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In continuum mechanics of materials with zero volumetric change, the material condition can be expressed by the strain deviatoric tensor instead of the strain tensor itself. To express the plasticity of the materials, the plasticity surface is constructed from the second and third strain invariants, i.e.,

$I_2 = \sqrt{-\frac{1}{2}tr(\varepsilon_{dev})^2 }$,

$I_3 = \det(\varepsilon_{dev})$.

It is obvious that the second invariant is not able to describe the tension-compression asymmetry of the material. Therefore, the third invariant is also included in the plasticity surface. Now the question is why the third invariant can express the tension-compression asymmetry. I mean, how the determinant of the strain deviatoric determines the tensile or compressive state of the material.

Thanks in advance

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If $E\equiv \varepsilon_{dev}$ describes a tension state then $-E$ describes a compression state. Now, $I_2(E)=I_2(-E)$ meaning, as you say, that $I_2$ cannot distinguish traction from compression. However, in 3D, $\det(-E)=-\det(E)$. Thus, $E$ and $-E$ have opposite third invariants. This difference can be exploited to build yield functions which do distinguish traction from compression.

I do not whether $I_3$ has a direct "physical meaning" or not. However, this is the only available choice for isotropic materials. Recall that for an isotropic material, the yield function $f$ must be an isotropic function of $E$; that is $$ f(E) = f(RER') $$ for any rotation $R$ and its inverse $R'$. In that case, $f$ can be written as a function of the invariants: $f(E)=f(I_1,I_2,I_3)$. But $I_1=0$ and $I_2$ is even, meaning that the only way to distinguish traction from compression in isotropic materials is by writing an $I_3$-dependent yield criterion.

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