Results for all solids

If there are two solids present, for our present purposes of averaging we may lump them together and act as if they are both generic ``solids,'' using subscript ``s'' to refer to the regions occupied. The averaging theorem (or in this case just the normal divergence theorem) states that

<u_s> = 1V_() u_s d^3x = 1V_E B<>n_su_sdS + 1V_I B<>n_su_sdS,   where $\Omega$ is the averaging volume and $V = \int_\Omega\, d^3x$,with $\partial E$ being the external boundary and $\partial I$ being the internal or pore boundary. The divergence of the average (found by taking the Frechet derivative with respect to the averaging volume) is

<u_s> = 1V_E B<>n_su_sdS.   The internal surface integral is easily interpreted as the negative of the change in porosity $\delta\phi$, since the displacement integrated along the bounding surface produces a volume change that is then normalized by the total volume V in the averaging volume $\Omega(\vecx)$, so

1V_I B<>n_su_sdS = - .   The left hand side of (divergencetheorem) is also easily interpreted as the total change in solid volume $\delta[V(1-\phi)]$ divided by the averaging volume V. Comparing these expressions shows that

<u_s> = (1-)VV.   Note that there has been no change in the averaging volume $\Omega$, but there has been movement of solid in or out of volume and/or a change in state of compression of the solid. In this regard, our notation is trying to reflect the fact that experimentally we normally start with a fixed volume of material V and then measure changes $\delta V$ in that volume.

Accounting for the volume occupied by the solid initially, we have $\left<{\bf u}_s\right\gt \equiv (1-\phi)\bar{\bf u}_s$, so

B<>u_s = VV + B<>u_s 1-.   We normally neglect the second term on the right hand side of (usrelation), since we assume that the scales of variation of the displacement field are much smaller than those for the porosity, so that $\left\vert \bar{\bf u}_s \cdot \nabla \phi\right\vert << \left\vert\nabla\cdot\bar{\bf u}_s\right\vert$.However, a completely general analysis must account for the presence of this term.