Theoretical analysis of desiccation crack spacing of a thin, long soil layer

Abstract

Soil desiccation cracking is important for a range of engineering applications, but the theoretical advancement of this process is less than satisfactory. In particular, it is not well understood how the crack spacing-to-depth ratio depends on soil material behaviour. In the past, two approaches, namely stress relief and energy balance, have been used to predict the crack spacing-to-depth ratio. The current paper utilises these two approaches to predict the approximate spacing-to-depth ratio of parallel cracks that form in long desiccating soil layers subjected to uniform tensile stress (or suction profile) while resting on a hard base. The theoretical developments have examined the formation of simultaneous and sequential crack patterns and have identified an important relationship between the stress relief and energy approaches. In agreement with experimental observations, it was shown that the spacing-to-depth ratio decreases with layer depth, and crack spacing generally increases with layer depth. The influence of the stiffness at the base interface indicated that decreasing the basal interface stiffness makes the crack spacing to increase in sequential crack formation. The experimental observations also show a decrease in cracking water content with the decrease in layer thickness, and this behaviour was explained on the basis of a critical depth concept.

Appendix

Approximate analytical solution for horizontal stress relief due to an isolated crack in an elastic layer with a hard base.

A solution is presented for a horizontal stress $\sigma_{xc}$ for the situation depicted in Fig. 1c. The governing equations can be described as follows. The sign convention uses compressive stress as positive and clockwise shear as positive. This is traditional in soil mechanics. For horizontal equilibrium,

$$\frac{{\partial \sigma_{xc} }}{\partial x} = - \frac{{\partial \tau_{xy} }}{\partial y}$$

(16)

where $\tau_{xy}$ is the shear stress in xy plane. Ignoring normal stress change in y direction and considering plane strain condition, $\sigma_{\text{xc}}$ can be expressed as:

$$\sigma_{\text{xc}} = - \frac{E}{{(1 - \nu^{2} )}}\frac{\partial u}{\partial x}$$

(17)

where u is the shear displacement in x direction. E and $\nu$ are the Young’s modulus and Poisson’s ratio of the soil, respectively. The shear stress is assumed to follow a linear stress distribution as follows:

$$\tau_{xy} = \left( {\frac{y}{d}} \right)\tau_{\text{b}}$$

(18)

where $\tau_{\text{b}}$ is the shear stress at the basal interface. Assuming a spring of shear stiffness of k to represent the basal interface characteristics, $\tau_{\text{b}}$ can be related to the shear displacement at the base u _b as:

$$\tau_{\text{b}} = ku_{\text{b}}$$

(19)

Ignoring the shear strain component associated with vertical displacement, the shear stress $\tau_{xy}$ can be expressed:

$$\tau_{xy} = - G\frac{\partial u}{\partial y}$$

(20)

Equations (16), (18) and (19) can be rearranged to give:

$$\frac{{\partial^{2} \sigma_{\text{xc}}^{{}} }}{{\partial x^{2} }} = - \frac{k}{d}\frac{{\partial u_{\text{b}} }}{\partial x}$$

(21)

Also, Eqs. (18), (19) and (20) can be rearranged to obtain:

$$\frac{\partial u}{\partial y} = - \frac{k}{G}u_{\text{b}} \frac{y}{d}$$

(22)

Equation (22) can be solved for u by using the boundary condition $u = u_{\text{b}} , \,y = d.$ Using this result and Eq. (17), it is possible to obtain the following expression from Eq. (21).

$$\frac{{\partial^{2} \sigma_{xc}^{{}} }}{{\partial x^{2} }} - \frac{1}{{d^{2} }}\frac{1}{{\left[ {\frac{1}{{\left( {1 - \nu } \right)}} + \frac{E}{kd}\frac{1}{{\left( {1 - \nu^{2} } \right)}}} \right]}}\sigma_{xc} = 0$$

(23)

Equation (23) can be solved subject to the boundary conditions: $\sigma_{xc} = \sigma_{0} {\text{ when }}x = 0$; ${\text{and }}\sigma_{xc} \to 0 {\text{ when }}x \to \infty \,$. The solution for $\sigma_{xc}^{y = 0}$ in non-dimensional form can be expressed as:

$$\frac{{\sigma_{xc}^{y = 0} }}{{\sigma_{0} }} = e^{{ - \frac{1}{{\sqrt {\frac{1}{{\left( {1 - v} \right)}} + \frac{E}{kd}\frac{1}{{\left( {1 - \nu^{2} } \right)}}} }}\left( {\frac{x}{d}} \right)}} .$$

(24)