Simulation of long-term thermo-mechanical response of clay using an advanced constitutive model

Abstract

There is extensive data to show that heating and cooling produces irrecoverable deformations in clays under fully drained conditions. The effects are most pronounced for normally and lightly overconsolidated clays that undergo significant compression. Most constitutive models have key limitations for predicting the thermo-mechanical response of clays through long-term (seasonal) cycles of heating and cooling. The Tsinghua ThermoSoil model (TTS; Zhang and Cheng in Int J Numer Anal Methods Geomech 41(4):527–554, 2017) presents a novel theoretical framework for simulating the coupled thermo-mechanical response of clays. The model uses a double-entropy approach to capture effects of energy dissipation at the microscopic particulate contact level on continuum behavior. This paper proposes a simple procedure for calibrating input parameters and illustrates this process using recent laboratory data for Geneva Clay (Di Donna and Laloui in Eng Geol 190:65–76, 2015). We then investigate capabilities of the TTS model in simulating familiar aspects of thermal consolidation of clays as well as the long-term, progressive accumulation of strains associated with seasonal heating and cooling processes for shallow geothermal systems installed in clays. The model predicts the existence of a long-term steady-state condition where there is no further accumulation of strain. This state depends on the consolidation stress and stress history but is independent of the imposed range of temperature, T_cyc. However, the value of T_cyc does affect the rate of accumulation of strain with thermal cycles. Simulations for normally consolidated Geneva Clay find steady-state strain conditions ranged from 2.0 to 3.7% accumulating within N = 10–50 thermal cycles.

Appendices

Appendix A

The TTS model accounts for strain rate dependence through parameter a shown in Eqs. 6b and 6c. Figure 18 presents hydrostatic compression of Geneva Clay, as simulated by the TTS model, assuming different strain rates. Using a = 0.5 results in a unique response (cf., solid lines overlap in Fig. 18), providing rate independence of the model. On the other hand, using a = 0.3 (dashed lines) results in different VCLs for different strain rates. An increase in rate of strain results in a decrease in density (increased void ratio) at a given effective consolidation stress (i.e., the void ratio is higher at $\dot{\varepsilon }_{v} = 0.05/\hbox{min}$ than at $\dot{\varepsilon }_{v} = 0.001/\hbox{min}$ at a given effective consolidation stress). The effect of strain rate is similar to the effect of temperature (cf., Fig. 6a) since they both affect the viscous deformation of soils, as reported previously in the literature (e.g., [35]).

Fig. 18

Effect of input constant, a, on rate dependence assumed in TTS model

Appendix B

Panagiotidou [42] studied the TTS model behavior at critical state assuming undrained triaxial shearing with axial strain $\dot{\varepsilon }_{a}$, under isothermal conditions (i.e., $\dot{T} = 0$). The resulting total volumetric and deviatoric strain rates are $\dot{\varepsilon }_{v} = 0$ and $\dot{\varepsilon }_{s} = \sqrt {{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern-0pt} 3}} \dot{\varepsilon }_{a}$, respectively. At critical state, the soil reaches steady-state conditions with constant deformations without change in volume or stresses $\dot{\varepsilon }_{vf} = \Delta p_{f}^{{\prime }} = \Delta q_{f} = 0$ and so all internal state variables are constant (cf., Table 1). Therefore, since the change of elastic strain is zero, the plastic strain is equal to the total applied strain:

$$\dot{\varepsilon }_{v}^{D} = \dot{\varepsilon }_{v} = 0$$

(16a)

$$\dot{\varepsilon }_{s}^{D} = \dot{\varepsilon }_{s} = \sqrt {{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern-0pt} 3}} \dot{\varepsilon }_{a}$$

(16b)

From Eq. 5, under isothermal conditions and assuming that $\dot{T}_{g} = 0$, the granular temperature at critical state becomes:

$$T_{gCS} = m_{2} \left( {\dot{\varepsilon }_{s} } \right)^{2}$$

(17)

From Eq. 7b, given that $\dot{\varepsilon }_{s}^{h} = 0$ and $\dot{\varepsilon }_{v}^{D} = 0$:

$$\varepsilon_{s}^{h} = h$$

(18)

For $\dot{\varepsilon }_{v}^{h} = 0$, it is deduced from Eq. 7a that:

$$\varepsilon_{v}^{h} = 0$$

(19)

For $\dot{\varepsilon }_{v}^{D} = 0$ and using Eq. 6a:

$$\varepsilon_{v}^{e} = \varepsilon_{v}^{h} = 0$$

(20)

From Eq. 6b and assuming rate independence (i.e., a = 0.5):

$$\begin{aligned} & \dot{\varepsilon }_{s}^{D} = \left( {T_{\rm g} } \right)^{\alpha } \left( {\varepsilon_{s}^{e} - \varepsilon_{s}^{h} } \right) = \left[ {m_{2} \left( {\dot{\varepsilon }_{s}^{D} } \right)^{2} } \right]^{\alpha } \left( {\varepsilon_{s}^{e} - \varepsilon_{s}^{h} } \right) \\ & \because \varepsilon_{s}^{e} = h + \frac{1}{{\sqrt {m_{2} } }} \\ \end{aligned}$$

(21)

The evolution of the stress components at critical state can then be calculated (cf., Eq. 11):

$$p_{f}^{{}} = 1.5B\xi \left( {c^{\prime } } \right)^{0.5} \left( {\varepsilon_{s}^{e} } \right)^{2}$$

(22a)

$$q_{f} = \sqrt 6 B\xi \left( {c^{\prime } } \right)^{1.5} \varepsilon_{s}^{e}$$

(22b)

Therefore, the slope of the critical state line, M, is given by the ratio of the deviatoric to the mean effective stress:

$$M = \frac{{q_{f} }}{{p_{f}^{\prime } }} = \frac{{\sqrt 6 c^{\prime } }}{{1.5\left( {h + \frac{1}{{\sqrt {m_{2} } }}} \right)}}$$

(23)