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Next: HIGHER ORDER TORSIONAL MODES Up: R. Clapp: STANFORD EXPLORATION Previous: BOUNDARY CONDITIONS

ELEMENTARY TORSIONAL MODES

The torsional mode of cylinder oscillation (which is trivial for a simple cylinder, porous or not) is determined here by a $3 \times 3$ system, of which 8 elements are in general nonzero. This system is therefore similar in size and difficulty to the cases studied earlier by Berryman (1983) for extensional waves in a simple fully saturated poroelastic cylinder. On the other hand, for extensional waves, the matrix determining the extensional wave dispersion relation for patchy saturation has 81 elements, of which 69 will in general be nonzero. This problem requires sufficiently different treatment from that for the torsional case that we set it aside to be studied fully in a future publication.

 
concentric
Figure 1
Cross-section of a circular cylinder, where $R_1 =
S^{1\over2}R_2$ is determined by the liquid saturation level S.

concentric
view

We assume that the cylinder has liquid saturation level S = (R1/R2)2, where R2 is the radius of the cylinder and r=R1 is the location of the liquid-gas interface (see Fig.1). The dispersion relation for torsional waves is then given by
   \begin{eqnarray}
\left\vert\begin{array}
{ccc}
m_{11}^*(R_2) & n_{11}^*(R_2) & 0...
 ...(R_1) & - n_{21}^*(R_1) & m_{21}(R_1)\end{array}\right\vert = 0,
 \end{eqnarray} (228)
where m11 and m21 are given by (46) and (utheta). The coefficients m11* and m12* have the same functional forms as m11 and m21, but the constants are those for the shell, rather than the inner cylinder. Similarly, n11* and n12* are just the same as m11* and m12* except that J0 and J1 are replaced everywhere by Y0 and Y1, respectively.

Now we notice immediately that there could be two elementary solutions of (detT), one with m11*(R2) = n11*(R2) = 0 (exterior condition) and another with m11(R1) = m21(R1) = 0 (interior condition). First, the interior condition is satisfied, for example, when ksr = 0 or, equivalently, when kz2 = ks2. This corresponds to a torsional mode of propagation having wave speed and attenuation determined exactly by the bulk shear wave in the interior region, but the interior region is not moving since ksr=0 also implies that $u_{\theta}=0$ from (utheta). Thus, the interior condition results in the drained outer shell twisting around a stationary inner liquid-saturated cylinder. Second, the exterior condition is similarly satisfied when ksr* = 0 or, equivalently, when kz2 = (ks*)2. This condition looks at first glance as if it might be spurious because ksr* = 0 suggests that $u_\theta$ at the exterior boundary might vanish identically, and then this would correspond to a trivial solution of the equations. However, looking closer, this is not the case, because at the external boundary  
 \begin{displaymath}
u_\theta = k_{sr}^*\left[J_1(j_s^*)\gamma_s^* + Y_1(j_s^*)\epsilon_s^*\right],
 \end{displaymath} (229)
so as $k_{sr}^* \to 0$, the first term on the right hand side of (exterior) does vanish, both because $k_{sr}^* \to 0$ and also because $J_1(j_s^*)\to 0$.But the second term does not vanish in this limit because $\vert Y_1(j_s^*)\vert\to 2/\pi k_{sr}^*R_2\to\infty$as $k_{sr}^* \to 0$,and the product gives the finite result: $2/\pi R_2$. So this condition is not spurious, and corresponds to a torsional wave propagating with the speed and attenuation of the bulk shear wave speed in the drained shell material.

Can both of these elementary modes be excited? If we assume for the moment that Gassmann's equations (1951) [also see Berryman (1999)] apply to the sample, then $\mu^* = \mu$ and the only changes in shear wave velocity in the two regions are those induced by the changes in mass. In this situation, the wave speed in the air/gas saturated region will be faster than that in the water/liquid saturated region, since liquid is more dense than gas. Thus, the real part of ks* is smaller than that of ks, and while the condition (ksr*)2 = 0 implies that the real part of ksr2 is positive, the condition ksr2 = 0 implies that the real part of (ksr*)2 is negative. Therefore, assuming (as we generally do here) that the attenuation in the system is relatively small, the condition kz = ks* leads to a propagating wave, while kz = ks leads to a strongly evanescent wave. Note that, if Gassmann's results do not apply to the system (say at ultrasonic frequencies), then the results of the preceeding paragraph may need to be reconsidered. In particular, if the shear modulus changes rapidly with the introduction of liquid saturant, it is possible that the shear wave speed for a liquid saturated porous material may be higher than that for the gas saturated case. In this situation, all the inequalities of the preceding paragraph would be reversed, and then the condition kz = ks leads to a propagating wave, while kz = ks* leads to a strongly evanescent wave.

Our conclusion then is that both modes can indeed be excited, but probably not simultaneously in the same system in the same frequency band. In a highly dispersive porous system and with broadband acoustic signal input, it could happen that both modes are propagating simultaneously in time, but in distinct/disjoint frequency bands.


next up previous print clean
Next: HIGHER ORDER TORSIONAL MODES Up: R. Clapp: STANFORD EXPLORATION Previous: BOUNDARY CONDITIONS
Stanford Exploration Project
11/11/2002