Author Topic: Holzapfel model : fibre orientation  (Read 10182 times)

bharathn

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Re: Holzapfel model : fibre orientation
« Reply #15 on: April 16, 2019, 10:21:48 AM »
If you mean the input file, then yes.

Thank you professor!

Prof. R.L. Taylor

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Re: Holzapfel model : fibre orientation
« Reply #16 on: April 17, 2019, 04:19:20 PM »
I note you have the fiber axes at 1 1 0 -- this would mean the specified a0(:) would be 0.701 0.701 0.0  -- to have unit length. Is this what you are using or did you modify what the 1 1 0 means.  When I changed to 1 0 0 and ran with cylindrical orientation the cylinder was circular.

bharathn

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Re: Holzapfel model : fibre orientation
« Reply #17 on: April 17, 2019, 10:40:29 PM »
Dear professor, I had tried the 100 (fibres in the radial direction), 010 (fibres in the circumferential direction) and the 110 (at an angle) configurations. When I use 100, I got the cylindrical results that I had attached earlier. The 010 configuration gave me the non-cylindrical displacement. You had suggested that even for 010, it should provide us with a cylindrical displacement. I had probably attached the input file after having tried the 110 configuration.

With regards to the a0 values, I am using the same system (110 ~ a0 = 0.7071 0.7071 0).

Did you get a cylindrical result when you used the 010 fibre direction?
« Last Edit: April 17, 2019, 11:41:10 PM by bharathn »

Prof. R.L. Taylor

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Re: Holzapfel model : fibre orientation
« Reply #18 on: April 18, 2019, 10:10:12 AM »
I do get a cylindrical shape as shown

bharathn

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Re: Holzapfel model : fibre orientation
« Reply #19 on: April 18, 2019, 12:08:56 PM »
Okay, then maybe there is something wrong with my formulation. Do you have any idea if I could use the file from FEAp Ver 8.5? Also, would it be possible to see the output file for the radial case? I would like to confirm where the problem lies.

From my outputs for the 100 and the 010 directions, I see the following discrepancies in the convergence of the residual:

Code: [Select]
######################### FOR 0 1 0 CONFIGURATION ############################

Computing solution at time  1.0000E-01: Total proportional load  1.0000E-01
   Residual norm =     2.9199336E+00    1.0000000E+00      t=     0.87     0.01
   End Triangular Decomposition                            t=     0.87     0.01
   Residual norm =     4.8878692E-01    1.6739659E-01      t=     0.88     0.01
   End Triangular Decomposition                            t=     0.88     0.01
   Residual norm =     1.9583752E-01    6.7069169E-02      t=     0.89     0.01
   End Triangular Decomposition                            t=     0.89     0.01
   Residual norm =     7.7890117E-02    2.6675304E-02      t=     0.90     0.01
   End Triangular Decomposition                            t=     0.90     0.01
   Residual norm =     3.1134057E-02    1.0662591E-02      t=     0.91     0.01
   End Triangular Decomposition                            t=     0.91     0.01
   Residual norm =     1.2430845E-02    4.2572356E-03      t=     0.92     0.02
   End Triangular Decomposition                            t=     0.92     0.02
   Residual norm =     4.9646143E-03    1.7002490E-03      t=     0.93     0.02
   End Triangular Decomposition                            t=     0.93     0.02
   Residual norm =     1.9825770E-03    6.7898015E-04      t=     0.94     0.02
   End Triangular Decomposition                            t=     0.94     0.02
   Residual norm =     7.9172505E-04    2.7114488E-04      t=     0.95     0.02
   End Triangular Decomposition                            t=     0.95     0.02
   Residual norm =     3.1616084E-04    1.0827672E-04      t=     0.96     0.02
   End Triangular Decomposition                            t=     0.96     0.02
   Residual norm =     1.2625065E-04    4.3237507E-05      t=     0.97     0.02
   End Triangular Decomposition                            t=     0.97     0.02
   Residual norm =     5.0413894E-05    1.7265425E-05      t=     0.98     0.02
   End Triangular Decomposition                            t=     0.98     0.02
   Residual norm =     2.0130661E-05    6.8942186E-06      t=     0.99     0.02
   End Triangular Decomposition                            t=     0.99     0.02
   Residual norm =     8.0381633E-06    2.7528583E-06      t=     1.00     0.02
   End Triangular Decomposition                            t=     1.00     0.02
   Residual norm =     3.2095678E-06    1.0991921E-06      t=     1.00     0.02
   End Triangular Decomposition                            t=     1.01     0.02
   Residual norm =     1.2815252E-06    4.3888847E-07      t=     1.02     0.02
   End Triangular Decomposition                            t=     1.02     0.02
   Residual norm =     5.1168007E-07    1.7523688E-07      t=     1.02     0.03
   End Triangular Decomposition                            t=     1.02     0.03
   Residual norm =     2.0429628E-07    6.9966070E-08      t=     1.03     0.03
   End Triangular Decomposition                            t=     1.03     0.03
   Residual norm =     8.1566904E-08    2.7934507E-08      t=     1.04     0.03
   End Triangular Decomposition                            t=     1.04     0.03
   Residual norm =     3.2565282E-08    1.1152748E-08      t=     1.05     0.03
   End Triangular Decomposition                            t=     1.05     0.03
The 010 configuration stops the iterations as soon as the residual norm hits 10e-08. However, the 100 configuration continues till it hits 10e-13. Shouldn't the residual be lower for the 010 case before the solver decides to move to the next load step?
Code: [Select]
######################### FOR 1 0 0 CONFIGURATION ############################

  Computing solution at time  1.0000E-01: Total proportional load  1.0000E-01
   Residual norm =     2.9199336E+00    1.0000000E+00      t=     0.89     0.02
   End Triangular Decomposition                            t=     0.89     0.02
   Residual norm =     1.5349910E+01    5.2569382E+00      t=     0.90     0.02
   End Triangular Decomposition                            t=     0.90     0.02
   Residual norm =     5.6304700E+00    1.9282870E+00      t=     0.91     0.02
   End Triangular Decomposition                            t=     0.91     0.02
   Residual norm =     4.6535173E+00    1.5937066E+00      t=     0.92     0.02
   End Triangular Decomposition                            t=     0.92     0.02
   Residual norm =     3.2524666E-02    1.1138838E-02      t=     0.93     0.02
   End Triangular Decomposition                            t=     0.93     0.02
   Residual norm =     5.3513725E-07    1.8327035E-07      t=     0.94     0.02
   End Triangular Decomposition                            t=     0.94     0.02
   Residual norm =     2.5086218E-14    8.5913660E-15      t=     0.95     0.02
   End Triangular Decomposition                            t=     0.95     0.02

Thank you so much once again!
« Last Edit: April 18, 2019, 03:03:21 PM by bharathn »

Prof. R.L. Taylor

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Re: Holzapfel model : fibre orientation
« Reply #20 on: April 18, 2019, 03:41:33 PM »
Who has the license for the copy you are using?  It may be possible to update what you have.

bharathn

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Re: Holzapfel model : fibre orientation
« Reply #21 on: April 19, 2019, 11:12:33 AM »
Dear professor, I am verifying that now and I will get back to you as soon as I find out!

bharathn

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Re: Holzapfel model : fibre orientation
« Reply #22 on: May 13, 2019, 02:49:44 AM »
Dear professor,

I have sent you a message with further details!