4. Integrating the behavioral relationship

A time discretization in finite differences is used to obtain the expressions of the components of the effective stress matrix. The calculation steps to arrive at the results are listed below.

The following notations are used:

\(\dot{{\sigma }_{{m}_{}}}=\frac{{\sigma }_{{m}_{n+1}}-{{\sigma }_{m}}_{n}}{\mathit{dt}}\)

Where \({{\sigma }_{m}}_{n}\) and \({{\sigma }_{m}}_{n+1}\) are respectively the average constraints effective at the previous time and at the current time and dt the time step.

Using the temporal discretization of \(\stackrel{´}{{\sigma }_{{m}_{}}}\) and \(\stackrel{´}{{s}_{\mathit{ij}}}\), we look for the expressions for \({\sigma }_{{m}_{n+1}}\) and \({s}_{i{j}_{n+1}}\):

\(\dot{{\epsilon }_{{v}_{n+1}}}=\frac{1}{{\eta }_{v}}{\sigma }_{{m}_{n+1}}+\frac{1}{K}\frac{{\sigma }_{{m}_{n+1}}-{{\sigma }_{m}}_{n}}{\mathit{dt}}=\left(\frac{1}{{\eta }_{v}}+\frac{1}{K\mathit{dt}}\right){\sigma }_{{m}_{n+1}}-\frac{{{\sigma }_{m}}_{n}}{K\mathit{dt}}\)

\(\dot{{e}_{i{j}_{n+1}}}=\frac{1}{{\eta }_{d}}{s}_{i{j}_{n+1}}^{\text{'}}+\frac{1}{2G}\frac{{s}_{i{j}_{n+1}}-{{s}_{\mathit{ij}}}_{n}}{\mathit{dt}}=\left(\frac{1}{{\eta }_{d}}+\frac{1}{2G\mathit{dt}}\right){s}_{i{j}_{n+1}}-\frac{{{s}_{\mathit{ij}}}_{n}}{2G\mathit{dt}}\)

Which is equivalent to:

\({\sigma }_{{m}_{n+1}}=\left(\dot{{\epsilon }_{{v}_{n+1}}}+\frac{{{\sigma }_{m}}_{n}}{K\mathit{dt}}\right)\frac{K\mathit{dt}{\eta }_{v}}{{\eta }_{v}+K\mathit{dt}}\)

\({s}_{i{j}_{n+1}}=\left(\dot{{e}_{i{j}_{n+1}}}+\frac{{{s}_{\mathit{ij}}}_{n}}{2G\mathit{dt}}\right)\frac{2G\mathit{dt}{\eta }_{d}}{{\eta }_{d}+2G\mathit{dt}}\)

Which is ultimately equivalent to:

\({\sigma }_{{m}_{n+1}}=\left(\mathit{dt}\dot{{\epsilon }_{{v}_{n+1}}}+\frac{{{\sigma }_{m}}_{n}}{K}\right)\frac{K}{1+\frac{\mathit{Kdt}}{{\eta }_{v}}}\)

\({s}_{i{j}_{n+1}}=\left(\mathit{dt}\dot{{e}_{i{j}_{n+1}}}+\frac{{{s}_{\mathit{ij}}}_{n}}{2G}\right)\frac{2G}{1+\frac{2\mathit{Gdt}}{{\eta }_{d}}}\)

that can be written in incremental form:

\({\sigma }_{{m}_{n+1}}^{\text{'}}=\left(\Delta {\epsilon }_{v}(t)+\frac{{{\sigma }_{m}^{\text{'}}}_{n}}{K}\right)\frac{K}{1+\frac{\mathit{Kdt}}{{\eta }_{v}}}\)

\({s}_{i{j}_{n+1}}^{\text{'}}=\left(\Delta {e}_{\mathit{ij}}\left(t\right)+\frac{{{s}_{\mathit{ij}}^{\text{'}}}_{n}}{2G}\right)\frac{2G}{1+\frac{2\mathit{Gdt}}{{\eta }_{d}}}\)