next up previous contents
Next: Example 3: Plasticity in Up: Examples Previous: Example 1: Backward facing   Contents

Example 2: Confined compression of fluid filled porous material

In this 1D example, a porous solid saturated with fluid is compressed by a force F on the solid phase. The length of the test specimen is L, plane strain conditions hold, the Young's modulus is E, Poisson's ratio is $\nu$, permeability kp, width of test specimen W, thickness T, area A=WT. Some non-dimensional variables are introduced

\begin{displaymath}
X = x/L \; \; \; \; \;
T = \frac{Hk_p}{L^2} t \; \; \; \; \;
P = \frac{Ap}{F}
\end{displaymath}


\begin{displaymath}
H=\frac{E(1-\nu)}{(1+\nu)(1-2\nu)}
\end{displaymath}

At X=1 free drainage holds (P=0) for T>0. At X=0 the boundary is not permeable ( $\frac{\partial P}{\partial X} = 0$) for T>0. Using the non-dimensional variables, the analytical solution reads (see [1])

\begin{displaymath}
P = \sum_{n=0}^{\infty} \frac{2}{M} \sin(MY) \exp (-M^2T)
\end{displaymath}


\begin{displaymath}
M = \frac{\pi}{2} (2n+1)
\end{displaymath}

As a typical result, P at T=0.22 is checked; at X=0.4 the exact solution for the non-dimensional pressure P is 0.6.

Now we choose: L=1, W=1, T=1, E=1.e1, $\nu=0$, kp=0.1 and F=1. Thus, for x=0.4 at t=0.22 we should find p=0.6. The numerical analysis (5 elements, time step size 0.01) gives p=0.607. Note that relaxation on the pressure in combination with iterations is used (because the compressibility is taken zero to model a fully incompressible fluid phase). The pressure distribution at T=0.22 is given below

\begin{figure}
\centerline{\epsfig{file=ps/ex2pres.ps,width=4cm}}\end{figure}


next up previous contents
Next: Example 3: Plasticity in Up: Examples Previous: Example 1: Backward facing   Contents
root
1998-11-16