Poroelastic measurements resulting in complete data sets for granular and other anisotropic porous media

Deducing drained moduli from undrained: Isotropic system with homogeneous grains

Rewriting Gassmann's formula in these terms, I first find that

$$K^u = K^d + \frac{(1-K^d/K^g)^2}{1/K_{susp} - K^d/(K^g)^2}.$$ (3)

Note that all explicit porosity ($\phi$) dependence is now imbedded in the modulus $K_{susp}$. Now if I simply multiply through by the denominator on the right hand side, then I find

$$K^u \left(\frac{1}{K_{susp}}-\frac{K^d}{(K^g)^2}\right) = 1 - 2\frac{K^d}{K^g} + \frac{K^d}{K_{susp}}.$$ (4)

Also notice that two terms of the form $(K^d/K^g)^2$ have cancelled from this expression. Once these convenient cancellations have occurred, $K^d$ appears only linearly in the resulting expression. The equation can therefore be solved immediately for drained modulus $K^d$ in terms of the undrained modulus $K^u$ and the other factors that are also assumed to be known (and in fact these other factors are usually easier to measure than either $K^u$ or $K^d$). Finally, I obtain:

$$K^d = \left(\frac{K^u}{K_{susp}} - 1\right)\left[1/K_{susp} - 2/K^g + K^u/(K^g)^2\right]^{-1}.$$ (5)

This result shows that the drained modulus can be deduced from measurements of the undrained modulus, together with knowledge of $\phi$, $K_f$, and $K^g$. Note that this result was first derived by Zhu and McMechan (1990), but apparently published only in an SEG conference proceedings.

Poroelastic measurements resulting in complete data sets for granular and other anisotropic porous media