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Deviatorial form — Ansys Learning Forum

# Deviatorial form

Member Posts: 31

I want to go to the deviator form, for this I need to fill in the elastic coefficient matrix. The matrix itself looks like this:

In one source I found information on determining elastic coefficients, but I'm not sure what to do exactly this way!

Also, ANSYS HELP says that the matrix must be positive definite and semmetric.

Also in ANSYS HELP says that the matrix can be either rigid or pliable.

How to determine the coefficients of elasticity for a rigid and flexible matrix?

• Member Posts: 11,369
edited March 29

The top matrix you show is the Anisotropic Elasticity material definition. That stiffness matrix is multiplied by the strains to get the stresses.

```https://ansyshelp.ansys.com/account/secured?returnurl=/Views/Secured/corp/v211/en/ans_mat/anel.html
```

The help page above says that you can define the material using the flexibility form (compliance) instead of stiffness.

1. Initialize the constant table (TB,ANEL or TB,ELAS,,,,AELS/AELF). Select the stiffness or flexibility form via the appropriate `TBOPT` value.

The bottom matrix you show is the Compliance (flexibility) matrix that is multiplied by the stresses to get the strains and the values you show are for an orthotropic material.

Ansys can define an orthotropic material by providing the nine material constants.

```https://ansyshelp.ansys.com/account/secured?returnurl=/Views/Secured/corp/v211/en/ans_mat/ELlinmatstressstrain.html
```

You can also use Engineering Data to enter Orthotropic elasticity.

If you want to define an Anisotropic material, you can do that in Engineering Data also.

Now you have to enter the 18 constants.

"Also in ANSYS HELP says that the matrix can be either rigid or pliable."

The words I found in the Help system are "stiffness" or "flexibility"

An isotropic material is defined by only two material constants and an equation to get the third. The values to use in the anisotropic material definition using the Stiffness form is given by equation 6.3.21 in this reference:

http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_I/BookSM_Part_I/06_LinearElasticity/06_Linear_Elasticity_03_Anisotropy.pdf

You can see that it is easier to take the nine orthotropic material constants and put them in the flexibility form of the anisotropic matrix rather than the stiffness form of the anisotropic matrix.

• Member Posts: 31

Thanks for the answer! So, I take the matrix coefficients for an orthotropic material:

And I substitute them into the matrix for anisotropic material:

At the same time, I select the type of matrix - Stiffness

If you select the Compliance parameter, then you need to raise the matrix to the power of minus one.

peteroznewman - did I understand you correctly?

• Member Posts: 31

I still can't figure out why the constant G (SHEAR MODULUS) is doubled in your source!

In this source, the constant G (SHEAR MODULUS) is not doubled!

http://solidmechanics.org/text/Chapter3_2/Chapter3_2.htm

• Member Posts: 11,369

Engineering Shear Strain is divided by 2 to get the tensorial shear strain components.

https://en.wikipedia.org/wiki/Deformation_(physics)