3.7 Multiphysics problems

3.7.1 Poro-elasticity

When rocks deform, most of the deformation transfers into changes of porosity. As a matter of fact, the solid phase can also deform and change volume. In that case the equation that links stresses and deformations $\uuline{\varepsilon} = \uuline{D} \: \uuline{\sigma}$ needs to be corrected according to Biot's effective stress:

$$\uuline{\sigma} = \uuline{S} - \alpha P_p \uuline{I}$$ (3.41)

where $\alpha$ is the Biot coefficient and $\uuline{I}$ is the identity tensor. In the case of linear elastic and isotropic porous media, the Biot coefficient is:

$$\alpha = 1 - \frac{K_{drained}}{K_{unj}}$$ (3.42)

where $K_{drained}$ is the drained bulk modulus of the porous solid and $K_{unj}$ is the unjacketed bulk modulus. The unjacketed bulk modulus is equal to the mineral bulk modulus $K_{unj} = K_{m}$ when all porosity is connected and the rock has a mono-mineral composition. The values of the Biot coefficient range from the value of porosity to 1. Most rocks and sediments subjected to large depths have a Biot coefficient that ranges from $\sim$ 0.4 to 0.95, decreasing in value as rocks get stiffer. In anisotropic media, the poroelasticity correction factor becomes a tensor $\uuline{\alpha}$. The corrections for poroelasticity $\alpha \neq 1$ become significant in tight rocks with high stiffness, and low and unconnected porosity.

3.7.2 Thermo-elasticity

Changes of temperature in solids change the equilibrium distance between molecules, and therefore induce strains. Imagine a solid heated up, but not allowed to dilate in vertical direction (Figure 3.28). The result is an increase of stress in vertical direction rather than a deformation in vertical direction.

Figure 3.28: Thermo-elasticity example of dilation thermal stress.

The coefficient of thermal dilation $\alpha_L$ quantifies strains as a function of changes of temperature $T$ at constant pressure $p$, and it is defined as

$$\alpha_L = \left. \cfrac{1}{L} \cfrac{ \mathrm{d} L}{ \mathrm{d} T} \right\vert _p$$ (3.43)

Typical linear thermal dilation coefficients of rock range from 5 to 10 $\times 10^{-6}$ 1/C$^{\circ }$. You may compare this range to the linear thermal dilation coefficient of steel $12\times 10^{-6}$ 1/C$^{\circ }$ and water $70\times 10^{-6}$ 1/C$^{\circ }$.

Under unconstrained (displacement) conditions, a negative change in temperature causes shrinkage and a positive change in temperature causes dilation. The elastic equations extended to consider thermal changes make explicit that stresses can be changed as a result of a change in temperature $\Delta T$ and/or as a result of a change of volumetric strains.

$$\left\lbrace \begin{array}{rcl} \sigma_{11} & = & (\lambda + ... ...igma_{12} & = & 2 \mu \: \varepsilon_{12}\\ \end{array}\right.$$ (3.44)

Eq. 3.44 does not include the effects of pore pressure. The coupled thermo-poro-elastic equations are covered in the Advanced Geomechanics course.

Thermal(-induced) stresses can cause reductions in fracture gradient when drilling with relatively cold drilling mud. Thermal(-induced) stresses can also cause enhanced fractured reactivation when injecting cold fluids in a hot reservoir, such as in deep geothermal energy recovery.

EXAMPLE 3.7: Derive an expression of the thermal swelling stress (in vertical direction) for the example shown in Figure 3.28.

SOLUTION
Let us assume the axis 3 in the vertical direction, then $\sigma_{11}=\sigma_{22}=0$, $\varepsilon_{11}=\varepsilon_{22} \neq 0$, $\varepsilon_{33}=0$. Then, for $\Delta T \neq 0$, a simplification of Eq. 3.44 results in,

$$\left\lbrace \begin{array}{rcl} 0 & = & (\lambda + 2 \mu) \: ... ...\varepsilon_{11} + 3 K \alpha_L \Delta T \\ \end{array}\right.$$

Solving for $\varepsilon_{11}$ and pluging it the $\sigma_{33}$ equation results in

$$\sigma_{33} = \left( \frac{6-\mu/K}{3+\mu/K} \right) K \alpha_L \Delta T \: \: \blacksquare$$