Author Topic: Thermal deformation due to temperature change  (Read 27553 times)

Prof. R.L. Taylor

  • Administrator
  • FEAP Guru
  • *****
  • Posts: 2649
Re: Thermal deformation due to temperature change
« Reply #30 on: June 06, 2023, 11:33:41 AM »
Yes, that looks correct. 

Ishihara

  • Jr. Member
  • **
  • Posts: 22
Re: Thermal deformation due to temperature change
« Reply #31 on: June 06, 2023, 07:22:43 PM »
thank you.
I would like to change it to "Norm(1:2) = Norm(1:2) - alp(1:2)*d(9)*thk" and verify it.

Ishihara

  • Jr. Member
  • **
  • Posts: 22
Re: Thermal deformation due to temperature change
« Reply #32 on: June 07, 2023, 01:14:20 AM »
As a result of multiplying the right side by "thk(=0.5)", the displacement is halved. This seems to indicate that the membrane strain is 0. Is membrane strain given by "Norm(1:2)"?

Prof. S. Govindjee

  • Administrator
  • FEAP Guru
  • *****
  • Posts: 1164
Re: Thermal deformation due to temperature change
« Reply #33 on: June 07, 2023, 04:00:27 AM »
In subroutine stre3d( ) in the file shl3ds.f, the membrane strains should be in eps(1:3).
When multiplied with dd( , ) you should get the membrane stress and after multiplication with thk you should get the membrane resultants in norm(1:3).

Ishihara

  • Jr. Member
  • **
  • Posts: 22
Re: Thermal deformation due to temperature change
« Reply #34 on: June 07, 2023, 05:41:31 AM »
Then, should I change the formula to be added to "norm(1:3)=eps(1:3)-alp(1:3)*d(9)*thk"?

Prof. S. Govindjee

  • Administrator
  • FEAP Guru
  • *****
  • Posts: 1164
Re: Thermal deformation due to temperature change
« Reply #35 on: June 07, 2023, 06:19:28 AM »
I have made a mistake.  It turns out that inmate.f multiplies the coefficients of thermal expansion by the moduli.  Thus alp( ) holds the stress coefficients.  What you need only do is have
Code: [Select]
!     Membrane forces
      norm(1) = (dd(1,1)*eps(1) + dd(1,2)*eps(2) + dd(1,4)*eps(3))*thk
      norm(2) = (dd(2,1)*eps(1) + dd(2,2)*eps(2) + dd(2,4)*eps(3))*thk
      norm(3) = (dd(4,1)*eps(1) + dd(4,2)*eps(2) + dd(4,4)*eps(3))*thk
!     Modify in-plane forces by constant temperature change
      norm(1:2) = norm(1:2) - alp(1:2)*d(9)*thk
The rest of the original code is correct.

Ishihara

  • Jr. Member
  • **
  • Posts: 22
Re: Thermal deformation due to temperature change
« Reply #36 on: June 07, 2023, 06:41:27 AM »
Originally, the thicker the structure, the smaller the amount of deformation.
Looking at this formula, it seems that the thicker the material, the greater the amount of deformation. Is it appropriate to multiply by thk...?

Also, I actually tried that method, but the amount of deformation was only halved (because thk=0.5), and it did not match the results of other analysis software.