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Triaxial testing geometry

One common example of this type of measurement uses triaxial testing [see Lockner and Stanchits (2002)], where a two-sided confining stress is defined as $ \sigma_{22} = \sigma_{33}$, and then the deviatoric stress is determined by

$\displaystyle \sigma_{dev} \equiv \left(\sigma_{11} - \sigma_{33}\right)/2.$ (51)

In this situation, the general equation relating undrained pressure to the confining stresses is given by:

$\displaystyle -p_f = B\sigma_m + 2\left(A-\frac{1}{3}\right)B\sigma_{dev},$ (52)

where the only new symbol is the first coefficient $ A$ of Skempton (1954).

1.2

TABLE 5.for the principal stiffness coefficients $ c_{ij}$ for $ i,j = 1,2,3$, as well as $ c_{44}$, of hexagonal minerals: cadmium (Cd), H$ _2$O ice, $ \beta$-quartz (SiO$ _2$), titanium (Ti), and zirconium (Zr). All data from Simmons and Wang (1971) [entry numbers: 52473, 52563, 52643, 52726, and 52798, respectively], but re-expressed in units of GPa.

Stiffness Cadmium (Cd) H$ _2$O Ice $ \beta$-Quartz Titanium (Ti) Zirconium (Zr)
$ c_{11}$ 115.30 13.85 116.6 163.9 137.0 
$ c_{33}$  51.20 14.99 110.4 181.6 160.7 
$ c_{12}$  39.24  7.07  16.7  91.3  75.6 
$ c_{13}$  40.22  5.81  32.8  68.9  65.4 
$ c_{44}$  20.40  3.19  36.1  47.2  30.1 

1.2

TABLE 6.for various measures of bulk modulus $ K$ (Voigt, Reuss, and three partial-sum moduli) for hexagonal minerals: cadmium (Cd), H$ _2$O ice, $ \beta$-quartz (SiO$ _2$), titanium (Ti), and zirconium (Zr). All data from Simmons and Wang (1971) [see TABLE 5], while the expressions in the main text were used for all the computations. All moduli in units of GPa.

Modulus Cadmium (Cd) H$ _2$O Ice $ \beta$-Quartz Titanium (Ti) Zirconium (Zr)
$ K_V$  57.89 8.90 56.47 107.51  94.17
$ K_R$  48.61 8.90 56.37 107.50  94.02
$ K_1 = K_2$ 143.07 8.94 53.97 109.00  89.58
$ K_3$  20.95 8.82 61.86 104.63 104.36

It is not difficult to show that, in terms of our previous definitions for the triaxial testing geometry, the coefficient $ A$ is given precisely by the ratio

$\displaystyle A \equiv \frac{\beta_1}{\beta_1+\beta_2+\beta_3}.$ (53)

For an isotropic system, $ A = 1/3$, so this contribution always vanishes in (54). This fact explains why I did not encounter this coefficient before in the analysis. Note that there is no assumption here that the poroelastic system itself is necessarily transversely isotropic. Only the prescribed equality of the two applied transverse stresses, $ \sigma_{22}$ and $ \sigma_{33}$, is assumed. Then, the formula (54) follows directly from the equations already presented.


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Next: Non-triaxial testing geometries Up: Deducing anisotropic drained constants Previous: Deducing anisotropic drained constants

2009-10-19