Benchmark Mohr Coulomb (MC) plastic material compressed by a constant velocity

Plastic Oedometer Test

In this benchmark, a square block (1 m x 1 m) of Mohr-Coulomb (MC) plastic material is compressed with a constant velocity.

Analytic Solution to the Mohr-Coulomb Oedometer Test

This is from Appendix A in Choi et al. (2013) doi:10.1002/jgrb.50148

In the setting depicted by the schematic diagram, the components of the strain increments are given by

$$ \Delta \epsilon_{xx} = \frac{v_{x} \Delta t}{L} $$

where, $L$ is the length of the side of the square domain, $\Delta t$ is time increment.

$$ \Delta \epsilon_{yy} = \Delta \epsilon_{zz} =0 $$

The corresponding stress increments in the elastic regime and stresses at any time t before yielding are given as

$$ \Delta \sigma_{xx}^{el} = \left( \lambda + 2 \mu \right) \Delta \epsilon_{xx} $$

$$ \Delta \sigma_{yy}^{el} = \Delta \sigma_{zz}^{el} = \lambda \Delta \epsilon_{xx} $$

while at the yielding, the stresses defined above make the following two yield functions simultaneously zero:

$$ f_{s,1} \left(\sigma_{xx}, \sigma_{yy} \right) = \sigma_{xx} - \frac{1+sin \phi}{1-sin \phi} \sigma_{yy} + 2C \sqrt \frac{1+sin \phi}{1-sin \phi} \ \ \ = \sigma_{xx} - N_{\phi} \sigma_{yy} + 2C \sqrt N_{\phi} $$

$$ f_{s,2} \left(\sigma_{xx}, \sigma_{zz} \right) = \sigma_{xx} - N_{\phi} \sigma_{zz} + 2C \sqrt N_{\phi} $$

Because of the inherent symmetry (i.e., $\sigma_{yy} = \sigma_{zz}$ ), from $f_{s,1}$ or $f_{s,2}$ = 0, for instance, we get the following expression for the time when yielding starts

$$ t_{yield} = \frac{CL \sqrt N_{\phi}}{\left( \left( \lambda + 2 \mu \right) - \lambda N_{\phi} \right) |v_{x}|} $$

By the same token, plastic flows need to be computed from two yield potentials,

$$ g_{s,1} \left(\sigma_{xx}, \sigma_{yy} \right) = \sigma_{xx} - \frac{1+sin \psi}{1-sin \psi} \sigma_{yy} = \sigma_{xx} - N_{\psi} \sigma_{yy} $$

$$ g_{s,2} \left(\sigma_{xx}, \sigma_{zz} \right) = \sigma_{xx} - N_{\psi} \sigma_{zz} $$

Then, plastic strain increments are given as

$$ \Delta \epsilon_{xx}^{pl} = \beta_1 \frac{\partial g_{s,1}}{\partial \sigma_{xx}} + \beta_2 \frac{\partial g_{s,2}}{\partial \sigma_{xx}} = \beta_1 + \beta_2 = 2 \beta $$

$$ \Delta \epsilon_{yy}^{pl} = \beta_1 \frac{\partial g_{s,1}}{\partial \sigma_{yy}} = -\beta_1 N_{\psi} = -\beta N_{\psi} $$

where the inherent symmetry is utilized again to identify $\beta_1$ with $\beta_2$ and denote them as $\beta$ . Collecting the previous results, we conclude that the stress increments after yielding are the following:

$$ \Delta \sigma_{xx} = \left( \lambda + 2 \mu \right) \Delta \epsilon_{xx}^{el} + \lambda \left( \Delta \epsilon_{yy}^{el} + \Delta \epsilon_{zz}^{el} \right) $$

$$ \Delta \sigma_{yy} = \left( \lambda + 2 \mu \right) \Delta \epsilon_{yy}^{el} + \lambda \left( \Delta \epsilon_{xx}^{el} + \Delta \epsilon_{zz}^{el} \right) $$

$$ \Delta \sigma_{zz} = \left( \lambda + 2 \mu \right) \Delta \epsilon_{zz}^{el} + \lambda \left( \Delta \epsilon_{xx}^{el} + \Delta \epsilon_{yy}^{el} \right) $$

By substituting the expressions for the elastic strain increments into the above equations,

$$ \Delta \epsilon_{xx}^{el} = \Delta \epsilon_{xx} - \Delta \epsilon_{xx}^{pl} = \frac{v_{x} \Delta t}{L} - 2 \beta $$

$$ \Delta \epsilon_{yy}^{el} = \Delta \epsilon_{yy} - \Delta \epsilon_{yy}^{pl} = \beta N_{\psi} $$

$$ \Delta \epsilon_{zz}^{el} = \Delta \epsilon_{zz} - \Delta \epsilon_{zz}^{pl} = \beta N_{\psi} $$

By substituting the expressions for the elastic strain increments into the above equations,

$$ \Delta \sigma_{xx} = \left( \lambda + 2 \mu \right) \frac{v_{x} \Delta t}{L} - 2 \beta + 2 \lambda \beta N_{\psi} $$

$$ \Delta \sigma_{yy} = \Delta \sigma_{zz} = \left( \lambda + 2 \mu \right) \beta N_{\psi} + \lambda \left( \frac{v_{x} \Delta t}{L} + \left( N_{\psi} - 2 \right) \beta \right) $$

Now, we solve the following incremental consistency condition for $\beta$ :

$$ \Delta f_{s,1} \left(\Delta \sigma_{xx}, \Delta \sigma_{yy} \right) = \Delta \sigma_{xx} - N_{\phi} \Delta \sigma_{yy} = 0 $$

Thus, the closed-form expression for $\beta$ is given as

$$ \beta = \frac{\left( \lambda + 2 \mu \right) - \lambda N_{\phi} }{2 \left( \lambda + \mu \right) N_{\phi} N_{\psi} + 2 \left( \lambda + 2 \mu \right) - 2 \left( N_{\phi} + N_{\psi} \right)\lambda} \frac{v_{x} \Delta t}{L} $$

Comparison with LHOGST solution

See the benchmark notebook