Question：calculation of the hardening parameter matrixc

[notleareM]

Hi Erlap,

usermaterials.f 605--613 ：

      hintmat1 = hardeningparam(4)
      do k = 1, int(nslip/3.)     
        do i = 1, 3
              do j = 1, 3
                  hintmat1(3*(k-1)+i, 3*(k-1)+j)=1.
              enddo
          enddo
      enddo

I would like to ask if the code here should be changed to the following form

      hintmat1 = hardeningparam(4)
      do k = 1, int(nslip/3.)     
        do i = 1, 3
              do j = 1, 3
                  if i==j hintmat1(3*(k-1)+i, 3*(k-1)+j)=1.
              enddo
          enddo
      enddo

Thank you for your attention to this matter.

Best regards

[EralpDemir]

Nope, that is correct.

Try this in matlab: hintmat1=1.4ones(12); for k = 1: 4 for i = 1: 3 for j = 1: 3 hintmat1(3(k-1)+i, 3*(k-1)+j)=1.; end end end