Thermal deformation due to temperature change

[Prof. R.L. Taylor]

Yes, that looks correct. 

[Ishihara]

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]

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]

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]

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]

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)*thkThe rest of the original code is correct.

[Ishihara]

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.