Anisotropic finite strain viscoelasticity based on the Sidoroff multiplicative decomposition and logarithmic strains

Abstract

In this paper a purely phenomenological formulation and finite element numerical implementation for quasi-incompressible transversely isotropic and orthotropic materials is presented. The stored energy is composed of distinct anisotropic equilibrated and non-equilibrated parts. The nonequilibrated strains are obtained from the multiplicative decomposition of the deformation gradient. The procedure can be considered as an extension of the Reese and Govindjee framework to anisotropic materials and reduces to such formulation for isotropic materials. The stress-point algorithmic implementation is based on an elastic-predictor viscous-corrector algorithm similar to that employed in plasticity. The consistent tangent moduli for the general anisotropic case are also derived. Numerical examples explain the procedure to obtain the material parameters, show the quadratic convergence of the algorithm and usefulness in multiaxial loading. One example also highlights the importance of prescribing a complete set of stress-strain curves in orthotropic materials.

Appendices

Appendix 1: Proof of Eqs. (114) and (115)

The stress tensor $\varvec{S}_{neq}^{|d}$ defined in Eq. (97) is “traceless” in the sense that

$$\begin{aligned} \varvec{S}_{neq}^{|d}:\varvec{C}^{d}=\varvec{\tau }_{neq}^{|d}:\varvec{I}=\varvec{\tau }_{neq} ^{|e}:\varvec{I}=\varvec{T}_{neq}^{|e}:\varvec{I}=0 \end{aligned}$$

(157)

where the results $\varvec{\tau }_{neq}^{|d}=\varvec{X}^{d}\varvec{S} _{neq}^{|d}\varvec{X}^{dT}=\varvec{X}^{e}\varvec{S}_{neq}^{|e}\varvec{X} ^{eT}=\varvec{\tau }_{neq}^{|e}$, $tr(\varvec{\tau }_{neq}^{|e})=tr(\varvec{T}_{neq}^{|e})$ (see Ref. [59]) and Eq. (66) have been used. Therefore, the expression of the second Piola–Kirchhoff stress tensor $\varvec{S}_{neq}$ that derives from the purely deviatoric strain energy function $\mathcal {W}_{neq}$, as given in Eq. (97), reduces to Eq. (114)

$$\begin{aligned} \varvec{S}_{neq}&=\varvec{S}_{neq}^{|d}:J^{-2/3}\left( \mathbb {I}-\frac{1}{3}\varvec{C} ^{d}\otimes \varvec{C}^{d-1}\right) \nonumber \\&=J^{-2/3}\varvec{S}_{neq}^{|d} \end{aligned}$$

(158)

where $d\varvec{A}^{d}/d\varvec{A}=d\varvec{C}^{d}/d\varvec{C}$ has been obtained differentiating the expression $\varvec{C}^{d}=J^{-2/3}\varvec{C}$, with $J^2=det(\varvec{C})$, with respect to $\varvec{C}$.

Differentiating Eq. (157) with respect to $\varvec{C}^{d}$

$$\begin{aligned} \varvec{0}&=\frac{d(\varvec{S}_{neq}^{|d}: \varvec{C}^{d})}{d\varvec{C}^{d} }=\varvec{S}_{neq}^{|d}: \frac{d\varvec{C}^{d}}{d\varvec{C}^{d}}+\varvec{C} ^{d}:\frac{d\varvec{S}_{neq}^{|d}}{d\varvec{C}^{d}}\nonumber \\&=\varvec{S}_{neq} ^{|d}+\varvec{C}^{d}:\frac{1}{2}\mathbb {C}_{neq}^{|d} \end{aligned}$$

(159)

Using the major symmetries of $\mathbb {C}_{neq}^{|d}$, cf. Eqs. (102) and (113), we arrive at the following relation

$$\begin{aligned} \varvec{S}_{neq}^{|d}+\frac{d\varvec{S}_{neq}^{|d}}{d\varvec{C}^{d} }:\varvec{C}^{d}=\varvec{0} \end{aligned}$$

(160)

After some algebraic manipulations, we identify this last result in the expression of the deviatoric constitutive tensor that derives from $\varvec{S}_{neq}$, which finally simplifies to Eq. (115)

$$\begin{aligned} \mathbb {C}_{neq}&=\frac{d\varvec{S}_{neq}}{d\varvec{A}}=2\varvec{S} _{neq}^{|d}\otimes \frac{d J^{-2/3}}{d\varvec{C}}+J^{-2/3} \frac{d\varvec{S}_{neq}^{|d}}{d\varvec{A}^{d}}:\frac{d\varvec{A}^{d} }{d\varvec{A}}\nonumber \\&=J^{-4/3}\mathbb {C}_{neq}^{|d} \end{aligned}$$

(161)

Appendix 2: General expressions for $\varvec{T}_{neq}^{|tr}$ and $d\varvec{T}_{neq}^{|tr}/d\,^{tr}\varvec{E}_{e}$

If the approximation of Eq. (108) is not considered adequate, we can compute the mapping tensor $\partial \varvec{E}_{e} /\partial \,^{tr}\varvec{E}_{e}$ with the viscous flow frozen, needed for the computation of the stresses in Eq. (106), and its gradient with respect to $\,^{tr}\varvec{E}_{e}$, needed for the computation of the consistent tangent moduli.

From the relation $\,^{tr}\varvec{X}_{e}=\varvec{X}^d\,^{tr}\varvec{X}_{v}^{-1}$, with $\,^{tr}\varvec{X}_v$ fixed by definition, see Fig. 3, we obtain

$$\begin{aligned} \,^{tr}\varvec{d}_{e}=sym(\,^{tr}{\dot{\varvec{X}}}_{e}\,^{tr}\varvec{X} _{e}^{-1})=sym({\dot{\varvec{X}}}^d\varvec{X}^{d-1})=\varvec{d}^d \end{aligned}$$

(162)

which represents the spatial counterpart, in rate-form, of the change of variable given in Eq. (99). Accordingly, the independent variables of the isochoric counterpart of Eq. (40) may be changed to give

$$\begin{aligned} \varvec{d}_{e}\left( \,^{tr}\varvec{d}_{e},\varvec{l}_{v}\right) =\,^{tr}\varvec{d}_{e}-sym\left( \varvec{X}_{e}\varvec{l}_{v}\varvec{X} _{e}^{-1}\right) \end{aligned}$$

(163)

or

$$\begin{aligned} \varvec{d}_{e}\left( \,^{tr}\varvec{d}_{e},\varvec{l}_{v}\right) = \mathbb {M}_{\,^{tr}d_{e}}^{d_{e}}\Big \vert _{\varvec{l}_v = \varvec{0}\varvec{}}:\,^{tr}\varvec{d} _{e}+ \mathbb {M}_{l_{v}}^{d_{e}}\Big \vert _{{\, ^{tr}\varvec{d}_e = \varvec{0}}}:\varvec{l}_{v} \end{aligned}$$

(164)

with $\left. \mathbb {M}_{\,^{tr}d_{e}}^{d_{e}}\right| _{\varvec{l}_v = \varvec{0}\varvec{}} =\mathbb {I}^{S}$. Hence we obtain —recall Eq. (54)

$$\begin{aligned} \varvec{\tau }_{neq}^{|e}:\left. \varvec{d}_{e}\right| _{\varvec{l}_v = \varvec{0}\varvec{}} =\varvec{\tau }_{neq}^{|tr}:\,^{tr}\varvec{d}_{e}=\varvec{\tau }_{neq}^{|d} :\varvec{d}^d=\left. \mathcal {\dot{W}}_{neq}\right| _{\varvec{l}_v = \varvec{0}\varvec{}} \end{aligned}$$

(165)

with

$$\begin{aligned} \varvec{\tau }_{neq}^{|tr}=\varvec{\tau }_{neq}^{|e}:\left. \mathbb {M} _{\,^{tr}d_{e}}^{d_{e}}\right| _{\varvec{l}_v = \varvec{0}\varvec{}}=\varvec{\tau }_{neq}^{|e} :\mathbb {I}^{S}=\varvec{\tau }_{neq}^{|e} \end{aligned}$$

(166)

Although $\varvec{\tau }_{neq}^{|e}=\varvec{\tau }_{neq}^{|tr}=\varvec{\tau }_{neq}^{|d}$ represent all them the same Kirchhoff stress tensor operating in the current isochoric configuration, we use different superscripts to emphasize the fact that this stress tensor may be obtained from different Lagrangian stress tensors defined in different configurations. In terms of Generalized Kirchhoff stresses, Eq. (166) reads

$$\begin{aligned} \varvec{T}_{neq}^{|tr}:\mathbb {M}_{\,^{tr}d_{e}}^{\,^{tr}\dot{E}_{e} }=\varvec{T}_{neq}^{|e}:\mathbb {M}_{d_{e}}^{\dot{E}_{e}}:\left. \mathbb {M}_{\,^{tr}d_{e}}^{d_{e}}\right| _{\varvec{l}_v = \varvec{0}\varvec{}} \end{aligned}$$

(167)

where, for example, $\mathbb {M}_{d_{e}}^{\dot{E}_{e}}$ is the fourth-order tensor that maps, on the one hand, the elastic deformation rate tensor $\varvec{d}_{e}$ to the material rate tensor ${\dot{\varvec{E}}}_{e}$ and, on the other hand, the stresses $\varvec{T}_{neq}^{|e}$ to the stresses $\varvec{\tau }_{neq}^{|e}$, compare to Eqs. (45) and (49). Hence

$$\begin{aligned} \varvec{T}_{neq}^{|tr}&=\varvec{T}_{neq}^{|e}:\mathbb {M}_{d_{e}}^{\dot{E}_{e}}:\left. \mathbb {M}_{\,^{tr}d_{e}}^{d_{e}}\right| _{\varvec{l}_v = \varvec{0}\varvec{}} :\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}d_{e}}\\&=\varvec{T}_{neq}^{|e}:\left. \mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\dot{E}_{e}}\right| _{\varvec{l}_v = \varvec{0}\varvec{}} =\varvec{T}_{neq}^{|e}:\left. \frac{\partial \varvec{E}_{e}}{\partial \,^{tr}\varvec{E}_{e}}\right| _{{\dot{\varvec{X}}}_v = \varvec{0}\varvec{}}\nonumber \end{aligned}$$

(168)

Taking into consideration that $\left. \mathbb {M}_{\,^{tr}d_{e}}^{d_{e} }\right| _{\varvec{l}_v = \varvec{0}\varvec{}}=\mathbb {I}^{S}$ we arrive at

$$\begin{aligned} \left. \frac{\partial \varvec{E}_{e}}{\partial \,^{tr}\varvec{E}_{e} }\right| _{{\dot{\varvec{X}}}_v = \varvec{0}\varvec{}}&=\mathbb {M}_{d_{e}}^{\dot{E}_{e}}:\mathbb {M}_{\,^{tr} \dot{E}_{e}}^{\,^{tr}d_{e}}\nonumber \\&=\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e}}:\left( \mathbb {M}_{d_{e}}^{\bar{d}_{e}}:\mathbb {M}_{\,^{tr}\bar{d}_{e}}^{\,^{tr} d_{e}}\right) :\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e} }\nonumber \\&=\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e}}:\mathbb {M}_{\,^{tr}\dot{E}_{e} }^{\,^{tr}\bar{d}_{e}} \end{aligned}$$

(169)

where we have defined the rotated deformation rate tensors

$$\begin{aligned} \varvec{\bar{d}}_{e}&:=\varvec{R}_{e}^{T}\odot \varvec{R}_{e}^{T} :\varvec{d}_{e}=\mathbb {M}_{d_{e}}^{\bar{d}_{e}}:\varvec{d}_{e} \end{aligned}$$

(170)

$$\begin{aligned} \,^{tr}\varvec{\bar{d}}_{e}&:=\,^{tr}\varvec{R}_{e}^{T}\odot \,^{tr} \varvec{R}_{e}^{T}:\,^{tr}\varvec{d}_{e}=\mathbb {M}_{\,^{tr}d_{e}} ^{\,^{tr}\bar{d}_{e}}:\,^{tr}\varvec{d}_{e} \end{aligned}$$

(171)

and we have used the fact that $\,^{tr}\varvec{R}_{e}=\varvec{R}_{e}$, so $\mathbb {M}_{d_{e}}^{\bar{d}_{e}}:\mathbb {M}_{\,^{tr}\bar{d}_{e}} ^{\,^{tr}d_{e}}=\mathbb {I}^{S}$. Thus, the general expression for Eq. (106) reads

$$\begin{aligned} \varvec{T}_{neq}^{|tr}=\varvec{T}_{neq}^{|e}:\left. \frac{\partial \varvec{E}_{e}}{\partial \,^{tr}\varvec{E}_{e}}\right| _{{\dot{\varvec{X}}}_v = \varvec{0}\varvec{}} =\varvec{T}_{neq}^{|e}:\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e}}:\mathbb {M} _{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}} \end{aligned}$$

(172)

which defines the mapping between the stress tensors $\varvec{T}_{neq}^{|e}$, defined in the updated intermediate configuration, and $\varvec{T}_{neq}^{|tr}$, defined in the trial (fixed) intermediate configuration. The reader is referred to Ref. [59], Eq. (35), to see the specific spectral form of the mapping tensors present in Eq. (172), where the Lagrangian basis and the stretches are to be adapted to each case. Note that if the deformation occurs about the preferred material directions, then the shear terms of these mapping tensors do not take place in the relation between $\varvec{T}_{neq}^{|tr}$ and $\varvec{T}_{neq}^{|e}$ (because they are coaxial), so from the spectral forms of $\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e}}$ and $\mathbb {M} _{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}}$ we obtain $\varvec{T}_{neq}^{|tr} =\varvec{T}_{neq}^{|e}$; recall Identity (107)$_{2}$. Furthermore, the approximation of Identity (108)$_{2}$ is also based on the specific spectral forms of $\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e}}$ and $\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}}$ and on the fact that the eigenvectors of $\varvec{E}_{e}$ and $\,^{tr}\varvec{E}_{e}$ are almost coincident for $\Delta t/\tau \ll 1$, as one may deduce from Eq. (81).

For the computation of the consistent tangent moduli $d\varvec{T}_{neq}^{|tr}/d\,^{tr}\varvec{E}_{e}$, to be used in Eq. (104), we must take into consideration that the trial logarithmic strains $\,^{tr}\varvec{E}_{e}$ and the updated logarithmic strains $\,_{\qquad 0}^{t+\Delta t}\varvec{E}_{e}$ are related in the algorithm through Eq. (81), so their increments relate through Eq. (111), see also Eq. (112). Then, taking derivatives in Eq. (172) —note that $\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e}}$ and $\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}}$ have major and minor symmetries

$$\begin{aligned} \frac{d\varvec{T}_{neq}^{|tr}}{d\,^{tr}\varvec{E}_{e} }&=\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}} :\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e}}:\frac{d\varvec{T}_{neq}^{|e} }{d\varvec{E}_{e}}:\dfrac{d\,_{\qquad 0}^{t+\Delta t}\varvec{E}_{e}}{d\,^{tr}\varvec{E}_{e}}\nonumber \\&\quad +\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}} :\left( \varvec{T}_{neq}^{|e}:\frac{d\mathbb {M}_{\bar{d}_{e}} ^{\dot{E}_{e}}}{d\varvec{E}_{e}}\right) :\dfrac{d \,_{\qquad 0}^{t+\Delta t}\varvec{E}_{e}}{d\,^{tr}\varvec{E} _{e}}\nonumber \\&\quad +\varvec{T}_{neq}^{|e}:\mathbb {M}_{\bar{d}_{e}} ^{\dot{E}_{e}}:\frac{d\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}}}{d\,^{tr}\varvec{E}_{e}} \end{aligned}$$

(173)

or, equivalently

$$\begin{aligned} \frac{d\varvec{T}_{neq}^{|tr}}{d\,^{tr}\varvec{E}_{e} }&=\left. \frac{\partial \varvec{E}_{e}}{\partial \,^{tr}\varvec{E} _{e}}\right| _{{\dot{\varvec{X}}}_v = \varvec{0}\varvec{}}^{T}:\frac{d^{2}\mathcal {W}_{neq} }{d\varvec{E}_{e}d\varvec{E}_{e}}:\dfrac{d\,_{\qquad 0}^{t+\Delta t}\varvec{E}_{e}}{d\,^{tr}\varvec{E}_{e}}\nonumber \\&\quad -\left. \frac{\partial \varvec{E}_{e}}{\partial \,^{tr}\varvec{E} _{e}}\right| _{{\dot{\varvec{X}}}_v = \varvec{0}\varvec{}}^{T}:\left( \varvec{\bar{\tau }}_{neq}^{|e} :\frac{d\mathbb {M}_{\dot{E}_{e}}^{\bar{d}_{e}}}{d \varvec{E}_{e}}\right) :\dfrac{d\,_{\qquad 0}^{t+\Delta t}\varvec{E}_{e}}{d\,^{tr}\varvec{E}_{e}}\nonumber \\&\quad +\varvec{\bar{\tau }}_{neq}^{|e} :\frac{d\mathbb {M}_{\,^{tr}\dot{E}_{e}}^{\,^{tr}\bar{d}_{e}} }{d\,^{tr}\varvec{E}_{e}} \end{aligned}$$

(174)

where the result obtained from taking derivatives with respect to $\varvec{E}_{e}$ in the identity $\mathbb {M}_{\bar{d}_{e}}^{\dot{E}_{e} }:\mathbb {M}_{\dot{E}_{e}}^{\bar{d}_{e}}=\mathbb {I}^{S}$ has been used and $\varvec{\bar{\tau }}_{neq}^{|e}$ stands for the Kirchhoff stress tensor $\varvec{\tau }_{neq}^{|e}$ rotated by $\varvec{R}_{e}^{T}$, i.e. $\varvec{\bar{\tau }}_{neq}^{|e}=\varvec{T}_{neq}^{|e}:\mathbb {M} _{\bar{d}_{e}}^{\dot{E}_{e}}$. Following customary arguments, the sixth-order tensors of the type $d\mathbb {M}_{\dot{E}}^{\bar{d}}/d \varvec{E}$ present in this last equation may be obtained from the comparison of the spectral decompositions of the material rate of $\mathbb {M}_{\dot{E} }^{\bar{d}}$ and the material rate of $\varvec{E}$, i.e.

$$\begin{aligned} \mathbb {\dot{M}}_{\dot{E}}^{\bar{d}}=\frac{d\mathbb {M}_{\dot{E}} ^{\bar{d}}}{d\varvec{E}}:{\dot{\varvec{E}}} \end{aligned}$$

(175)

see Refs. [49, 59, 72] for similar derivations and further details.

Fig. 14

Uniaxial test over an incompressible orthotropic specimen in which the preferred material axes $\{1,2\}$ are not aligned with the test axes $\{x,y\}$

Appendix 3: Interpretation of off-axis shearing effects

From the third example above we infer that two different orthotropic materials subjected to the same off-axis finite deformation and with the same orientation of the preferred material axes may undergo angular distortions of opposite sign. Based on the fact that finite logarithmic strains extend the small strains meaning to the large strains setting [46] and on the fact that in that example we are using strain energy functions based on the same invariants used in infinitesimal orthotropic elasticity, we can explain these different mechanical responses from the infinitesimal theory and then extend the results to the case of Example 3.

Consider as an example the uniaxial test of Fig. 14 performed over a perfectly incompressible orthotropic material with the preferred material direction 1 oriented at $\alpha =30 {{}^o}$ with respect to the test axis x. We consider a plane strain state, so the out-of-plane strains vanish, i.e. $\varepsilon _{31}=\varepsilon _{32} =\varepsilon _{33}=0$. The in-plane contribution of the (deviatoric) strain energy function is expressed in terms of the components of the infinitesimal strain tensor $\varvec{\varepsilon }$ in the preferred material axes as

$$\begin{aligned} \mathcal {W}\left( \varvec{\varepsilon },\varvec{a}_{1},\varvec{a}_{2}\right) =\mu _{11}\varepsilon _{11}^{2}+\mu _{22}\varepsilon _{22}^{2}+\mu _{12}\left( \varepsilon _{12}^{2}+\varepsilon _{21}^{2}\right) \end{aligned}$$

(176)

The stresses in principal material directions are

$$\begin{aligned} \sigma _{11}&=2\mu _{11}\varepsilon _{11}+p \end{aligned}$$

(177)

$$\begin{aligned} \sigma _{22}&=2\mu _{22}\varepsilon _{22}+p=-2\mu _{22}\varepsilon _{11}+p \end{aligned}$$

(178)

$$\begin{aligned} \sigma _{12}&=2\mu _{12}\varepsilon _{12}=2\mu _{12}\varepsilon _{21} =\sigma _{21} \end{aligned}$$

(179)

where the incompressibility constraint $\varepsilon _{22}=-\varepsilon _{11}$ has been used and p is the initially unknown hydrostatic pressure. Since $\sigma _{12}<0$, Eq. (179) yields $\varepsilon _{12}<0$. From the Mohr’s circle of in-plane stresses shown in Fig. 15 we get the relation $\sigma _{11}=3\sigma _{22}$ (note that the axes $\{x,y\}$ are the principal directions of stresses because $\sigma _{xy}=0$). Combining Eq. (177), Identity (178)$_{2}$ and the relation $\sigma _{11}=3\sigma _{22}$ we arrive at

$$\begin{aligned} \sigma _{11}=3\left( \mu _{11}+\mu _{22}\right) \varepsilon _{11} \end{aligned}$$

(180)

The sign of the angular distortion $\gamma _{xy}=2\varepsilon _{xy}$ undergone by the specimen may be obtained by direct comparison of the Mohr’s representations of stresses and strains, see Fig. 15. On the one hand, in the Mohr’s circle of stresses we have $-\sigma _{12}/\left( \sigma _{11}/3\right) =\tan \left( 2\times 30 {{}^o} \right) =\tan \left( 60 {{}^o} \right)$. On the other hand, from the Mohr’s circle in the strain space we obtain $-\varepsilon _{12}/\varepsilon _{11}=\tan \left( 2\theta \right)$. These angles are related by Eqs. (179) and (180 ), i.e.

$$\begin{aligned} \frac{-\sigma _{12}}{\sigma _{11}/3}=\frac{2\mu _{12}}{\mu _{11}+\mu _{22}} \frac{-\varepsilon _{12}}{\varepsilon _{11}} \end{aligned}$$

(181)

Hence we distinguish three different possibilities

$$\begin{aligned} 2\mu _{12}&=\mu _{11}+\mu _{22}\text { }\Rightarrow \text { }2\theta =60 {{}^o} \text { }\Rightarrow \text { }\gamma _{xy}=0 \end{aligned}$$

(182)

$$\begin{aligned} 2\mu _{12}&>\mu _{11}+\mu _{22}\text { }\Rightarrow \text { }2\theta <60 {{}^o} \text { }\Rightarrow \text { }\gamma _{xy}>0 \end{aligned}$$

(183)

$$\begin{aligned} 2\mu _{12}&<\mu _{11}+\mu _{22}\text { }\Rightarrow \text { }2\theta >60 {{}^o} \text { }\Rightarrow \text { }\gamma _{xy}<0 \end{aligned}$$

(184)

which, note, satisfactorily explain the different behaviors obtained for the instantaneous (equilibrated plus non-equilibrated) and equilibrated responses in the Example 3 above. Finally, we remark that the condition $2\mu _{12} =\mu _{11}+\mu _{22}$ does not imply isotropic behavior in the plane 12 (although $\gamma _{xy}=0$). Evidently, if the material is isotropic in the plane 12, then $\mu _{11} =\mu _{22}=\mu _{12}$ and the condition $2\mu _{12}=\mu _{11}+\mu _{22}$ is also satisfied, as one would expect.

Fig. 15

Mohr’s circles for stresses (left) and strains (right) associated to the uniaxial test under plane strain of Figure 14 with $\alpha =30{{}^o}$. In the Mohr’s circle of stresses we use $\sigma _{xy} =\sigma _{yy}=0$ (boundary conditions). In the Mohr’s circle of strains we use $\varepsilon _{yy}=-\varepsilon _{xx}$ (incompressibility). Subscript n means “normal” and subscript t means “tangential”

Interestingly, the strain components $\varepsilon _{xx}$ and $\varepsilon _{xy}$ obtained for the orientations of $\alpha =30 {{}^o}$ and $\alpha =60 {{}^o}$ for the same uniaxial stress $\sigma _{xx}$ relate through

$$\begin{aligned} \varepsilon _{xx}^{60 {{}^o} }&=\varepsilon _{xx}^{30 {{}^o} } \end{aligned}$$

(185)

$$\begin{aligned} \gamma _{xy}^{60 {{}^o} }&=-\gamma _{xy}^{30 {{}^o} } \end{aligned}$$

(186)

which, again, let us explain the symmetric responses in the finite deformation context shown in Figures 8 and 9 of Ref. [59] (compare the cases $\alpha =30 {{}^o}$ and $\alpha =60 {{}^o}$ of each figure). Note that the reference configurations for $\alpha =30 {{}^o}$ and $\alpha =60 {{}^o}$ are different (i.e. they are not a reflection from each other). The symmetric responses are just a consequence of the plane strain condition, the incompressible behavior and the symmetry of the strain energy terms $\omega _{ij}(E_{ij})=\omega _{ij}(-E_{ij})$ considered in that paper (as in the small strains case).

Another interesting view of this phenomenon may be obtained through the skew part of the Mandel stress tensor, used in Refs. [50, 63, 73] in the context of plasticity to account for the update of the principal material directions. This tensor, work-conjugate to spins and which may be interpreted as fictitious angular moments per unit volume (couple-stress), accounts for the lack of commutativity due to elastic anisotropy and is obtained from the elastic strains and stored energy function as

$$\begin{aligned} \varvec{T}_{w}:=\varvec{ET}-\varvec{TE} \end{aligned}$$

(187)

For this particular case, using Eq. (176) and small strains

$$\begin{aligned} {\varvec{\sigma }}_{w}:=\varvec{\varepsilon \sigma }-\varvec{\sigma \varepsilon }=\left[ \begin{array}{c@{\quad }c@{\quad }c} 0 &{} -\sigma _{w21} &{} 0\\ \sigma _{w21} &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array} \right] \end{aligned}$$

(188)

where

$$\begin{aligned} \sigma _{w21}=2\varepsilon _{11}\varepsilon _{12}\left( \mu _{11}+\mu _{22} -2\mu _{12}\right) \end{aligned}$$

(189)

Note that there is a change of sign if either $\mu _{11}+\mu _{22}-2\mu _{12}$ (material dependent) changes sign or if $\varepsilon _{11}\varepsilon _{12}$ (load dependent) changes sign. Furthermore, for in-axis (axial) loading $\varepsilon _{12}=0$ or pure shear loading $\varepsilon _{11}=\varepsilon _{22}=0$, the tensor $\varvec{\sigma }_{w}$ vanishes. Obviously for the isotropic case all $\mu _{ij}$ are coincident and $\varvec{\sigma }_{w}$ also vanishes.