In this manuscript, the derivation of the Lagrange multiplier in MTest is presented. The primary objective is to compute the pore pressure under undrained conditions through the Lagrange multiplier. The undrained condition leads to an effectively incompressible material response, characterised by zero volumetric strain. The Lagrange multiplier corresponding to the zero-volumetric-strain constraint is obtained by minimising the free energy function and can be interpreted as a pore pressure. The resulting pore pressure is then compared with the pore pressure computed using OpenGeoSys.
In general, this approach can be used to interpret Lagrange multipliers associated with constraints physically.
In this document, the computation of the Lagrange multiplier \(\lambda\) associated with the and its relation to the pore pressure \(p\) developing in undrained compression tests is presented. The undrained condition refers to a loading process in which no fluid is allowed to enter or leave the porous medium, so the total fluid mass (or fluid content) in the system remains constant during deformation. As a result, changes in stress are accompanied by changes in pore pressure rather than fluid flow, leading to effective material properties that differ from drained ones. Hooke’s law for an isotropic elastic body, representing the strain-effective stress relations in the theory of poro-elasticity taking into account the effect of the pore pressure on total stress is given as follows (only normal components) : \[ \begin{aligned} \varepsilon_{xx} &= \frac{\sigma_{xx}}{E} - \frac{1-\nu}{E}\left( \sigma_{yy} + \sigma_{zz} \right) + \frac{\Delta\, p}{3H}, \\[1.5ex] \varepsilon_{yy} &= \frac{\sigma_{yy}}{E} - \frac{1-\nu}{E}\left( \sigma_{xx} + \sigma_{zz} \right) + \frac{\Delta\,p}{3H}, \\[1.5ex] \varepsilon_{zz} &= \frac{\sigma_{zz}}{E} - \frac{1-\nu}{E}\left( \sigma_{xx} + \sigma_{yy} \right) + \frac{\Delta\,p}{3H}, \end{aligned} \] where \(\dfrac{1}{H}\) is an additional parameter that denotes the poroelastic expansion coefficient, and determined as: \[ \cfrac{1}{H}= \cfrac{3 \left( 1 - 2\nu\right) \alpha}{2 \left( 1 + \nu\right) G} \quad \mathrm{with} \quad G = \frac{E}{2 \left( 1 + \nu\right)} \label{H_coeff} \]
Moreover, the variation in water content is given as a function of the volumetric strain as follows : \[ \zeta = \alpha\, \mathrm{tr} \boldsymbol{\varepsilon}+ \cfrac{\Delta p}{Q}, \label{variation_in_water_content} \] where \[ \cfrac{1}{Q} = \cfrac{1}{R} - \cfrac{\alpha}{H}. \] It is a measure of the amount of water which can be forced into the soil under pressure while the volume of the soil is kept constant. \(Q\) is called the Biot’s modulus in the literature .
In a more standard notation, these constitutive relations are written as: \[ \boldsymbol{\sigma} = \boldsymbol{\sigma}^{\mathrm{eff}}-\alpha p \boldsymbol{I} \quad \mathrm{with} \quad \boldsymbol{\sigma}^{\mathrm{eff}} = K\,\mathrm{tr}\left( \boldsymbol{\varepsilon }\right) + 2\,G\,\boldsymbol{\varepsilon}^{\mathrm{dev}} \quad \mathrm{and} \quad \boldsymbol{\varepsilon}^{\mathrm{dev}} - \cfrac{\mathrm{tr}(\boldsymbol{\varepsilon})}{3}\,\boldsymbol{I}, \] by considering an effective free energy function of the form: \[ \psi(\boldsymbol{\varepsilon}) = \cfrac{1}{2}\,K\, \mathrm{tr}^{2}\left(\boldsymbol{\varepsilon}\right) + G\,\boldsymbol{\varepsilon}^{\mathrm{dev}}:\boldsymbol{\varepsilon}^{\mathrm{dev}}, \] where the bulk and the shear moduli are calculated as: \[ K = \cfrac{E}{3\,(1-2\,\nu)} \quad \mathrm{with} \quad G=\cfrac{E}{2\,\left( 1+\nu \right)}. \]
We now consider an unconfined uniaxial compression test controlled by \(\varepsilon_{yy}=-\delta\,t\) under undrained conditions and an incompressible fluid. This condition amounts to an isochoric test, so we need to enforce \(\epsilon:=\mathrm{tr}\varepsilon=0\). To obtain the Lagrange multiplier associated with material incompressibility, the stationary point of the energy complemented by two constraints and the work of the external stresses is sought: \[ \varepsilon_{xx} + \varepsilon_{yy} + \varepsilon_{zz} = 0, \quad \mathrm{and} \quad \sigma_{xx} = \sigma_{zz}. \] The Lagrangian of the system reads: \[ \mathcal{L} = \frac{1}{2}K\,\mathrm{tr}^{2}(\boldsymbol{\varepsilon}) + G\,\boldsymbol{\varepsilon}^{\mathrm{dev}}:\boldsymbol{\varepsilon}^{\mathrm{dev}} - \lambda_1\,\mathrm{tr}(\boldsymbol{\varepsilon}) - \lambda_2 \left( \sigma_{xx} - \sigma_{zz} \right) - \boldsymbol{\sigma} : \boldsymbol{\varepsilon}, \] where \(\lambda_1\), \(\lambda_2\) are the Lagrange multipliers associated with the volumetric strain constraint and the stress constraint, respectively. The stationarity conditions are obtained by taking the first variations of the Lagrangian with respect to the unknown strain components and the Lagrange multipliers.
We introduce the deviatoric projector \[ \mathbf{A} = \mathbf{e}_x \otimes \mathbf{e}_x - \mathbf{e}_z \otimes \mathbf{e}_z, \qquad \sigma_{xx} - \sigma_{zz} = \boldsymbol{\sigma} : \mathbf{A}. \]
The variation with respect to the strain tensor yields \[ \delta_\varepsilon \mathcal{L} = \left( K\,\mathrm{tr}(\boldsymbol{\varepsilon})\,\mathbf{I} + 2G\,\boldsymbol{\varepsilon}^{\mathrm{dev}} - \lambda_1\,\mathbf{I} - 2G\,\lambda_2\,\mathbf{A} - \boldsymbol{\sigma} \right) : \delta \boldsymbol{\varepsilon} = 0, \] which must hold for all \(\delta\boldsymbol{\varepsilon}\) and thus implies \[ K\,\mathrm{tr}(\boldsymbol{\varepsilon})\,\mathbf{I} + 2G\,\boldsymbol{\varepsilon}^{\mathrm{dev}} - \lambda_1\,\mathbf{I} - 2G\,\lambda_2\,\mathbf{A} - \boldsymbol{\sigma} = \mathbf{0}. \]
The variations with respect to the Lagrange multipliers give the constraints \[ \delta_{\lambda_1} \mathcal{L} = -\mathrm{tr}(\boldsymbol{\varepsilon})\,\delta \lambda_1 = 0, \qquad \delta_{\lambda_2} \mathcal{L} = -(\sigma_{xx} - \sigma_{zz})\,\delta \lambda_2 = 0, \] so that \[ \mathrm{tr}(\boldsymbol{\varepsilon}) = 0, \qquad \sigma_{xx} = \sigma_{zz}. \]
From the stationarity condition with respect to \(\boldsymbol{\varepsilon}\) we can express the (given) stress tensor as \[ \boldsymbol{\sigma} = K\,\mathrm{tr}(\boldsymbol{\varepsilon})\,\mathbf{I} + 2G\,\boldsymbol{\varepsilon}^{\mathrm{dev}} - \lambda_1\,\mathbf{I} - 2G\,\lambda_2\,\mathbf{A}. \] Using the result of the first constraint, \(\mathrm{tr}(\boldsymbol{\varepsilon}) = 0\), this simplifies to \[ \boldsymbol{\sigma} = 2G\,\boldsymbol{\varepsilon}^{\mathrm{dev}} - \lambda_1\,\mathbf{I} - 2G\,\lambda_2\,\mathbf{A}. \]
For a uniaxial compression test with \(\varepsilon_{yy}\) as the control variable, the volumetric constraint \[ \mathrm{tr}(\boldsymbol{\varepsilon}) = \varepsilon_{xx} + \varepsilon_{yy} + \varepsilon_{zz} = 0 \] together with the symmetry condition \(\varepsilon_{xx} = \varepsilon_{zz}\) yields \[ \varepsilon_{xx} = \varepsilon_{zz} = -\frac{1}{2}\,\varepsilon_{yy}. \]
Inserting this strain state into the stationarity condition \[ \sigma_{ij} = 2G\,\varepsilon_{ij}^{\mathrm{dev}} - \lambda_1\,\delta_{ij} - 2G\,\lambda_2\,A_{ij}, \qquad A_{ij} = \mathrm{diag}(1,0,-1), \] gives the componentwise relations \[\begin{align} \sigma_{xx} &= 2G\,\varepsilon_{xx} - \lambda_1 - 2G\,\lambda_2,\\ \sigma_{yy} &= 2G\,\varepsilon_{yy} - \lambda_1,\\ \sigma_{zz} &= 2G\,\varepsilon_{zz} - \lambda_1 + 2G\,\lambda_2. \end{align}\]
The second constraint, \(\sigma_{xx} = \sigma_{zz}\), then reduces to \[ -2G\,\lambda_2 = +2G\,\lambda_2 \qquad\Longrightarrow\qquad \lambda_2 = 0. \]
For a uniaxial test the lateral stresses vanish, \[ \sigma_{xx} = \sigma_{zz} = 0. \] Using \(\lambda_2=0\) and \(\varepsilon_{xx} = -\tfrac12\varepsilon_{yy}\) in \(\sigma_{xx} = 2G\varepsilon_{xx} - \lambda_1\) gives \[ 0 = 2G\!\left(-\frac{1}{2}\varepsilon_{yy}\right) - \lambda_1, \qquad\Longrightarrow\qquad \lambda_1 = -G\,\varepsilon_{yy}. \]
Thus, the Lagrange multipliers for the uniaxial compression case are \[ \boxed{ \lambda_1 = -G\,\varepsilon_{yy}, \qquad \lambda_2 = 0. } \]
Finally, the resulting axial stress follows as \[ \sigma_{yy} = 2G\,\varepsilon_{yy} - \lambda_1 = 3G\,\varepsilon_{yy}. \]
The Lagrange multiplier is then scaled by the factor of \(K+\cfrac{4\,G}{3}\), which is the first component of the tangent operator. Finally, we get: \[ \lambda^{*}_1 = \cfrac{\lambda_1}{K + 4\,G/3} \]
The Lagrange multiplier \(\lambda_1\) represents the pore pressure under undrained conditions, i.e. \(p=\lambda_1\).
The comparison between the pore pressure obtained with OGS and the Lagrange multiplier computed with MTest is presented in Figure 1.