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[Jan Stránský <question670047@xxxxxxxxxxxxxxxxxxxxx>]

Question #670047 on Yade changed:
https://answers.launchpad.net/yade/+question/670047

    Status: Open => Answered

Jan Stránský proposed the following answer:
> I see there are 6 components for stress and strain so I guess they
represent

stress tensor has 3 normal components (11,22,33) and 3 shear components (23,31,12) [1]
Strain is deformation conterparts. 3 normal strain (stretching along 3 axes) and 3 shear strains (skewing in 3 planes)

Shear parts are not important for this stuff, so you can just take the first 3 rows and first 3 columns of the matrix stress=stiffness*strain laws
Than s11,s22,s33 are stresses along x,y,z axes and e11,e22,e33 are "stretch ratios" (Delta L)/L along x,y,z axes

> but how could I get e22 and e33

this is just theoretical point how Poisson's ratio is defined, as a
lateral strain in uniaxial stress load. How to compute it in DEM is the
tricky part*. We avoid it using a different simulation (uniaxial
strain** of triaxial loading) evaluating one more elastic constant (e.g.
bulk mudulus).

*: see my comments in #1

**:
uniaxial stress = one stress is nonzero and the lateral stresses are zero -> all normal strain are in general nonzero
uniaxial strain = one strain is nonzero and the lateral strains are zero -> all normal stresses are in general nonzero

concerning changing stresses and strain, in the elastic regime the
stresses and strains are changing, but not independently, e.g. for
uniaxial stress, s11=E*e11, in general
stressTensor=stiffnessMatrix*strainTensor

Jan

[1] https://en.wikipedia.org/wiki/Cauchy_stress_tensor

PS: I will give short answers, do not hesitate to ask what you do not
understand. We can also change to personal communication..

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