2. Analytical solution

First, let’s introduce the following notations:

(2.1)\[\begin{split} \ {\ begin {array} {c} A=K+\ frac {4} {3} G\\ B=K-\ frac {2} {3} G\\ C=2\ left (K+\ frac {G} {G} {3}\ right)\ end {array}\end{split}\]

With \(K=\frac{E}{3\left(1-2\mathrm{\nu }\right)}\) and \(G=\frac{E}{2\left(1+\mathrm{\nu }\right)}\) the compressibility and shear modules, respectively.

Let \(C\) be the Hooke elasticity tensor, we will have with hypothesis \({\mathrm{ϵ}}_{\mathit{yy}}={\mathrm{ϵ}}_{\mathit{xx}}\):

(2.2)\[\begin{split} C\ mathrm {.} d\ mathrm {} =\ {\ begin {array} =\ {\ begin {array} {c}} Bd {\ mathrm {zz}} +Cd {\ mathrm {}}} =\ {\ mathit {xx}} _ {\ mathit {xx}} _ {\ mathit {zz}}} +Cd {\ mathrm {}} _ {\ mathit {xx}}}\\ Ad {\ mathrm {E.G.}} _ {\ mathit {zz}} +2Bd {\ mathrm {}}} _ {\ mathit {xx}}\ end {array}} _ {\ mathrm {xx}}\ end {array}}\end{split}\]

To simplify, we note the vertical constraint at the time \(\text{+}\) \({\mathrm{\sigma }}^{\text{+}}={\mathrm{\sigma }}_{\mathit{zz}}^{\text{+}}\), so that the Rankine criterion is written:

(2.3)\[ {\ mathrm {\ sigma}}} ^ {\ text {+}}\ the {\ mathrm {\ sigma}} _ {t}\]

In addition, we have:

\[\]

: label: eq-4

{begin {array} {c} {mathrm {sigma}}} ^ {text {pred}} = {mathrm {sigma}} ^ {text {-}} +nmathrm {.} +nmathrm {.}.} Cmathrm {.}} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {.} dmathrm {sigma}} ^ {text {-}} +nmathrm {.} Cmathrm {.} left (dmathrm {}} -dmathrm {lambda} nright) = {mathrm {sigma}} ^ {text {pred}} -underset {mathrm {thrm {delta}} {delta}} {mathrm {sigma}} {mathrm {sigma}}} _ {sigma}}} {underset {} {dmathrm {lambda}} nmathrm {delta}} {mathrm {delta}} {mathrm {sigma}}} _ {sigma}} _ {sigma}}} {underset {}} {dmathrm {lambda}} nmathrm {delta}} {.} Cmathrm {.} n}}}end {array}

With \(n=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\) and where:

\[\]

: label: eq-5

dmathrm {lambda} =frac {{⟶ {mathrm {sigma}}} ^ {mathit {pred}} - {mathrm {sigma}}} - {sigma}} _ {t} ⟩} __ {t} ⟩}

According to the associated flow law, we also have:

(2.4)\[ \ {\ begin {array} {c} d {\ mathrm {}}} _ {\ mathit {zz}}} ^ {P} =d\ mathrm {\ lambda} =d {\ mathrm {}}} _ {v}}}} _ {v}}}} _ {v} ^ {P}} _ {v} ^ {P}} =\ frac {2} {3} d\ mathrm {\ lambda}}} =\ frac {2} {3} d\ mathrm {\ lambda}}} _ {v}}} _ {v} ^ {P}} _ {v} ^ {P}} =\ frac {2} {3} d\ mathrm {array}\]

Like \(n=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\), we get:

(2.5)\[ \ mathrm {\ Delta} {\ mathrm {\ sigma}} _ {C} =d\ mathrm {\ lambda} A\]

Combining equations (), (), and () gives us constraint \({\mathrm{\sigma }}_{\mathit{zz}}^{\text{+}}\). Equations () and () give us the norm of deviatoric plastic deformation \({e}^{P}\).

Now let’s try to get the expression for horizontal elastic deformation \({\mathrm{ϵ}}_{\mathit{xx}}^{\text{élas}}\).

Laterally, we have condition \({\mathrm{\sigma }}_{\mathit{xx}}^{\text{+}}={P}_{0}\), which is:

(2.6)\[ {\ mathrm {\ sigma}} _ {\ mathit {xx}}} ^ {\ text {pred}} -\ mathrm {\ delta} {\ mathrm {\ sigma}}} _ {\ mathit {xx}}} _ {\ mathit {xx}}, C} = {P} _ {0}\]

With \(\mathrm{\Delta }{\mathrm{\sigma }}_{\mathit{xx},C}=d\mathrm{\lambda }B\)

Using equation (), we then obtain:

(2.7)\[ {\ mathrm {\ sigma}} _ {\ mathit {xx}}} ^ {\ mathit {xx}}} ^ {\ mathrm {-}} _ {\ mathit {xx}} ^ {\ text {alas}}} ^ {\ text {alas}}} +Bd {\ mathit {xx}} -Bd\ mathrm {\ lambda}} ^ {\ text {alas}}} ^ {\ text {alas}}} +Bd {\ mathrm {xx}} ^ {\ text {alas}}} +Bd {\ mathrm {xx}} ^ {\ text {alas}}} +Bd {\ mathrm {elas}}} +Bd {\ mathrm {\ alas}}} ^ {\ text {alas}}} +Bd\]

Hence the horizontal elastic deformation increment:

(2.8)\[ d {\ mathrm {}} _ {\ mathit {xx}}} ^ {\ text {xx}}} ^ {\ text {elas}}} =\ frac {{P} _ {\ mathrm {\ sigma}}} _ {\ mathit {xx}}} _ {\ mathit {xx}}} ^ {\ mathit {xx}}} ^ {\ text {-}} ^ {\ text {-}}} +B\ left (d\ mathrm {\ sigma}}} _ {\ mathit {zz}}\ right)} {C}\]