flow and Darcy’s law for the flow rate of water under axisymmetric condition, the equation can be expressed by [19]
(6)∂uw∂t=−Cw∂ua∂t−Cvrw∂2uw∂r2+1r∂uw∂r−Cvzw∂2uw∂z2.The relevant parameters are as follows:(7)Cw=m1kw−m2wm2w,Cvrw=kwrγwm2w,Cvzw=kwzγwm2w.,where kwr and kwz are the coefficients of permeability for the water phase in the radial and vertical domains (m/s);γw is the unit weight of water, and γw=9.8kN/m3.Similarly, based on the assumption of the continuity of airflow and Fick’s Law for the flow rate of air under axisymmetric condition, the equation can be written as [19]
(8)∂ua∂t=−Ca∂uw∂t−Cvra∂2ua∂r2+1r∂ua∂r−Cvza∂2ua∂z2.The relevant parameters areCa=−m2a/m2a−m1ka+uatm1−Sr0n0/ua0¯2, Cvra=−karRT/gua0¯Mm2a−m1ka+uatm1−Sr0n0/ua0¯2, and Cvza=−kazRT/gua0¯Mm2a−m1ka+uatm1−Sr0n0/uao¯2. kar and kaz are the coefficients of permeability for the air phase in the radial and vertical domains(m/s); R is the universal air constant, and R=8.314J/mol·K; the absolute temperature T is assumed 293.6 K in this paper; g is the acceleration of gravity, and g=9.8m/s2; ua0¯ (kPa) is the absolute initial excess pore-air pressure and ua0¯=ua0+uatm, ua0 (kPa) is the value of the initial excess pore-air pressure, and uatm (kPa) is the atmospheric pressure; M is the average molar mass of air, and M=0.029kg/mol; Sr0 and n0 are the initial saturation and porosity, respectively.


### 2.3. Initial and Boundary Condition

(1)
Initial conditions0<z<H(9a)uwr,z,0=uw0,(9b)uar,z,0=ua0,where uw0 and ua0 are the initial excess pore-water and pore-air pressures, respectively.
(2)
Boundary conditionsrw<r<re


### 2.4. Radial Boundary

On the inner surface, the boundary is considered permeable to the air and water phases, respectively. The outer boundary is considered as the impermeable drainage boundary of the air and water phases.(10)uwrw,z,t=uarw,z,t=0,∂uare,z,t∂r=∂uwre,z,t∂r.


### 2.5. Vertical Boundary

Considering the complex surrounding environment in practical engineering, the following four boundary conditions cases are taken into account to meet the consolidation development of unsaturated soil under various boundary conditions.Case 1.
The top boundary is considered permeable to the air and water phases, respectively, and the bottom boundary is considered the impermeable drainage boundary for both the air and the water phases.(11a)uwr,0,t=uar,0,t=0,(11b)∂uwr,H,t∂z=∂uar,H,t∂z=0.Case 2.
At the top surface, the boundary is regarded as permeable to the air and water phases, respectively. Similarly, the bottom boundary is considered permeable to the air and water phases, respectively.(11c)uwr,0,t=uar,0,t=0,(11d)uwr,H,t=uar,H,t=0.Case 3.
The boundary of the top surface is deemed to permeable to the air phase, while impermeable to the water phase, and the bottom boundary is deemed to impermeable to both the air and water phase.(11e)∂uwr,0,t∂z=uar,0,t=0,(11f)∂uwr,H,t∂z=∂uar,H,t∂z=0.Case 4.
The top boundary is consistent with Case3, while the bottom boundary is regarded as permeable to the air and water phase, respectively.
(11g)∂uwr,0,t∂z=uar,0,t=0,(11h)uwr,H,t=uar,H,t=0.


## 2.1. Basic Assumptions and Model Introduction

Based on the axisymmetric consolidation theory for unsaturated soils proposed by Dakshanamurthy et al. [13], the basic assumptions are as follows:
(1)
The flow of water and air phases are independent and continuous(2)
The interface between soil particles and water maintains equilibrium under external forces(3)
Soil particles and water phases are incompressible, and the total volume change of unsaturated soil during consolidation is equal to the sum of water and air volumes(4)
The coefficients of volumetric change and permeability related to the water and air phase of the unsaturated soil are constant(5)
The soil element is isotropic(6)
The deformation of sand well foundation occurs only in the vertical direction(7)
The derivation is based on the free strain assumption(8)
The smear effect and well resistance are not considered(9)
The saturation keep constant during the consolidation processIn practice, the permeability coefficients for two phases and saturation change all the time during consolidation. With a purpose to simplify calculations, for the sake of focusing on the effect of boundary conditions on axisymmetric consolidation for unsaturated soil, the permeability coefficients are assumed to be constant. In addition, the smear effect is also ignored in this paper. The elastic deformation of unsaturated soil is mainly concerned in the consolidation theory proposed by Fredlund and Morgenstern [1] in this paper. Although plastic deformation does exist in actual engineering, it is ignored in this paper for the time being.The consolidation model is shown in Figure1; a vertical drain with the radius rw is set in an unsaturated soil layer where the thickness is H. A soil element is adopted to illustrate the radial and vertical flow of air and water phases. re is the outer radius of the unsaturated soil column. kar,kwr,kaz, and kwz are the permeability coefficients of air and water phases flowing in radial and vertical directions, respectively. To a certain extent, the model is to stimulate the drainage and exhaust of unsaturated soils under axisymmetric conditions. The selection of the coordinate axis position is shown in the figure below. In addition, in order to focus on the influence of the top and bottom boundary conditions, the smear effect and well resistance in the radial direction are ignored.Figure 1
A simplified model of axisymmetric consolidation in unsaturated soils.


## 2.2. Governing Equations

According to the assumption of the incompressibility of the solid phase and liquid phase, the total volume change of an unsaturated soil element is equal to the sum of the volume changes in the air and water phase in the element.(1)εv=ΔVvV0=ΔVwV0+ΔVaV0,where ΔVv is change of pore volume in soil and ΔVw and ΔVa are the volume changes of water phase and air phase in soil, respectively. V0 is the initial total volume of an unsaturated soil element.The constitutive relations of unsaturated soils under an axisymmetric condition can be depicted as follows(2)ΔVvV0=m1ksΔσz+σr+σθ3−ua+m2sΔua−uw,(3)ΔVwV0=m1kwΔσz+σr+σθ3−ua+m2wΔua−uw,(4)ΔVaV0=m1kaΔσz+σr+σθ3−ua+m2aΔua−uw,where m1ks, m1kw, and m1ka (kPa−1) are the coefficients of volume change for the soil element, water, and air phase corresponding to the change of net normal stress (Δσz+σr+σθ/3−ua), respectively. m1ks=m1kw+m1ka. m2s, m2w, and m2a (kPa−1) are the coefficient of volume change for the soil element, water, and air phase corresponding to the change of matric suction (Δua−uw), respectively. m2s=m2w+m2a. ua and uw (kPa) are, respectively, excess pore-air pressure and excess pore-water pressure. σz, σr, and σθ are total stress in z-direction, r-direction, and θ-direction, respectively.The first-order partial derivative of Equation (2) on time t can be given as
(5)∂εv∂t=∂∆Vv/V0∂t=m1ks∂σz+σr+σθ/3−ua∂t+m2s∂ua−uw∂t.On the basis of the assumption of the continuity of water flow and Darcy’s law for the flow rate of water under axisymmetric condition, the equation can be expressed by [19]
(6)∂uw∂t=−Cw∂ua∂t−Cvrw∂2uw∂r2+1r∂uw∂r−Cvzw∂2uw∂z2.The relevant parameters are as follows:(7)Cw=m1kw−m2wm2w,Cvrw=kwrγwm2w,Cvzw=kwzγwm2w.,where kwr and kwz are the coefficients of permeability for the water phase in the radial and vertical domains (m/s);γw is the unit weight of water, and γw=9.8kN/m3.Similarly, based on the assumption of the continuity of airflow and Fick’s Law for the flow rate of air under axisymmetric condition, the equation can be written as [19]
(8)∂ua∂t=−Ca∂uw∂t−Cvra∂2ua∂r2+1r∂ua∂r−Cvza∂2ua∂z2.The relevant parameters areCa=−m2a/m2a−m1ka+uatm1−Sr0n0/ua0¯2, Cvra=−karRT/gua0¯Mm2a−m1ka+uatm1−Sr0n0/ua0¯2, and Cvza=−kazRT/gua0¯Mm2a−m1ka+uatm1−Sr0n0/uao¯2. kar and kaz are the coefficients of permeability for the air phase in the radial and vertical domains(m/s); R is the universal air constant, and R=8.314J/mol·K; the absolute temperature T is assumed 293.6 K in this paper; g is the acceleration of gravity, and g=9.8m/s2; ua0¯ (kPa) is the absolute initial excess pore-air pressure and ua0¯=ua0+uatm, ua0 (kPa) is the value of the initial excess pore-air pressure, and uatm (kPa) is the atmospheric pressure; M is the average molar mass of air, and M=0.029kg/mol; Sr0 and n0 are the initial saturation and porosity, respectively.


## 2.3. Initial and Boundary Condition

(1)
Initial conditions0<z<H(9a)uwr,z,0=uw0,(9b)uar,z,0=ua0,where uw0 and ua0 are the initial excess pore-water and pore-air pressures, respectively.
(2)
Boundary conditionsrw<r<re


## 2.4. Radial Boundary

On the inner surface, the boundary is considered permeable to the air and water phases, respectively. The outer boundary is considered as the impermeable drainage boundary of the air and water phases.(10)uwrw,z,t=uarw,z,t=0,∂uare,z,t∂r=∂uwre,z,t∂r.


## 2.5. Vertical Boundary

Considering the complex surrounding environment in practical engineering, the following four boundary conditions cases are taken into account to meet the consolidation development of unsaturated soil under various boundary conditions.Case 1.
The top boundary is considered permeable to the air and water phases, respectively, and the bottom boundary is considered the impermeable drainage boundary for both the air and the water phases.(11a)uwr,0,t=uar,0,t=0,(11b)∂uwr,H,t∂z=∂uar,H,t∂z=0.Case 2.
At the top surface, the boundary is regarded as permeable to the air and water phases, respectively. Similarly, the bottom boundary is considered permeable to the air and water phases, respectively.(11c)uwr,0,t=uar,0,t=0,(11d)uwr,H,t=uar,H,t=0.Case 3.
The boundary of the top surface is deemed to permeable to the air phase, while impermeable to the water phase, and the bottom boundary is deemed to impermeable to both the air and water phase.(11e)∂uwr,0,t∂z=uar,0,t=0,(11f)∂uwr,H,t∂z=∂uar,H,t∂z=0.Case 4.
The top boundary is consistent with Case3, while the bottom boundary is regarded as permeable to the air and water phase, respectively.
(11g)∂uwr,0,t∂z=uar,0,t=0,(11h)uwr,H,t=uar,H,t=0.


## 3. Semianalytical Solutions

### 3.1. Mathematical Method

In this part, Laplace transform and finite Hankel transform and their inverse transforms are applied to solving the consolidation equations of unsaturated soil with vertical drain wells. These two mathematical methods are briefly described below.


#### 3.1.1. Finite Hankel Transform

The finite Hankel transform of a functionfx, defined for all real number a≤x≤b, is the function Hλn. The transform can be defined as
(12)Hλn=∫abxfxJ0λnxYλna−J0λnaYλnxdx,where J0λn and Y0λn are the zero-order of the first and second kind of Bessel functions, respectively. λn is the nth positive root of the following equation:
(13)J0λnaY1λnb−J1λnbY0λna=0,where J1λn and Y1λn are the one-order of the first and second king of Bessel functions, respectively. It also needs to point out that fa and f′b are constants in this case.


#### 3.1.2. Laplace Transform

The complex valued functionft is introduced. If there is an infinite integral ∫0∞fte−stdt which converges to Fs in the region of the complex plan s, then it is called
(14)Fs=∫0∞fte−stdt,where s is a complex number frequency parameter.


### 3.2. General Semianalytical Solutions

According to the definition of finite Hankel transform and the radial boundaries, implementing the zero-order finite Hankel transform on Equations (6) and (8)
(15)∂uw,H∂t=−Cw∂ua,H∂t+λn,w2Cvrwuw,H−Cvzw∂2uw,H∂z2,(16)∂ua,H∂t=−Ca∂uw,H∂t−λn,a2Cvraua,H−Cvza∂2ua,H∂z2,where ui,H=H0ui=∫rwre∫uiB0λdrn,i, H0∂2ui/∂r2+1/r∂ui/∂r=−λn,i2ui,Hi=w,a.The inverse finite Hankel transform can be written as(17)ui=H−1ui,H=∑n=1∞ui,HB02λn,irB0λn,ir,where B02λn,ir=1/2re2Y0λn,irwJ0λn,ire−J0λn,irwY0λn,ire2+Y0λn,irwJ1λn,ire−J0λn,irwY1λn,ire2−2/π2λn,i2, B0λn,ir=J0λn,irY0λn,irw−J0λn,irwY0λn,ir, λn,i is the positive root of the transcendental equation,J0λn,irwY1λn,ire−J1λn,ireY0λn,irw=0, and J0λn,irw and Y0λn,irw are the value of zero-order of first and second kind of Bessel functions atr=rw, respectively. J1λn,ire and Y1λn,ire are the value of one-order of first and second kind of Bessel functions at r=re.Conducting the Laplace transform to Equations (15) and (16) gives
(18)suw,HL−uw,H0=−Cwsua,HL−ua,H0+λn,w2Cvrwuw,HL−Cvzw∂2uw,HL∂z2,(19)sua,HL−ua,H0=−Casuw,HL−uw,H0+λn,a2vraa,HLvza∂2ua,HL/∂z2,where uw,H0=∫rerwuw0Bλn,irrdr, ua,H0=∫rerwua0Bλn,irrdr, uw,HL, and ua,HL are the results of finite Hankel and Laplace transform of excess pore-water and pore-air pressures.Rearrange Equations (18) and (19) to get
(20)uw,HL=−CvzasCa∂2ua,HL∂z2−s−λn,a2vrasCaa,HLua,H0+Cauw,H0/sCa,21)ua,HL=−CvzwsCw∂2uw,HL∂z2−s−λn,w2vrwsCww,HLuw,H0+Cwua,H0/sCw.Taking the first and second derivatives of Equation (20) with respect to z leads to
(22)∂ua,HL∂z=−CvzwsCw∂3uw,HL∂z3−s−λn,w2CvrwsCw∂uw,HL∂z,(23)∂2ua,HL∂z2=−CvzwsCw∂4uw,HL∂z4−s−λn,w2CvrwsCw∂2uw,HL∂z2.Submitting Equations (21) and (23) into (20) results in
(24)a1∂4uw,HL∂z4+a2∂2uw,HL∂z2+a3uw,HL+a4=0,wherea1=CvzaCvzw,a2=Cvzas−λn,w2Cvrw+Cvzws−λn,a2Cvra,a3=s−λn,w2Cvrws−λn,a2Cvra−s2CaCw, anda4=sCwCa−s+λn,a2Cvrauw,H0+λn,a2CvraCwua,H0. The solution to the inhomogeneous ordinary differential Equation (24) has the form of
(25)uw,HL=Aeξz+Be−ξz+Ceηz+De−ηz−a4a3,where ξ=−a2+a22−4a1a3/2a1, η=−a2−a22−4a1a3/2a1, and A, B, C, and D are functions defined by the boundary conditions.Taking the first-order and second-order derivative of Equation (25) with respect to z gives
(26)∂uw,HL∂z=Aξeξz−Bξe−ξz+Cηeηz−Dηe−ηz,(27)∂2uw,HL∂z2=Aξ2eξz+Bξ2e−ξz+Cη2eηz+Dη2e−ηz.Submitting Equations (25) and (27) into (21) gives
(28)ua,HL=−CvzwsCwA1ξ2eξz+B1ξ2e−ξz+C1η2eηz+D1η2e−ηz−s−λn,w2CvrwsCwA1eξz+B1e−ξz+C1eηz+D1e−ηz−a4a3+uw,H0+Cwua,H0sCwReorganizing and redefining the intermediate variables of Equation (28) leads to
(29)ua,HL=Ab1eξz+Bb1e−ξz+Cb2eηz+Db2e−ηz+b3,where b1=λn,w2Cvrw−ξ2Cvzw−s/sCw, b2=λn,w2−η2Cvzw−s/sCw, and b3=s−λn,w2Cvrw/sCwa4/a3+uw,H0+Cwua,H0/sCw.Similarly, taking the first-order derivative of Equation (29) with respect to z produces
(30)∂ua,HL∂z=Ab1ξeξz−Bb1ξe−ξz+Cb2ηeηz−Db2ηe−ηz.As a summary, Equations (25) and (29) are the general solutions of Equations (6) and (8) in the domain of finite Hankel transform and Laplace transform.Implementing finite Hankel transform and Laplace transform on the vertical boundaries, i.e., Equations (11a)–(11h), then a set of semianalytical solutions are given under the following boundary conditions, respectively.Case 1.
Top boundary:(31a)ua,HLλn,z=0,s=uw,HLλn,z=0,s=0.
Bottom boundary:(31b)∂ua,HLλn,z=H,s∂z=∂uw,HLλn,z=H,s∂z=0.Case 2.
Top boundary:(31c)ua,HLλn,z=0,s=uw,HLλn,z=0,s=0.
Bottom boundary(31d)ua,HLλn,z=H,s=uw,HLλn,z=H,s=0.Case 3.
Top boundary:(31e)ua,HLλn,z=0,s=∂uw,HLλn,z=0,s∂z=0.
Bottom boundary:(31f)∂ua,HLλn,z=H,s∂z=∂uw,HLλn,z=H,s∂z=0.Case 4.
Top boundary:(31g)ua,HLλn,z=0,s=∂uw,HLλn,z=0,s∂z=0.
Bottom boundary:(31h)ua,HLλn,z=H,s=uw,HLλn,z=H,s=0.Submitting Equations (25), (26), (29), and (30) into Equations (31a)–(31h), respectively, yieldsCase 1.
(32a)A1=−a4b2+a3b3a3b1−b21+e2ξH,(32b)B1=−a4b2+a3b3e2ξHa3b1−b21+e2ξH,(32c)C1=a4b1+a3b3a3b1−b21+e2ηH,(32d)D1=a4b1+a3b3e2ηHa3b1−b21+e2ηH.
Submitting Equations (32a)–(32d) into Equations (25) and (29) produces
(33a)ua,HL=−b1ϕ11CoshξH−zCoshξH+b2ϕ12CoshηH−zCoshηH+b3,(33b)uw,HL=−ϕ11CoshξH−zCoshξH+ϕ12CoshηH−zCoshηH−a4a3,where ϕ11=a4b2+a3b3/a3b1−b2, and ϕ12=a4b1+a3b3/a3b1−b2. Submitting Equations (33a) and (33b) into Equation (17) leads to
(34a)ua,L=∑n=1∞−b1ϕ11coshξH−z/coshξH+b2ϕ12coshηH−z/coshηH+b3B02λn,arB0λn,ar,(34b)uw,L=∑n=1∞−ϕ11coshξH−z/coshξH+ϕ12coshηH−z/coshηH−a4/a3B02λn,wrB0λn,wr.Case 2.
(35a)A2=−a4b2+a3b3a3b1−b21+eξH,B2=−a4b2+a3b3eξHa3b1−b21+eξH,C2=a4b1+a3b3a3b1−b21+eηH,D2=a4b1+a3b3eηHa3b1−b21+eηH.
Submitting Equations (34a) and (34b) into Equations (24) and (28) produces
(36a)ua,HL=b1ϕ21eηH−z+ezηa3b1−b2−b2ϕ22eξH−z+ezξa3b1−b2+b3,(36b)uw,HL=ϕ21eηH−z+ezηa3b1−b2−ϕ22eξH−z+ezξa3b1−b2−a4a3,where ϕ21=ϕ11=a4b2+a3b3/a3b1−b2, and ϕ22=ϕ12=a4b1+a3b3/a3b1−b2. Submitting Equations (36a) and (36b) into Equation (17) leads to
(37a)ua,L=∑n=1∞b1ϕ21eηH−z+ezη/1+eηH−b2ϕ22eξH−z+ezξ/1+eξH+b3B02λn,arB0λn,ar,(37b)uw,L=∑n=1∞ϕ21eηH−z+ezη/1+eηH−ϕ22eξH−z+ezξ/1+eξH−a4/a3B02λn,wrB0λn,wrCase 3.
(38a)A3=−b3e−η+ξHe2ηH−1η4b1ηcoshξHsinhηH−b2ξcoshηHsinhξH,(38b)B3=−b3eξHη2b1ηcoshξH−b2ξcothηHsinhξH,(38c)C3=b3e−η+ξHe2ξH−1ξ4b1ηcoshξHsinhηH−b2ξcoshηHsinhξH,(38d)D3=−b3eηHξ2b2ξcoshηH−b1ηcothξHsinhηH.
Submitting Equations (38a)–(38d) into Equations (25) and (29) produces
(39a)ua,HL=b1ϕ31eηz−η+ξH+b2ϕ32eηH−z+b3,(39b)uw,HL=ϕ31eηz−η+ξH+ϕ32eηH−z−a4a3,where ϕ31=b3ξe2Hξ−1/4b1ηcoshξHsinhηH−b2ξcoshηHsinhξH, and ϕ32=−b3ξ/2b2ξcoshηH−b1ηcothξHsinhηH. Submitting Equations (39a) and (39b) into Equation (17) produces
(40a)ua,L=∑n=1∞b1ϕ31eηz−η+ξH+b2ϕ32eηH−z+b3B02λn,arB0λn,ar,(40b)uw,L=∑n=1∞ϕ31eηz−η+ξH+ϕ32eηH−z−a4/a3B02λn,wrB0λn,wr.Case 4.
(41a)A4=e−ηHτ1+2a4b1b2eη−ξHη+ν1+e−ξHκ1+ο1eξHξ+eη+ξHξ+η−eηHη4a3b1−b2ψ,(41b)B4=e−ηHτ1−2a4b1b2eη+ξHη+ν1+κ1+ο1eηHξ+eη+ξHη+ξ−eξHη4a3b1−b2ψ,(41c)C4=e−ξHτ2+2a4b1b2eξ−ηHξ+ν2−e−ηHκ2+ο2eξHξ−eη+2ξHη−ξ+eξHη4a3b1−b2ψ,(41d)D4=e−ξHτ2−2a4b1b2eη+ξHξ+ν2−κ2+ο2eηHξ−eη+ξHξ−η−eξHη4a3b1−b2ψ,where τ1=a4b22e2ηH−1ξ, τ2=a4b12e2ξH−1η, κ1=−a3b1b3e2ηH+1eξH−1η, κ2=a3b2b3e2ξH+1eηH−1ξ, ο1=a3b2b3eηH−1, ο2=a3b1b3eξH−1, ν1=a4b1b21+e2ηHη, ν2=a4b1b21+e2ξHξ, and ψ=ηb1coshηHsinhξH−ξb2coshξHsinhηH. Submitting Equations (41a)–(41d) into Equations (25) and (29) produces
(42a)ua,HL=b1ϕ41+b2ϕ42ψ+b3,(42b)uw,HL=ϕ41+ϕ42ψ−a4a3,where ϕ41=ϕ11χ1+ηb2ϕ12χ3+b3χ2, ϕ42=−ϕ11SinhH−zη+ϕ12χ4+b3ξχ5, χ1=b2ξcoshzξsinhHη−b1ηcoshHηsinhzξ, χ2=coshHξcoshHηsinhzξ−coshzξcoshHηsinhHξ, χ3=coshHξsinhzξ−coshzξsinhHη, χ4=b1ηcoshzηsinhHξ−b2ξcoshHξsinhzη, and χ5=coshHξsinhH−zη. Submitting Equations (42a) and (42b) into Equation (17) produce
(43a)ua,L=∑n=1∞b1ϕ41+b2ϕ42/ψ+b3B02λn,arB0λn,ar,(43b)uw,L=∑n=1∞ϕ41+ϕ42/ψ−a4/a3B02λn,wrB0λn,wr.Equations (33a) and (33b), (36a) and (36b), ((39a)()) and (39b), and (42a) and (42b) are the semianalytical solution of excess pore-air pressure and excess pore-water pressure in the domains of finite Hankel and Laplace transform. And Equations (34a) and (34b), (37a) and (37b), (40a) and (40b), and (43a) and (43b) are excess pore-air pressure and pore-water pressure in the domain of Laplace transform under the vertical homogeneous and mixed boundary conditions.


### 3.3. Semianalytical Solution for Settlement

The deformation of unsaturated soils can be calculated by Equation (5). When the applied external load is a constant one, applying the Laplace transform on Equation (5) leads to
(44)εv,L=m2s−m1ksua,L−m2suw,L+m1ks−m2sua0+m2suw0s.By submitting Equations (34a) and (34b), (37a) and (37b), (40a) and (40b), and (43a) and (43b) into Equation (44), the settlement of unsaturated soils in the Laplace transform domain can be expressed as
(45)wr,s=∫0hεv,Lr,z,sdz,A series of semianalytical solutions aboutwr,s in the Laplace transform domain are shown below in Equations (46)–(49) for the consolidation equations of unsaturated soils subjected to instantaneous loading.Case 1.
(46)wr,s=m2s−m1ks∑n=1∞−b1ϕ11tanhξH/ξ+b2ϕ12tanhηH/η+hb3B02λn,arB0λn,ar−m2s∑n=1∞−ϕ11tanhξH/ξ+ϕ12tanhηH/η+ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s.Case 2.
(47)wr,s=m2s−m1ks∑n=1∞2b1ϕ21eηH−1/η1+eηH−2b2ϕ22eξH−1/ξ1+eξH+hb3B02λn,arB0λn,ar−m2s∑n=1∞2ϕ21eηH−1/η1+eηH−2ϕ22eξH−1/ξ1+eξH−ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s.Case 3.
(48)wr,s=m2s−m1ks∑n=1∞b1ϕ31e−η+ξHeηH−1/η+b2ϕ32eηH−1/η+hb3B02λn,arB0λn,ar−m2s∑n=1∞b1ϕ31e−η+ξHeηH−1/η+b2ϕ32eηH−1/η−ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s.Case 4.
(49)wr,s=m2s−m1ks∑n=1∞−4b1b3Λ1+b1b2Λ2/ψξ+−b2Λ3+b1b2Λ4/ψηB02λn,arB0λn,ar−m2s∑n=1∞−4b3Λ1+b2Λ2/ψξ+−Λ3+b1Λ4/ψη−ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s,where Λ1=ηcoshHηsinhHξ/22, Λ2=ϕ11ξ−ϕ12ηsinhHξsinhHη, Λ3=ξϕ11+b2ϕ12−b3coshHξcoshHη−1, and Λ4=ηϕ12sinhHξsinhHη.Then the final solutions in the time domain can be obtained by using Laplace inverse transformation on Equations (34a) and (34b), (37a) and (37b), (40a) and (40b), (43a) and (43b), and (46)–(49) by Crump’s method [32] with the self-compiled program, the detail is shown in the appendix. The analytical solutions of excess pore pressures and soil settlement are expressed as follows:
(50)uar,z,tj≈eαtjτ12ua,Lr,z,a−∑k=1∞Reua,Lr,z,a+kπiτcoskπtjτ−Imua,Lr,z,a+kπiτsinkπtjτ,uwr,z,tj≈eαtjτ12uw,Lr,z,a−∑k=1∞Reuw,Lr,z,a+kπiτcoskπtjτ−Imuw,Lr,z,a+kπiτsinkπtjτ,wr,tj≈eαtjτ12wr,a−∑k=1∞Rewr,a+kπiτcoskπtjτ−Imwr,a+kπiτsinkπtjτ.On the basis of Crump’s method, the numerical computation programs are compiled to get the specific results of excess pore pressures and settlement.


## 3.1. Mathematical Method

In this part, Laplace transform and finite Hankel transform and their inverse transforms are applied to solving the consolidation equations of unsaturated soil with vertical drain wells. These two mathematical methods are briefly described below.


### 3.1.1. Finite Hankel Transform

The finite Hankel transform of a functionfx, defined for all real number a≤x≤b, is the function Hλn. The transform can be defined as
(12)Hλn=∫abxfxJ0λnxYλna−J0λnaYλnxdx,where J0λn and Y0λn are the zero-order of the first and second kind of Bessel functions, respectively. λn is the nth positive root of the following equation:
(13)J0λnaY1λnb−J1λnbY0λna=0,where J1λn and Y1λn are the one-order of the first and second king of Bessel functions, respectively. It also needs to point out that fa and f′b are constants in this case.


### 3.1.2. Laplace Transform

The complex valued functionft is introduced. If there is an infinite integral ∫0∞fte−stdt which converges to Fs in the region of the complex plan s, then it is called
(14)Fs=∫0∞fte−stdt,where s is a complex number frequency parameter.


## 3.1.1. Finite Hankel Transform

The finite Hankel transform of a functionfx, defined for all real number a≤x≤b, is the function Hλn. The transform can be defined as
(12)Hλn=∫abxfxJ0λnxYλna−J0λnaYλnxdx,where J0λn and Y0λn are the zero-order of the first and second kind of Bessel functions, respectively. λn is the nth positive root of the following equation:
(13)J0λnaY1λnb−J1λnbY0λna=0,where J1λn and Y1λn are the one-order of the first and second king of Bessel functions, respectively. It also needs to point out that fa and f′b are constants in this case.


## 3.1.2. Laplace Transform

The complex valued functionft is introduced. If there is an infinite integral ∫0∞fte−stdt which converges to Fs in the region of the complex plan s, then it is called
(14)Fs=∫0∞fte−stdt,where s is a complex number frequency parameter.


## 3.2. General Semianalytical Solutions

According to the definition of finite Hankel transform and the radial boundaries, implementing the zero-order finite Hankel transform on Equations (6) and (8)
(15)∂uw,H∂t=−Cw∂ua,H∂t+λn,w2Cvrwuw,H−Cvzw∂2uw,H∂z2,(16)∂ua,H∂t=−Ca∂uw,H∂t−λn,a2Cvraua,H−Cvza∂2ua,H∂z2,where ui,H=H0ui=∫rwre∫uiB0λdrn,i, H0∂2ui/∂r2+1/r∂ui/∂r=−λn,i2ui,Hi=w,a.The inverse finite Hankel transform can be written as(17)ui=H−1ui,H=∑n=1∞ui,HB02λn,irB0λn,ir,where B02λn,ir=1/2re2Y0λn,irwJ0λn,ire−J0λn,irwY0λn,ire2+Y0λn,irwJ1λn,ire−J0λn,irwY1λn,ire2−2/π2λn,i2, B0λn,ir=J0λn,irY0λn,irw−J0λn,irwY0λn,ir, λn,i is the positive root of the transcendental equation,J0λn,irwY1λn,ire−J1λn,ireY0λn,irw=0, and J0λn,irw and Y0λn,irw are the value of zero-order of first and second kind of Bessel functions atr=rw, respectively. J1λn,ire and Y1λn,ire are the value of one-order of first and second kind of Bessel functions at r=re.Conducting the Laplace transform to Equations (15) and (16) gives
(18)suw,HL−uw,H0=−Cwsua,HL−ua,H0+λn,w2Cvrwuw,HL−Cvzw∂2uw,HL∂z2,(19)sua,HL−ua,H0=−Casuw,HL−uw,H0+λn,a2vraa,HLvza∂2ua,HL/∂z2,where uw,H0=∫rerwuw0Bλn,irrdr, ua,H0=∫rerwua0Bλn,irrdr, uw,HL, and ua,HL are the results of finite Hankel and Laplace transform of excess pore-water and pore-air pressures.Rearrange Equations (18) and (19) to get
(20)uw,HL=−CvzasCa∂2ua,HL∂z2−s−λn,a2vrasCaa,HLua,H0+Cauw,H0/sCa,21)ua,HL=−CvzwsCw∂2uw,HL∂z2−s−λn,w2vrwsCww,HLuw,H0+Cwua,H0/sCw.Taking the first and second derivatives of Equation (20) with respect to z leads to
(22)∂ua,HL∂z=−CvzwsCw∂3uw,HL∂z3−s−λn,w2CvrwsCw∂uw,HL∂z,(23)∂2ua,HL∂z2=−CvzwsCw∂4uw,HL∂z4−s−λn,w2CvrwsCw∂2uw,HL∂z2.Submitting Equations (21) and (23) into (20) results in
(24)a1∂4uw,HL∂z4+a2∂2uw,HL∂z2+a3uw,HL+a4=0,wherea1=CvzaCvzw,a2=Cvzas−λn,w2Cvrw+Cvzws−λn,a2Cvra,a3=s−λn,w2Cvrws−λn,a2Cvra−s2CaCw, anda4=sCwCa−s+λn,a2Cvrauw,H0+λn,a2CvraCwua,H0. The solution to the inhomogeneous ordinary differential Equation (24) has the form of
(25)uw,HL=Aeξz+Be−ξz+Ceηz+De−ηz−a4a3,where ξ=−a2+a22−4a1a3/2a1, η=−a2−a22−4a1a3/2a1, and A, B, C, and D are functions defined by the boundary conditions.Taking the first-order and second-order derivative of Equation (25) with respect to z gives
(26)∂uw,HL∂z=Aξeξz−Bξe−ξz+Cηeηz−Dηe−ηz,(27)∂2uw,HL∂z2=Aξ2eξz+Bξ2e−ξz+Cη2eηz+Dη2e−ηz.Submitting Equations (25) and (27) into (21) gives
(28)ua,HL=−CvzwsCwA1ξ2eξz+B1ξ2e−ξz+C1η2eηz+D1η2e−ηz−s−λn,w2CvrwsCwA1eξz+B1e−ξz+C1eηz+D1e−ηz−a4a3+uw,H0+Cwua,H0sCwReorganizing and redefining the intermediate variables of Equation (28) leads to
(29)ua,HL=Ab1eξz+Bb1e−ξz+Cb2eηz+Db2e−ηz+b3,where b1=λn,w2Cvrw−ξ2Cvzw−s/sCw, b2=λn,w2−η2Cvzw−s/sCw, and b3=s−λn,w2Cvrw/sCwa4/a3+uw,H0+Cwua,H0/sCw.Similarly, taking the first-order derivative of Equation (29) with respect to z produces
(30)∂ua,HL∂z=Ab1ξeξz−Bb1ξe−ξz+Cb2ηeηz−Db2ηe−ηz.As a summary, Equations (25) and (29) are the general solutions of Equations (6) and (8) in the domain of finite Hankel transform and Laplace transform.Implementing finite Hankel transform and Laplace transform on the vertical boundaries, i.e., Equations (11a)–(11h), then a set of semianalytical solutions are given under the following boundary conditions, respectively.Case 1.
Top boundary:(31a)ua,HLλn,z=0,s=uw,HLλn,z=0,s=0.
Bottom boundary:(31b)∂ua,HLλn,z=H,s∂z=∂uw,HLλn,z=H,s∂z=0.Case 2.
Top boundary:(31c)ua,HLλn,z=0,s=uw,HLλn,z=0,s=0.
Bottom boundary(31d)ua,HLλn,z=H,s=uw,HLλn,z=H,s=0.Case 3.
Top boundary:(31e)ua,HLλn,z=0,s=∂uw,HLλn,z=0,s∂z=0.
Bottom boundary:(31f)∂ua,HLλn,z=H,s∂z=∂uw,HLλn,z=H,s∂z=0.Case 4.
Top boundary:(31g)ua,HLλn,z=0,s=∂uw,HLλn,z=0,s∂z=0.
Bottom boundary:(31h)ua,HLλn,z=H,s=uw,HLλn,z=H,s=0.Submitting Equations (25), (26), (29), and (30) into Equations (31a)–(31h), respectively, yieldsCase 1.
(32a)A1=−a4b2+a3b3a3b1−b21+e2ξH,(32b)B1=−a4b2+a3b3e2ξHa3b1−b21+e2ξH,(32c)C1=a4b1+a3b3a3b1−b21+e2ηH,(32d)D1=a4b1+a3b3e2ηHa3b1−b21+e2ηH.
Submitting Equations (32a)–(32d) into Equations (25) and (29) produces
(33a)ua,HL=−b1ϕ11CoshξH−zCoshξH+b2ϕ12CoshηH−zCoshηH+b3,(33b)uw,HL=−ϕ11CoshξH−zCoshξH+ϕ12CoshηH−zCoshηH−a4a3,where ϕ11=a4b2+a3b3/a3b1−b2, and ϕ12=a4b1+a3b3/a3b1−b2. Submitting Equations (33a) and (33b) into Equation (17) leads to
(34a)ua,L=∑n=1∞−b1ϕ11coshξH−z/coshξH+b2ϕ12coshηH−z/coshηH+b3B02λn,arB0λn,ar,(34b)uw,L=∑n=1∞−ϕ11coshξH−z/coshξH+ϕ12coshηH−z/coshηH−a4/a3B02λn,wrB0λn,wr.Case 2.
(35a)A2=−a4b2+a3b3a3b1−b21+eξH,B2=−a4b2+a3b3eξHa3b1−b21+eξH,C2=a4b1+a3b3a3b1−b21+eηH,D2=a4b1+a3b3eηHa3b1−b21+eηH.
Submitting Equations (34a) and (34b) into Equations (24) and (28) produces
(36a)ua,HL=b1ϕ21eηH−z+ezηa3b1−b2−b2ϕ22eξH−z+ezξa3b1−b2+b3,(36b)uw,HL=ϕ21eηH−z+ezηa3b1−b2−ϕ22eξH−z+ezξa3b1−b2−a4a3,where ϕ21=ϕ11=a4b2+a3b3/a3b1−b2, and ϕ22=ϕ12=a4b1+a3b3/a3b1−b2. Submitting Equations (36a) and (36b) into Equation (17) leads to
(37a)ua,L=∑n=1∞b1ϕ21eηH−z+ezη/1+eηH−b2ϕ22eξH−z+ezξ/1+eξH+b3B02λn,arB0λn,ar,(37b)uw,L=∑n=1∞ϕ21eηH−z+ezη/1+eηH−ϕ22eξH−z+ezξ/1+eξH−a4/a3B02λn,wrB0λn,wrCase 3.
(38a)A3=−b3e−η+ξHe2ηH−1η4b1ηcoshξHsinhηH−b2ξcoshηHsinhξH,(38b)B3=−b3eξHη2b1ηcoshξH−b2ξcothηHsinhξH,(38c)C3=b3e−η+ξHe2ξH−1ξ4b1ηcoshξHsinhηH−b2ξcoshηHsinhξH,(38d)D3=−b3eηHξ2b2ξcoshηH−b1ηcothξHsinhηH.
Submitting Equations (38a)–(38d) into Equations (25) and (29) produces
(39a)ua,HL=b1ϕ31eηz−η+ξH+b2ϕ32eηH−z+b3,(39b)uw,HL=ϕ31eηz−η+ξH+ϕ32eηH−z−a4a3,where ϕ31=b3ξe2Hξ−1/4b1ηcoshξHsinhηH−b2ξcoshηHsinhξH, and ϕ32=−b3ξ/2b2ξcoshηH−b1ηcothξHsinhηH. Submitting Equations (39a) and (39b) into Equation (17) produces
(40a)ua,L=∑n=1∞b1ϕ31eηz−η+ξH+b2ϕ32eηH−z+b3B02λn,arB0λn,ar,(40b)uw,L=∑n=1∞ϕ31eηz−η+ξH+ϕ32eηH−z−a4/a3B02λn,wrB0λn,wr.Case 4.
(41a)A4=e−ηHτ1+2a4b1b2eη−ξHη+ν1+e−ξHκ1+ο1eξHξ+eη+ξHξ+η−eηHη4a3b1−b2ψ,(41b)B4=e−ηHτ1−2a4b1b2eη+ξHη+ν1+κ1+ο1eηHξ+eη+ξHη+ξ−eξHη4a3b1−b2ψ,(41c)C4=e−ξHτ2+2a4b1b2eξ−ηHξ+ν2−e−ηHκ2+ο2eξHξ−eη+2ξHη−ξ+eξHη4a3b1−b2ψ,(41d)D4=e−ξHτ2−2a4b1b2eη+ξHξ+ν2−κ2+ο2eηHξ−eη+ξHξ−η−eξHη4a3b1−b2ψ,where τ1=a4b22e2ηH−1ξ, τ2=a4b12e2ξH−1η, κ1=−a3b1b3e2ηH+1eξH−1η, κ2=a3b2b3e2ξH+1eηH−1ξ, ο1=a3b2b3eηH−1, ο2=a3b1b3eξH−1, ν1=a4b1b21+e2ηHη, ν2=a4b1b21+e2ξHξ, and ψ=ηb1coshηHsinhξH−ξb2coshξHsinhηH. Submitting Equations (41a)–(41d) into Equations (25) and (29) produces
(42a)ua,HL=b1ϕ41+b2ϕ42ψ+b3,(42b)uw,HL=ϕ41+ϕ42ψ−a4a3,where ϕ41=ϕ11χ1+ηb2ϕ12χ3+b3χ2, ϕ42=−ϕ11SinhH−zη+ϕ12χ4+b3ξχ5, χ1=b2ξcoshzξsinhHη−b1ηcoshHηsinhzξ, χ2=coshHξcoshHηsinhzξ−coshzξcoshHηsinhHξ, χ3=coshHξsinhzξ−coshzξsinhHη, χ4=b1ηcoshzηsinhHξ−b2ξcoshHξsinhzη, and χ5=coshHξsinhH−zη. Submitting Equations (42a) and (42b) into Equation (17) produce
(43a)ua,L=∑n=1∞b1ϕ41+b2ϕ42/ψ+b3B02λn,arB0λn,ar,(43b)uw,L=∑n=1∞ϕ41+ϕ42/ψ−a4/a3B02λn,wrB0λn,wr.Equations (33a) and (33b), (36a) and (36b), ((39a)()) and (39b), and (42a) and (42b) are the semianalytical solution of excess pore-air pressure and excess pore-water pressure in the domains of finite Hankel and Laplace transform. And Equations (34a) and (34b), (37a) and (37b), (40a) and (40b), and (43a) and (43b) are excess pore-air pressure and pore-water pressure in the domain of Laplace transform under the vertical homogeneous and mixed boundary conditions.


## 3.3. Semianalytical Solution for Settlement

The deformation of unsaturated soils can be calculated by Equation (5). When the applied external load is a constant one, applying the Laplace transform on Equation (5) leads to
(44)εv,L=m2s−m1ksua,L−m2suw,L+m1ks−m2sua0+m2suw0s.By submitting Equations (34a) and (34b), (37a) and (37b), (40a) and (40b), and (43a) and (43b) into Equation (44), the settlement of unsaturated soils in the Laplace transform domain can be expressed as
(45)wr,s=∫0hεv,Lr,z,sdz,A series of semianalytical solutions aboutwr,s in the Laplace transform domain are shown below in Equations (46)–(49) for the consolidation equations of unsaturated soils subjected to instantaneous loading.Case 1.
(46)wr,s=m2s−m1ks∑n=1∞−b1ϕ11tanhξH/ξ+b2ϕ12tanhηH/η+hb3B02λn,arB0λn,ar−m2s∑n=1∞−ϕ11tanhξH/ξ+ϕ12tanhηH/η+ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s.Case 2.
(47)wr,s=m2s−m1ks∑n=1∞2b1ϕ21eηH−1/η1+eηH−2b2ϕ22eξH−1/ξ1+eξH+hb3B02λn,arB0λn,ar−m2s∑n=1∞2ϕ21eηH−1/η1+eηH−2ϕ22eξH−1/ξ1+eξH−ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s.Case 3.
(48)wr,s=m2s−m1ks∑n=1∞b1ϕ31e−η+ξHeηH−1/η+b2ϕ32eηH−1/η+hb3B02λn,arB0λn,ar−m2s∑n=1∞b1ϕ31e−η+ξHeηH−1/η+b2ϕ32eηH−1/η−ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s.Case 4.
(49)wr,s=m2s−m1ks∑n=1∞−4b1b3Λ1+b1b2Λ2/ψξ+−b2Λ3+b1b2Λ4/ψηB02λn,arB0λn,ar−m2s∑n=1∞−4b3Λ1+b2Λ2/ψξ+−Λ3+b1Λ4/ψη−ha4/a3B02λn,arB0λn,ar+hm1ks−m2sua0+m2suw0s,where Λ1=ηcoshHηsinhHξ/22, Λ2=ϕ11ξ−ϕ12ηsinhHξsinhHη, Λ3=ξϕ11+b2ϕ12−b3coshHξcoshHη−1, and Λ4=ηϕ12sinhHξsinhHη.Then the final solutions in the time domain can be obtained by using Laplace inverse transformation on Equations (34a) and (34b), (37a) and (37b), (40a) and (40b), (43a) and (43b), and (46)–(49) by Crump’s method [32] with the self-compiled program, the detail is shown in the appendix. The analytical solutions of excess pore pressures and soil settlement are expressed as follows:
(50)uar,z,tj≈eαtjτ12ua,Lr,z,a−∑k=1∞Reua,Lr,z,a+kπiτcoskπtjτ−Imua,Lr,z,a+kπiτsinkπtjτ,uwr,z,tj≈eαtjτ12uw,Lr,z,a−∑k=1∞Reuw,Lr,z,a+kπiτcoskπtjτ−Imuw,Lr,z,a+kπiτsinkπtjτ,wr,tj≈eαtjτ12wr,a−∑k=1∞Rewr,a+kπiτcoskπtjτ−Imwr,a+kπiτsinkπtjτ.On the basis of Crump’s method, the numerical computation programs are compiled to get the specific results of excess pore pressures and settlement.


## 4. Verification

The parameters used below refer to the literature [15, 19, 35], as shown below: h=5m, rw=0.5m, re=4.5m, n0=50%, Sr0=80%, kwr=2×10−10m/s, kwz=10−10m/s, m1ks=−2.5×10−4kPa−1, m1kw=−0.5×10−4kPa−1, m2s=−1×10−4kPa−1, m2w=−2×10−4kPa−1, q0=100kPa, ua0=20kPa, uw0=40kPa, uatm=101.3kPa, and γw=9.807kN/m3. According to Equations (6) and (8), the consolidation coefficients are calculated, and the results are as follows: Ca=−0.101; Cvra=−1.431×10−4; Cvza=−7.156×10−5; Cw=1.500×10−4; Cvrw=−1.020×10−7; and Cvzw=5.098×10−8. The initial excess pore pressures are computed by the method from Wang et al. [29].In order to verify the semianalytical solutions of the axisymmetric consolidation equation considering different boundaries, a comparative example based on analytical solutions proposed by Ho et al. [19] is shown in Figures 2 and 3. It can be seen that the results of Case 1 in this paper is consistent with those of the existing literature at ka/kw=1, 10, and 100 in excess pore pressures and settlement, indicating that the new method is more general, and other solutions for Case 2, Case 3, and Case 4 are correct in this paper.Figure 2
The comparison on (a) excess pore-air pressure and (b) excess pore-water pressure at different ratios ofka/kw between the proposed semianalytical solutions and the analytical solutions from Ho and Shuai [17].
(a)(b)Figure 3
The comparison on settlement at different ratios ofka/kw between the proposed semianalytical solutions and the analytical solutions from Ho and Shuai [17].


## 5. Case Study

Based on the solutions obtained above, there will be two parts to analyze the effects of different boundary conditions and related parameters on consolidation. These studies are conducted to investigate the influence of the dissipation of excess pore pressure and normalized settlement (w∗) at different permeability coefficient ratios of ka/kw and kr/kz, respectively. In addition, the contours of excess pore pressures at different time t are demonstrated to investigate the distribution characteristics of excess pore pressures with the depth. The researches of consolidation characteristics of these parameters under Case 1 and Case 3 will be placed in Section 5.1, and those under Case 2 and Case 4 will be set in Section 5.2. Furthermore, the study is verified and investigated when r=3 m, z=2.5 m, and the ratio of ka/kw and kr/kz are constants, which are 10 and 0.1, respectively. Specifically, the values of ka/kw change from 1 to 100, as ka varies while kw keeps constant. Similarly, the value of kr/kz change from 0.1 to 10, as krchanges while kz is changeless.


### 5.1. Consolidation under Case1 and Case 3

Figure4 shows the dissipation curve of excess pore pressures with changing ratios of ka/kw under Case 1 and Case 3. It can be found that both the excess pore-air pressure and the excess pore-water pressure have a plateau at the beginning and then decrease rapidly. Different from air pressure, the dissipation of excess water pressure will experience a plateau in the intermediate stage, too, and then rapidly decrease and finally stabilize. The dissipation of excess pore-air pressure accelerates as the ratio of ka/kw increases. The reason is that the change of ka/kw is induced by the change of ka as mentioned above, and it is also the main reason affecting the dissipation speed of excess pore-air pressure. So, the larger the air permeability is, the faster the excess pore-air pressure dissipates. It is clear that the excess pore-air pressure has exactly the same dissipation curve in Case 1 and Case 3, while the excess pore-water pressure starts to show a difference in the second stage. As we can see from Figure 4(b), the dissipation routes of excess pore-water pressure are divided into two stages: the first stage is caused by the dissipation of excess pore-air pressure, and then the excess pore-water pressure dissipates along the top and bottom boundary in the second stage. As for further comment, to the second period, the dissipation curves start to go through different routes, and the values of ka/kw has less influence than the boundary conditions. From the data in Figure 5, the rule of ka/kw on normalized settlement can be described as follows, the curve subsides faster at bigger values of ka/kw, and the settlement curves converge in the second stage. Like the dissipation of excess pore-water pressure, the settlement curves of Case 1 and Case 3 follow essentially the same path in the first stage. Compared with the results of Case 1, the excess pore-water pressure and settlement grow more slowly because the top boundary of Case 3 is impermeable to water when the bottom boundary conditions are the same.Figure 4
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofka/kw under Case 1 and Case 3.
(a)(b)Figure 5
Normalized settlement curves at different values ofka/kw under Case 1 and Case 3.The impacts of the ratios ofkr/kz on excess pore pressures can be seen in Figure 6. In this part, the values of kr/kz are distributed from 0.1 to 10, and the values of kaz and kwz are fixed at 10-9 and 10-10 (m/s). Obviously, like the analysis about the ratio of ka/kw above, the changes in excess pore pressures are induced by different values of kr/kz, but the dissipation pattern is similar. It should be noted that the values of kr/kz will not change the relative value of permeability coefficients ka and kw, so as kr/kz surges, the entire dissipation process will be in an arbitrary state of dissipation. Unlike the uniform distribution of the curves in Figure 4(a), when kr/kz=0.1 and 1, these two curves are adjacent, while when the value is 10, the dissipation of excess pore-air pressure accelerates evidently. It seems that when the ratio of kr/kz is very small, the variation of the value has little effect on the excess pore-air pressure, and when it rises to a certain level, the dissipation rate will be apparently changed. From Figure 6(b), we can see that with the mixed boundary (Case 3) as a control group, the homogeneous boundary (Case 1) has an overestimation of the dissipation process. In the second stage, the excess pore-water pressure dissipation bending lines of Case 3 is evenly distributed, while the dissipation curves of Case 1 are exactly the same as the excess pore-air pressure. This can be interpreted as in the second period where the boundary conditions play a major role, when the water phase is impermeable at top boundary, the excess pore-water pressure dissipates uniformly in the radial direction. Additionally, the effects of different values of kr/kz on normalized settlement with Case 1 and Case 3 are depicted in Figure 7. Due to the combined impacts of excess pore-air pressure and pore-water pressure, the consolidation rate of a homogeneous boundary at kr/kz=10 is still more rapid than that of mixed boundaries.Figure 6
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofkr/kz under Case 1 and Case 3.
(a)(b)Figure 7
Normalized settlement curves at different values ofkr/kz under Case 1 and Case 3.The dissipation patterns of excess pore-air pressures with various depths under Case1 and Case 3 are shown in Figure 8. Different from the previous parameters (ka/kwand kr/kz), the final dissipation time for several depths is the same. When the depth is 1 (m), it is the closest to the top surface, and the air is most easily discharged, that is, it starts to disappear first. It can be observed from Figure 8(b) that in the second stage, the patterns of excess pore-water pressure with various depths in Case 1 are similar to that in the first stage, while the bending lines of excess pore-water pressure of Case 3 are combined into a dissipation curve. This is because when both the top and bottom boundaries are impermeable to the water phase, the distance for the dissipation of excess pore-water pressure at any depth to the middle sand well is the same. This phenomenon also verifies that the boundary condition plays a dominant role in the second period of dissipation of excess pore-water pressure mentioned above. This is also the significance of this paper to study its influence on consolidation from different boundary conditions.Figure 8
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different depths under Case1 and Case 3.
(a)(b)


### 5.2. Consolidation under Case2 and Case 4

It can be found that the dissipation rates of excess pore-air pressure of Case2 and Case 4 are much faster than Case 1 and Case 3 from the final dissipation time. This is because the boundary conditions of Case 2 and Case 4 are the same for the bottom surface and are permeable to the air phase and water phase. With the increase of ka/kw, the excess pore-air pressure presents an order of magnitude increase, as shown in Figure 9(a). Due to the additional seepage channel (the bottom surface), the difference in the dissipation of excess pore-water pressure in the second period of Figure 9(b) is smaller than that of in Case 1 and Case 3. In all these four cases, Case 2 has the fastest settlement rate because its boundary conditions are all permeable to the air and water phase at the top and bottom surface. Taking normalized settlement as the research object, the homogeneous boundary overestimates the speed of consolidation. Under the premise of increasing seepage channels, only the top boundary of Case 2 and Case 4 has different permeability conditions for the air phase, which has few influences on settlement, as shown in Figure 10.Figure 9
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofka/kw under Case 2 and Case 4.
(a)(b)Figure 10
Normalized settlement curves at different values ofka/kw under Case 2 and Case 4.Figure11 demonstrates the dissipation curves of excess pore pressures with variable ratios of kr/kz under Case 2 and Case 4. No matter how the kr/kz changes from 0.1 to 10, the dissipation curves of Case 2 and Case 4 in the excess pore-air pressure are the same. These three curves of Case 2 and Case 4 are nearer to each other, so the value of kr/kz has less impact on the consolidation process than the other two cases. It is known by comparison from the second stage of excess pore-water pressure in Figure 6(b) and Figure 11(b) that the consolidation under Case 3 has the slowest settlement rate because it is only permeable for the air phase at the top surface. To further explanation, in the second stage of excess pore-water pressure in Figure 11(b), the value of kr/kz for Case 2 and Case 4 does not change patently when the value is relatively small. In four cases, when kr/kz=10, the final time of excess pore-water pressure dissipation is close to the same. This can be explained as when kr/kz is large to a certain value (10 in this paper), the effects of the change of boundary conditions on the dissipation rate of excess pore-water pressure are obviously weakened. From the final settlement time of Figures 7 and 12, it can be seen that the slowest in Case 4 settled at t=109s and the slowest in Case 3 settled at t=1010s. So the settlement in Case 2 and Case 4 are still faster than that in Case 1 and Case 3, in general. But if it is divided into mixed (Case 3 and Case 4) and homogeneous (Case 1 and Case 2) boundaries, there is no doubt that the mixed boundary has an inhibitory effect on the consolidation process.Figure 11
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofkr/kz under Case 2 and Case 4.
(a)(b)Figure 12
Normalized settlement curves at different values ofkr/kz under Case 2 and Case 4.Figure13 illustrates the variations of excess pore pressures with different depths. When the depth reaches one and four meters, Case 2 and Case 4 share identical dissipation curves in excess pore-air pressure, which is dissimilar to Case 1 and Case 3. The reason could be concluded that when the bottom boundary is permeable to air, z=1m and z=4m are the same distance to the top and bottom boundaries, respectively. As the depth gets to 2.5 meters, the distance from the top or bottom surface is longer than the other two depths, so more reaction time is required. What is shown in Figure 13(a) is that the excess pore-air pressure starts to dissipate at t=2×103s. The excess pore-water pressure of Case 4 changed from two curves to three curves in the second period. This is because the top boundary is impermeable to water phase and the bottom boundary is permeable to the water phase, so water near the bottom surface will be simpler to drain. This is not the same as the case of a homogeneous boundary. From the final consolidation time in Figures 8 and 13, it can be noticed that when the research factor is depth, the consolidation speed of Case 2 is the quickest, reflecting the promotion of homogeneous boundary.Figure 13
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different depths under Case2 and Case 4.
(a)(b)


## 5.1. Consolidation under Case1 and Case 3

Figure4 shows the dissipation curve of excess pore pressures with changing ratios of ka/kw under Case 1 and Case 3. It can be found that both the excess pore-air pressure and the excess pore-water pressure have a plateau at the beginning and then decrease rapidly. Different from air pressure, the dissipation of excess water pressure will experience a plateau in the intermediate stage, too, and then rapidly decrease and finally stabilize. The dissipation of excess pore-air pressure accelerates as the ratio of ka/kw increases. The reason is that the change of ka/kw is induced by the change of ka as mentioned above, and it is also the main reason affecting the dissipation speed of excess pore-air pressure. So, the larger the air permeability is, the faster the excess pore-air pressure dissipates. It is clear that the excess pore-air pressure has exactly the same dissipation curve in Case 1 and Case 3, while the excess pore-water pressure starts to show a difference in the second stage. As we can see from Figure 4(b), the dissipation routes of excess pore-water pressure are divided into two stages: the first stage is caused by the dissipation of excess pore-air pressure, and then the excess pore-water pressure dissipates along the top and bottom boundary in the second stage. As for further comment, to the second period, the dissipation curves start to go through different routes, and the values of ka/kw has less influence than the boundary conditions. From the data in Figure 5, the rule of ka/kw on normalized settlement can be described as follows, the curve subsides faster at bigger values of ka/kw, and the settlement curves converge in the second stage. Like the dissipation of excess pore-water pressure, the settlement curves of Case 1 and Case 3 follow essentially the same path in the first stage. Compared with the results of Case 1, the excess pore-water pressure and settlement grow more slowly because the top boundary of Case 3 is impermeable to water when the bottom boundary conditions are the same.Figure 4
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofka/kw under Case 1 and Case 3.
(a)(b)Figure 5
Normalized settlement curves at different values ofka/kw under Case 1 and Case 3.The impacts of the ratios ofkr/kz on excess pore pressures can be seen in Figure 6. In this part, the values of kr/kz are distributed from 0.1 to 10, and the values of kaz and kwz are fixed at 10-9 and 10-10 (m/s). Obviously, like the analysis about the ratio of ka/kw above, the changes in excess pore pressures are induced by different values of kr/kz, but the dissipation pattern is similar. It should be noted that the values of kr/kz will not change the relative value of permeability coefficients ka and kw, so as kr/kz surges, the entire dissipation process will be in an arbitrary state of dissipation. Unlike the uniform distribution of the curves in Figure 4(a), when kr/kz=0.1 and 1, these two curves are adjacent, while when the value is 10, the dissipation of excess pore-air pressure accelerates evidently. It seems that when the ratio of kr/kz is very small, the variation of the value has little effect on the excess pore-air pressure, and when it rises to a certain level, the dissipation rate will be apparently changed. From Figure 6(b), we can see that with the mixed boundary (Case 3) as a control group, the homogeneous boundary (Case 1) has an overestimation of the dissipation process. In the second stage, the excess pore-water pressure dissipation bending lines of Case 3 is evenly distributed, while the dissipation curves of Case 1 are exactly the same as the excess pore-air pressure. This can be interpreted as in the second period where the boundary conditions play a major role, when the water phase is impermeable at top boundary, the excess pore-water pressure dissipates uniformly in the radial direction. Additionally, the effects of different values of kr/kz on normalized settlement with Case 1 and Case 3 are depicted in Figure 7. Due to the combined impacts of excess pore-air pressure and pore-water pressure, the consolidation rate of a homogeneous boundary at kr/kz=10 is still more rapid than that of mixed boundaries.Figure 6
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofkr/kz under Case 1 and Case 3.
(a)(b)Figure 7
Normalized settlement curves at different values ofkr/kz under Case 1 and Case 3.The dissipation patterns of excess pore-air pressures with various depths under Case1 and Case 3 are shown in Figure 8. Different from the previous parameters (ka/kwand kr/kz), the final dissipation time for several depths is the same. When the depth is 1 (m), it is the closest to the top surface, and the air is most easily discharged, that is, it starts to disappear first. It can be observed from Figure 8(b) that in the second stage, the patterns of excess pore-water pressure with various depths in Case 1 are similar to that in the first stage, while the bending lines of excess pore-water pressure of Case 3 are combined into a dissipation curve. This is because when both the top and bottom boundaries are impermeable to the water phase, the distance for the dissipation of excess pore-water pressure at any depth to the middle sand well is the same. This phenomenon also verifies that the boundary condition plays a dominant role in the second period of dissipation of excess pore-water pressure mentioned above. This is also the significance of this paper to study its influence on consolidation from different boundary conditions.Figure 8
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different depths under Case1 and Case 3.
(a)(b)


## 5.2. Consolidation under Case2 and Case 4

It can be found that the dissipation rates of excess pore-air pressure of Case2 and Case 4 are much faster than Case 1 and Case 3 from the final dissipation time. This is because the boundary conditions of Case 2 and Case 4 are the same for the bottom surface and are permeable to the air phase and water phase. With the increase of ka/kw, the excess pore-air pressure presents an order of magnitude increase, as shown in Figure 9(a). Due to the additional seepage channel (the bottom surface), the difference in the dissipation of excess pore-water pressure in the second period of Figure 9(b) is smaller than that of in Case 1 and Case 3. In all these four cases, Case 2 has the fastest settlement rate because its boundary conditions are all permeable to the air and water phase at the top and bottom surface. Taking normalized settlement as the research object, the homogeneous boundary overestimates the speed of consolidation. Under the premise of increasing seepage channels, only the top boundary of Case 2 and Case 4 has different permeability conditions for the air phase, which has few influences on settlement, as shown in Figure 10.Figure 9
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofka/kw under Case 2 and Case 4.
(a)(b)Figure 10
Normalized settlement curves at different values ofka/kw under Case 2 and Case 4.Figure11 demonstrates the dissipation curves of excess pore pressures with variable ratios of kr/kz under Case 2 and Case 4. No matter how the kr/kz changes from 0.1 to 10, the dissipation curves of Case 2 and Case 4 in the excess pore-air pressure are the same. These three curves of Case 2 and Case 4 are nearer to each other, so the value of kr/kz has less impact on the consolidation process than the other two cases. It is known by comparison from the second stage of excess pore-water pressure in Figure 6(b) and Figure 11(b) that the consolidation under Case 3 has the slowest settlement rate because it is only permeable for the air phase at the top surface. To further explanation, in the second stage of excess pore-water pressure in Figure 11(b), the value of kr/kz for Case 2 and Case 4 does not change patently when the value is relatively small. In four cases, when kr/kz=10, the final time of excess pore-water pressure dissipation is close to the same. This can be explained as when kr/kz is large to a certain value (10 in this paper), the effects of the change of boundary conditions on the dissipation rate of excess pore-water pressure are obviously weakened. From the final settlement time of Figures 7 and 12, it can be seen that the slowest in Case 4 settled at t=109s and the slowest in Case 3 settled at t=1010s. So the settlement in Case 2 and Case 4 are still faster than that in Case 1 and Case 3, in general. But if it is divided into mixed (Case 3 and Case 4) and homogeneous (Case 1 and Case 2) boundaries, there is no doubt that the mixed boundary has an inhibitory effect on the consolidation process.Figure 11
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different values ofkr/kz under Case 2 and Case 4.
(a)(b)Figure 12
Normalized settlement curves at different values ofkr/kz under Case 2 and Case 4.Figure13 illustrates the variations of excess pore pressures with different depths. When the depth reaches one and four meters, Case 2 and Case 4 share identical dissipation curves in excess pore-air pressure, which is dissimilar to Case 1 and Case 3. The reason could be concluded that when the bottom boundary is permeable to air, z=1m and z=4m are the same distance to the top and bottom boundaries, respectively. As the depth gets to 2.5 meters, the distance from the top or bottom surface is longer than the other two depths, so more reaction time is required. What is shown in Figure 13(a) is that the excess pore-air pressure starts to dissipate at t=2×103s. The excess pore-water pressure of Case 4 changed from two curves to three curves in the second period. This is because the top boundary is impermeable to water phase and the bottom boundary is permeable to the water phase, so water near the bottom surface will be simpler to drain. This is not the same as the case of a homogeneous boundary. From the final consolidation time in Figures 8 and 13, it can be noticed that when the research factor is depth, the consolidation speed of Case 2 is the quickest, reflecting the promotion of homogeneous boundary.Figure 13
Dissipation curves of (a) excess pore-air pressure and (b) excess pore-water pressure at different depths under Case2 and Case 4.
(a)(b)


## 6. Conclusion

This paper provides a new series of semianalytical solutions of axisymmetric consolidation in unsaturated soils with homogeneous and mixed boundaries under the free strain conditions by applying Laplace transform and the infinite Hankel transform. Based on the proposed solutions, some computations are conducted to describe the impact of the four kinds of boundary conditions on the consolidation behavior. The main conclusions are drawn as follows:(1)
Considering the continuity assumption and flow laws of air and water phases under the axisymmetric condition, a semianalytical solution to the consolidation of a finite thickness of unsaturated soil layer with homogeneous and mixed boundaries are presented. Because of the fewer constraints on the boundary conditions, the new method incorporating the finite Hankel transform and Laplace transform is more versatile to solve the axisymmetric consolidation problem under all kinds of boundary conditions(2)
The consolidation under the homogeneous drainage boundary is overestimated. Especially, there is no influence on the dissipation of excess pore-air pressure considering these two drainage boundaries, while the mixed drainage boundary restrains the dissipation of excess pore-water pressure and the growth of the settlement. In actual projects, according to the seepage conditions of the top and bottom layers, the sufficient time for soil settlement is needed. For the most unfavorable situation with Case3, the construction should be carried out at least 1010 s later(3)
The time needed to complete the whole consolidation process for the mixed drainage boundary (Case3 and Case 4) is longer than that with homogeneous drainage boundaries (Case 1 and Case 2). Since the difference of boundary conditions between the two control groups is the same, with the change of ka/kw, the ratio of time for reaching the final settlements under Case 1 and Case 3 are 90% those under Case 2 and Case 4 are 88.9%, respectively(4)
The impacts ofkr/kz on the dissipation of excess pore pressures and settlement change for mixed drainage boundaries (Case 1 and Case 4) are more significant than those with homogeneous drainage boundaries (Case 1 and Case 2). The dissipation patterns of excess pore air/water pressure with the depth are similar for the single drainage boundary (Case 1) and the mixed drainage boundary (Case 4), while excess pore-water pressure at the second stage develops along the same line under the other mixed drainage boundary (Case 3).



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