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Determining off-diagonal $ \beta_i$ coefficients

I will now provide several results for the $ \beta_i$ coefficients, and then follow the results with a general proof of their correctness.

In many useful and important cases, the coefficients $ \beta_i$ are determined by

$\displaystyle \beta_i = s_{i1}^d + s_{i2}^d + s_{i3}^d - \frac{1}{3K_R^g}.$ (10)

Again, $ K_R^g$ is the Reuss average of the grain modulus, since the local grain modulus is not necessarily assumed uniform here as discussed previously. Equation (10) holds as written for homogeneous grains, such that $ K_R^g \equiv K^g$ . It also holds true for the case when $ K_R^g$ is determined instead (Wood, 1955; Hashin, 1962; Reuss, 1929) by

$\displaystyle \frac{1}{K_R^g} \equiv \sum_{m=1,\ldots,n}\frac{v_m}{K_m},$ (11)

where $ v_m$ is the volume fraction (out of all the solid material present, so that $ \sum_m v_m = 1$ ). However, when the grains themselves are uniform but anisotropic, I need to allow again for this possibility, and this can be accomplished by defining three directional grain bulk moduli determined by:

$\displaystyle \frac{1}{3\overline{K}_i^g} \equiv s_{i1}^g + s_{i2}^g + s_{i3}^g = s_{1i}^g + s_{2i}^g + s_{3i}^g,$ (12)

for $ i = 1,2,3$ . The second equality follows because the compliance matrix is always symmetric. I call these quantities in (12) the ``partial grain-compliance sums,'' and the $ \overline{K}_i^g$ are the directional grain bulk moduli. Note that the factors of three have again been correctly accounted for because in agreement with (5):

$\displaystyle \sum_{i=1,2,3} \frac{1}{3\overline{K}_i^g} = \frac{1}{K_R^g},$ (13)

I can also simplify and symmetrize our notation somewhat by introducing a similar concept for the drained constants, so that

$\displaystyle \frac{1}{3\overline{K}_i^d} \equiv s_{i1}^d + s_{i2}^d + s_{i3}^d = s_{1i}^d + s_{2i}^d + s_{3i}^d,$ (14)

for $ i = 1,2,3$ . Then, the formula for (10) is replaced by

$\displaystyle \beta_i = \frac{1}{3\overline{K}_i^d} - \frac{1}{3\overline{K}_i^g}.$ (15)

If the three contributions represented by (12) for $ i = 1,2,3$ happen to be equal, then clearly each equals one-third of the sum (13).

The preceding results are for perfectly aligned grains. If the grains are instead perfectly randomly oriented, then it is clear that the formulas in (10) hold as before, but now $ K_R^g$ is determined instead by (5).

All of these statements about the $ \beta_i$ 's are easily proven by considering the situation when $ \sigma_{11} = \sigma_{22} = \sigma_{33} = -p_c = -p_f$ , where $ p_c$ is uniform external confining pressure, and $ p_f$ is the internal fluid pressure. Because then, from (1), I have (since by my assumption $ p_c = p_f$ ):

$\displaystyle -e_{ii} = <tex2html_comment_mark>32 \frac{1}{3\overline{K}_i^d}p_...
...(s_{i1}^g + s_{i2}^g + s_{i3}^g\right)p_f \equiv \frac{p_f}{3\overline{K}_i^g},$ (16)

in the most general of the three cases discussed, and holding true for each value of $ i = 1,2,3$ . This is a statement about the strain $ e_{ii}$ that would be observed in this situation, as it must be the same if these anisotropic (or inhomogeneous) grains were immersed in the fluid, while measurements were taken of the strains observed in each of the three directions $ i = 1,2,3$ , during variations of the fluid pressure $ p_f$ . I consider this proof to be a thought experiment for determining these coefficients, in the same spirit as those proposed originally by Biot and Willis (1957) and Biot (1962) for the isotropic and homogeneous case.

next up previous [pdf]

Next: The coefficients and effective Up: BASICS OF ANISOTROPIC POROELASTICITY Previous: Orthotropic poroelasticity