How to degrade material properties for TYPE=ENGINEERING CONSTANT?

[frankluo]

Hi, Thanks for your attension.

I am trying to degrade a kind of orthotropic material properties at every Gauss point using ABAQUS.

Based on the example of ABAQUS, which is just for ISOTROPIC case, the key sentence is like below:
*ELASTIC, TYPE=ISOTROPIC, DEPENDENCIES=1
E, v, . fv1
E1, v1, , fv2

I understand it(ABAQUS) will interpolate material properties for me if the user defined variable is between fv1 and fv2.

However, I would like to accomplish it in TYPE = ENGINEERING CONSTANT and my objective is to use fv variable to degrage the engineering constants respectively. As for different material constant, every fv is different. For example,

E1, E2, E3, v12, v23, v13, G12, G23, G13 will be degraded as following:

E1/fv1, E2/fv2, E3/fv3, v12/fv4, v23/fv5, v13/fv6, G12/fv7, G23/fv8, G13/fv9.

It will dependent on 9 user defined variables. I will use:

*ELASTIC, TYPE=ENGINEERING CONSTANTS, DEPENDENCIES=9
E1, E2, E3, v12, v23, v13, G12, G23,
G13, , fv1, fv2, fv3, fv4, fv5, fv6, fv7, fv8, fv9
E1new, E2new, E3new, v12new, v23new, v13new, G12new, G23new,
G13new, , fv1n, fv2n, fv3n, fv4n, fv5n, fv6n, fv7n, fv8n, fv9n


My question is: If I would like to use similar scheme (ISOTROPIC), how can I know which variable is corresponding to which material constant? Or the way I used is totally wrong. I am looking forward to hearing suggestion from Advance users. Thanks a lot.

Frank
UMBC

[Jorgen]

I have never used the approach that you are suggesting, but it might actually work. You need to be very careful, though, when you create your table of interpolation values. The ABAQUS manuals will have the details. Note that your table of interpolation values will likely be very large since you have so many dependencies.

- Jorgen

[frankluo]

Dr. Jorgen

Thanks for your quick response. However, I am still confused with this implementation. I make an assumption that all the material properties are degraded in linear relationship. For example, the final properties will be one tenth of the original properties. I use this key words:

*ELASTIC, TYPE=ENGINEERING CONSTANTS, DEPENDENCIES=9
E1, E2, E3, v12, v23, v13, G12, G23,
G13, , 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0
E1*0.1, E2*0.1, E3*0.1, v12*0.1, v23*0.1, v13*0.1, G12*0.1, G23*0.1,
G13*0.1, , 10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 10.0

For every Gauss integration station, I apply the degrading law to get all the degrading coefficients (fv1, fv2, ...., fv9). I know all the (fv1, fv2, ...., fv9) is between 1.0~10.0. Can I use above key words to realize this objective? I still don't understand the mapping scheme inside of ABAQUS. ABAQUS will correspondingly interpolate the material properties for me, or not? How can ABAQUS know which coefficient is for specific material constant? I mean that fv1 for E1, fv2 for E2, .... fv9 for G13. Do you think what I am thinking is correct or I misunderstand the scheme in ABAQUS? Thanks a lot for your kind help.

Or make my question much simpler, I use TYPE=ISOTROPIC case.

For example, the material constant E is dependent on fv1 (1.0~10.0), possion's ratio id on fv2(1.0~5.0), the key words is as following:
*ELASTIC, DEPENDENCIES=2
E, v, , 1.0, 1.0
E*0.1, v*0.2, , 10.0, 5.0

For instance, we get the degrading coefficient is fv1=3.0, fv2=2.0, so I hope the degrading material properties is E/3.0, v/2.0 respectively. Do you think I will get the expected results? I am concerned how ABAQUS knows first variable is for E and second variable is for v. Thanks.


Frank

[Jorgen]

The way ABAQUS implements the field variable dependencies is through a look-up table. The values in this table needs to be arranged in a very specific way. I don't think your example above looks right. I suggest that you go through the examples in the ABAQUS manuals to see how it is done.

Also, shouldn't your fv1, fv2, etc, coefficients be between 0 and 1?

- Jorgen