According to the User subroutine Abaqus manual,
"If thermal expansion is included in the same material definition, the strains passed into UMAT are the mechanical strains only (that is, the thermal strains computed based upon the thermal expansion coefficient have been subtracted from
the total strains)." -quoted as in manual.
So what happens if I don't include Thermal co-efficient in the material definition. What will Abaqus provide me in the STRAN and DSTRAN variable if there is a temperature change included in the analysis?
If you do not have thermal coefficient defined in ABAQUS, simply, DSTRAN is the STRAIN INCREMENT, dEpsilon, and STRAN is the TOTAL STRAIN.
As reported in the manual, if thermal expansion is included in the same material definition, these are the mechanical strain increments (the total strain increments minus the thermal strain increments).
So, if you have temperature change, but the thermal coefficient is not defined, the total strain increment coincides with the "mechanical" strain increment.
@Andrea Thanks for the reply.
I do not have any mechanical loading in my analysis. I just have temperature change.
By your logic, am I right in saying that the total strain will coincide with Thermal Strain.
And how does Abaqus know DSTRAN as there is no mechanical loading?
DSTRAN in your case can derive ONLY from mechanical loading (pressure, tractions) or imposed displacement: DSTRAN is the gradient of the increment of displacements in your model.
Thank you for taking time to clarify my doubt. May be you can also help me out on another question related to UMAT pertaining to rotation of STATE variables.Attached is the link.
https://www.researchgate.net/post/Is_anyone_familiar_with_rotation_of_STATE_variables_in_UMAT
If your problem is purely thermal problem excluding the mechanical part, you should just use umatht and formulate the thermal problem in accordance with the second law of thermodynamics. Additionally, I mention that DSTRAN is the total strain increment which includes mechanical and thermal strains DSRAN = DSTRAN_elastic + (DSRAN_thermal = alpha*dtemp). In the case of state variable you should enter the neumber of dependent variable while you define the material type in abaqus number of Drpvar is equal to nstatev in UMAT.
I think you are right. Without any mechanical loading I am getting DSTRAN value. I have just assigned initial temperature and thermal expansion coefficient. Then cooled it to room temperature.
STRAN(NTENS) -Real array-Total strain increment
DSTRAN(NTENS)-eps-Real array-Mechanical strain increment
- Badges
- Science topic