Semi-analytical solutions for soil consolidation induced by drying

Abstract

Drying induced consolidation is ubiquitous in nature, yet no theoretical model is available to delineate this phenomenon. Fundamentally, this phenomenon is a multi-physical process involving that the soil–water interaction energy lowers (becomes more negative), thereby soil’s suction stress, elastic modulus, and hydraulic conductivity change. Therefore, the drying induced consolidation is a phenomenon reflecting soil’s mechanical, hydraulic, and hydro-mechanical properties altered by soil–water interaction. Here, a theoretical framework is developed for drying induced consolidation. Richards equation is selected to describe moisture flow subject to drying. The impact of soil–water interaction on soil mechanical properties is quantified via an elastic modulus characteristic curve equation and a unified effective stress equation, respectively. Gardner’s model is selected for representing soil water retention and hydraulic properties, facilitating the semi-analytical solutions for drying induced consolidation. The accuracy of solutions is verified by comparing with numerical solutions of moisture flow and soil shrinkage curve data. It reveals that the adsorption induced effective stress can reach up to hundreds of kPa and contributes 96.7% of the total consolidation settlement for clayey soils. In addition, the elastic modulus variation is commonly overlooked in existing consolidation models, but this overlook can lead to either underestimation up to 0.66 times or overestimation up to 6.86 times of consolidation settlement.

Appendices

Appendix 1

Derivations of Analytical Solution for Settlement

To obtain soil layer settlement under drying, the suction-stress-based effective stress is implemented in the classical elasticity. The stress state of a soil element and coordinates are illustrated in Fig. 1. The gravity can be neglected in drying induced consolidation, implying zero body force, and thus the force equilibrium equations can be expressed as:

$$\frac{{\partial \sigma_{x} }}{\partial x} + \frac{{\partial \tau_{yx} }}{\partial y} + \frac{{\partial \tau_{zx} }}{\partial z} = 0$$

(A1)

$$\frac{{\partial \sigma_{y} }}{\partial y} + \frac{{\partial \tau_{zy} }}{\partial z} + \frac{{\partial \tau_{xy} }}{\partial x} = 0$$

(A2)

$$\frac{{\partial \sigma_{z} }}{\partial z} + \frac{{\partial \tau_{xz} }}{\partial x} + \frac{{\partial \tau_{yz} }}{\partial y} = 0$$

(A3)

The stress–strain relationships for the homogeneous and isotropic soil layer can be obtained by Hooke’s law:

$${\text{d}} \varepsilon_{x} = \frac{1}{E}\left[ {{\text{d}} \sigma_{x} ^{\prime} - \nu \left( {{\text{d}} \sigma_{y} ^{\prime} + {\text{d}} \sigma_{z} ^{\prime}} \right)} \right] = \frac{1}{E}\left[ {{\text{d}} \sigma_{x} - \nu \left( {{\text{d}} \sigma_{y} + {\text{d}} \sigma_{z} } \right)} \right] - \frac{1 - 2\nu }{E}{\text{d}} \sigma^{s}$$

(A4)

$${\text{d}} \varepsilon_{y} = \frac{1}{E}\left[ {{\text{d}} \sigma_{y} ^{\prime} - \nu \left( {{\text{d}} \sigma_{z} ^{\prime} + {\text{d}} \sigma_{x} ^{\prime}} \right)} \right] = \frac{1}{E}\left[ {{\text{d}} \sigma_{y} - \nu \left( {{\text{d}} \sigma_{z} + {\text{d}} \sigma_{x} } \right)} \right] - \frac{1 - 2\nu }{E}{\text{d}} \sigma^{s}$$

(A5)

$${\text{d}} \varepsilon_{z} = \frac{1}{E}\left[ {{\text{d}} \sigma_{z} ^{\prime} - \nu \left( {{\text{d}} \sigma_{x} ^{\prime} + {\text{d}} \sigma_{y} ^{\prime}} \right)} \right] = \frac{1}{E}\left[ {{\text{d}} \sigma_{z} - \nu \left( {{\text{d}} \sigma_{x} + {\text{d}} \sigma_{y} } \right)} \right] - \frac{1 - 2\nu }{E}{\text{d}} \sigma^{s}$$

(A6)

Substituting Eqs. (A4)-(A6) into Eqs. (A1)-(A3) gives following equations which can be solved by displacement constraints:

$$\frac{E}{{2\left( {1 + \nu } \right)}}\left( {\frac{1}{1 - 2\nu }\frac{{\partial \varepsilon_{v} }}{\partial x} + \nabla^{2} u_{x} } \right) + \frac{{\partial \sigma^{s} }}{\partial x} = 0$$

(A7)

$$\frac{E}{{2\left( {1 + \nu } \right)}}\left( {\frac{1}{1 - 2\nu }\frac{{\partial \varepsilon_{v} }}{\partial y} + \nabla^{2} u_{y} } \right) + \frac{{\partial \sigma^{s} }}{\partial y} = 0$$

(A8)

$$\frac{E}{{2\left( {1 + \nu } \right)}}\left( {\frac{1}{1 - 2\nu }\frac{{\partial \varepsilon_{v} }}{\partial z} + \nabla^{2} u_{z} } \right) + \frac{{\partial \sigma^{s} }}{\partial z} = 0$$

(A9)

where ε_v is the volumetric strain for soil skeleton and can be expressed as ε_v = ε_x + ε_y + ε_z; and u_x, u_y, and u_z are the displacements at x,y,z directions, respectively. For the infinite soil layer in Fig. 1, the displacement constraints are:

$$u_{x} = u_{y} = 0, \, u_{z} = u_{z} \left( z \right)$$

(A10)

The displacement constraint of u_x = u_y = 0 is automatically satisfied by Eq. (A7)-(A8), whereas Eq. (A9) should satisfy the following equation:

$$\frac{E}{{2\left( {1 + \nu } \right)}}\left( {\frac{1}{1 - 2\nu }\frac{{{\text{d}}^{2} u_{z} }}{{{\text{d}} z^{2} }} + \frac{{{\text{d}}^{2} u_{z} }}{{{\text{d}} z^{2} }}} \right) + \frac{{{\text{d}} \sigma^{s} }}{{{\text{d}} z}} = 0$$

(A11)

Considering the soil layer is sufficiently thin with a thickness of ΔH, the suction stress within the soil layer is homogeneous. Consequently, the vertical displacement u_z of the thin soil layer ΔH can be expressed as follows:

$$\Delta u_{z} = Az + B$$

(A12)

where A is an undetermined constant, and B is a constant which can be determined by the lower boundary, i.e., $\Delta$ u_z(z = $0$) = 0.

The soil layer shrinks freely in the vertical direction, indicating zero vertical stress, i.e., σ_z = 0. Therefore, the constant A can be determined, and the incremental form of vertical strain and displacement can be obtained:

$${\text{d}} \varepsilon_{z} = - \frac{{\left( {1 + \nu } \right)\left( {1 - 2\nu } \right)}}{1 - \nu } \cdot \frac{{{\text{d}} \sigma^{s} }}{E}$$

(A13)

$${\text{d}} u_{z} = - \frac{{\left( {1 + \nu } \right)\left( {1 - 2\nu } \right)}}{1 - \nu } \cdot \frac{{{\text{d}} \sigma^{s} }}{E} \cdot z$$

(A14)

Appendix 2

Derivations of Analytical Solutions with Gravitational Potential

To obtain the analytical solutions, the one-dimensional Richards equation of Eq. (12) is first transformed into a one-dimensional non-homogeneous differential equation of parabolic type. Then, the eigenfunction method is utilized to derive the solutions for UM-LM and UF-LM conditions.

The governing equation of Eq. (12) can be simplified with the aid of two dimensionless variables for displacement and time: χ = βz and T = βK_zst/(θ_s−θ_r):

$$\frac{{\partial^{2} \Theta }}{{\partial \chi^{2} }} + \frac{\partial \Theta }{{\partial \chi }} = \frac{\partial \Theta }{{\partial T}}$$

(B1)

An intermediate variable φ(χ,T) = Θ∙exp(χ/2 + T/4) is introduced to transform Eq. (B1) into a differential equation of parabolic type (e.g., [20, 49]). This parabolic equation can be solved with homogenized boundary conditions. Herein, this homogenization is achieved by decomposing φ(χ,T) into two undetermined transient functions $\Gamma_{g} (\chi ,T)$ and V_g(χ,T), where V_g(χ,T) is utilized to homogenize the boundary conditions (e.g., [8, 36, 50]). Accordingly, Eq. (B1) can be expressed as:

$$\frac{{\partial^{2} \Gamma_{g} }}{{\partial \chi^{2} }} = \frac{{\partial \Gamma_{g} }}{\partial T} + \left( {\frac{{\partial V_{g} }}{\partial T} - \frac{{\partial^{2} V_{g} }}{{\partial \chi^{2} }}} \right)$$

(B2)

Introducing $\delta_{g} (\chi ,T)$ to represent the last term above, Eq. (B2) can be expressed as

$$\frac{{\partial^{2} \Gamma_{g} }}{{\partial \chi^{2} }} = \frac{{\partial \Gamma_{g} }}{\partial T} + \delta_{g} (\chi ,T)$$

(B3)

The transformation above reduces the Richards equation Eq. (12) to a one-dimensional non-homogeneous differential equation of parabolic type Eq. (B3). Then, the particular solutions can be obtained with Eq. (B3) for given boundary conditions.

Derivations for condition UM-LM

In this condition, the boundary and initial conditions can be expressed as:

$$\Theta (z = H,t) = g(T)$$

(B4)

$$\Theta (z = 0,t) = 1$$

(B5)

$$\Theta (z,t = 0) = \Theta_{i} \left( \chi \right)$$

(B6)

The function V_g(χ,T) can be determined by:

$$V_{g} (\chi ,T) = \frac{1}{L}\left[ {(L - \chi ){\text{e}}^{\frac{T}{4}} + \chi {\text{e}}^{{\frac{L}{2} + \frac{T}{4}}} g(T)} \right]$$

(B7)

where L = βH. The solution of Eq. (B3) in this condition can be found via the eigenfunction method:

$$\Gamma_{g} = \sum\limits_{m = 1}^{\infty } {T_{mg} (T) \cdot \sin \frac{m\pi \chi }{L} \quad m = 1,2,3, \ldots }$$

(B8)

where

$$T_{mg} (T) = {\text{e}}^{{ - (\frac{m\pi }{L})^{2} T}} [\int_{0}^{T} {f_{mg} (T) \cdot {\text{e}}^{{(\frac{m\pi }{L})^{2} T}} {\text{d}} T} + T_{mg} {(0)}]$$

(B9)

$$f_{mg} (T) = \frac{2}{L}\int_{0}^{L} { - \delta_{g} (\chi ,T)\sin \frac{m\pi \chi }{L}{\text{d}} \chi }$$

(B10)

$$T_{mg} (0){ = }\frac{2}{L}\int_{0}^{L} {\left[ {\Theta_{i} (\chi ) \cdot {\text{e}}^{{\frac{\chi }{2}}} - V_{g} (\chi ,0)} \right]\sin \frac{m\pi \chi }{L}{\text{d}} \chi }$$

(B11)

Derivations for condition UF-LM

In this condition, the upper boundary condition can be written as:

$$\left. {\left( {\frac{\partial }{\partial \chi }\Theta + \Theta } \right)} \right|_{\chi = L} = q(T)$$

(B12)

The lower boundary and initial conditions are identical to UM-LM condition in (B1). The function V_g(χ,T) can be determined as:

$$V_{g} (\chi ,T) = \frac{{2\chi^{2} }}{{4L + L^{2} }}\left[ {q(T){\text{e}}^{{(\frac{L}{2} + \frac{T}{4})}} - \frac{1}{2}{\text{e}}^{\frac{T}{4}} } \right] + {\text{e}}^{\frac{T}{4}}$$

(B13)

The solution of Eq. (B3) in this condition can be found to be as follows:

$$\Gamma_{g} = \sum\limits_{m = 1}^{\infty } {T_{mg} (T) \cdot \sin \frac{{\upsilon_{m} \chi }}{L} \quad m = 1,2,3,...}$$

(B14)

where

$$T_{mg} (T) = {\text{e}}^{{ - \left( {\frac{{\upsilon_{m} }}{L}} \right)^{2} \cdot T}} \left[ {\int_{0}^{T} {f_{mg} (T){\text{e}}^{{\left( {\frac{{\upsilon_{m} }}{L}} \right)^{2} T}} {\text{d}} T} + T_{mg} {(0)}} \right]$$

(B15)

$$f_{mg} (T) = 2\left[ {L + \frac{1}{{\frac{1}{2} + 2\left( {\upsilon_{m} /L} \right)^{2} }}} \right]^{ - 1} \int_{0}^{L} { - \delta_{g} (\chi ,T)\sin \frac{{\upsilon_{m} \chi }}{L}{\text{d}} \chi }$$

(B16)

$$T_{mg} (0){ = }2\left[ {L + \frac{1}{{\frac{1}{2} + 2\left( {\upsilon_{m} /L} \right)^{2} }}} \right]^{ - 1} \int_{0}^{L} {\left[ {\Theta_{i} (\chi ){\text{e}}^{{\frac{\chi }{2}}} - V_{g} (\chi ,0)} \right]\sin \frac{{\upsilon_{m} }}{L}\chi {\text{d}} \chi }$$

(B17)

$$- 2\frac{{\upsilon_{m} }}{L} = \tan \upsilon_{m} \,$$

(B18)