Learning Forum

[javat33489]

Hello. Friends I have a problem. I have a strain/stress graph. Now I need to find the deformation parameters for Johnson-Cook (A, B, C, n, m). There are no tutorials on how to do this. There are a few videos on youtube but it's not clear. Could you tell me how to do it consistently? The points. Do you have an excel spreadsheet?

[Ashish Khemka]

Please see if the following link helps:

Johnson-Cook Failure for Copper ÔÇö Ansys Learning Forum

Regards Ashish Khemka

[javat33489]

Thanks for the answer, but it doesn't work for me. I tested steel on the stand. I have a graph, I made a strenght and stress curve from this graph. Now, using these data, I need to find data for the Johnson-Cook deformation. Like here: https://www.youtube.com/watch?v=vRTgY1HikUQ&pp=ugMICgJydRABGAE%3D But this video is not clear. I would like a more detailed description of how to do this. Or maybe there is a ready-made Excel table where you could drive in the chart parameters and get Johnson-cook data

[javat33489]

https://www.youtube.com/watch?v=vRTgY1HikUQ

[javat33489]

Friends, I do not understand the parameter T. This is the temperature, but how to find it correctly?
(T-T0)/(Tp-T0)
Friends, I do not understand the parameter T. This is the temperature, but how to find it correctly? (T-T0)/(Tp-T0) what is T?

[javat33489]

Several temperatures are needed, but what are these temperatures? Where can I get them?

[Chris Quan]

The reference paper of Johnson-Cook Strength & Failure models can be found from: Engng. Frac. Mech. Vol 21. No. 1. pp 31-48. 1985Johnson + Cook. You can get this paper to find how they conduct the experiments to obtain the material data for the Johnson-Cook models.
In terms of temperature, T is the current temperature calculated from the local plastic work done, T0 is the initial reference temperature (stress-free), and Tp is the temperature for melting. When the deformed material has done a lot of plastic work, it will increase the temperature in the heavily deformed material. The increase of the temperature will further reduce the yield stress of the material. This is also called as "thermal softening".

[javat33489]

Okay, I'm confused. But I still have a question. If I tested the sample only once at room temperature, then T = 273K, T0 = 273K, then (T-T0)/(Tm-T0) = 0 because 273-273 = 0. What to do with T if the tests were carried out once and at room temperature? T=T0? What T to use?
The second question, when plotting to find the parameter B, I get too small B:
according to the schedule it is 3.3, EXP3.3 is very little?
watch the video how to find parameter B i did the same
https://www.youtube.com/watch?v=vRTgY1HikUQ

[Chris Quan]

Yes, it is true that T_H is equal to zero due to T = T_0 at the beginning of the simulation.
Thus, the third term of the Johnson-Cook model is equal to 1, which means no thermal softening at this temperature. See attached picture below.
Thermal softening starts when the temperature in an element is elevated and has exceeded the initial temperature.

[javat33489]

now everything is clear thanks!
it remains to understand why I have such a small coefficient B

[Assala Trabelsi]

Could you please tell me what are these Johnson-Cook parameter for the Copper 

Thank you in advance 

[jaideep18]

I cannot find the value for static damping in the Damping control under the Analysis setting for Ti6Al4V grade 23 material. Could you please help me with that. Thank you