Author Topic: Hill model  (Read 3465 times)

cri

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Hill model
« on: July 18, 2018, 09:07:49 AM »
Dear Prof. Taylor and dear all,

I'm trying to run an analysis by using the Hill plastic model. I adopted an orthotropic elastic material (with angle between the principal material axes 1 and the x1-axes equal to 0°) and I set the plastic parameters so as to obtain the yield value for uniaxial test along the x1-direction equal to 2. I attached my input file, as the analysis does not seem to work. I'm probably doing something wrong since it's the first time I use this material model in FEAP.

Thanks in advance

Yaakov

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Re: Hill model
« Reply #1 on: July 18, 2018, 10:05:49 AM »
cri,

it could be the error for command "orth vec" in ver84: solution please see this post: http://feap.berkeley.edu/forum/index.php?topic=1577.msg5789#msg5789

I would suggest you, you may directly use angle command to solve your problem, as your problem setting is in case of 2-D. (0-360 Degree or e.g. 0-90 Degree: Cubic symmetry)

otherwise, for hill's model, you can find some very useful posts using "Search" function in FEAP Forum, key words: 'orth', 'hill', etc

Best,
yaakov
« Last Edit: July 18, 2018, 10:12:43 AM by Yaakov »
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cri

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Re: Hill model
« Reply #2 on: July 18, 2018, 11:27:24 AM »
Dear Yaakov, thank for your answer.
I already checked the 'angle comand' and the 'vect orth' comands by considering different values of the rotation angle. Both comands work if a linear elastic analysis is performed. However I read in the user-manual that 'if the axes differ from the material coordinate axes a VECTor ORTHotropic record should be specified' for orthotropic plasticity. Furthermore, according your suggest, I tried to replace the 'vect orth' comand with the 'angle' one (attached file), but I had the same problem when the Hill plasticity model is activated.

Best,
Cri

Yaakov

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Re: Hill model
« Reply #3 on: July 18, 2018, 12:08:01 PM »
cri,

okay, I did not use hill's model. but perhaps we can discuss a little more here:

1. for Zero Degree, do we really need to define the 'angle' in the input file?

orthotropic properties: think about Zero Degree: Global Stiffness  C_global  = P[angle] . C_local . Tranpose [P[angle]] , if angle is equal to 0 Degree, what does that mean in this specific case?

for cubic symmetry materials: 0 or 90 Degree .... C_global  := C_local.

2. you can just find the related solution from the old post: http://feap.berkeley.edu/forum/index.php?topic=1599.msg5903#msg5903

3. VECTor ORTHotropic is very efficient for the case of 3-D. But for 2-D Problem we can use euler angle, why? As  'vect orth' constructs the 'material angle' (Anisotropy).
To check these arguments, you may read many related papers.

Best,
Yaakov
« Last Edit: July 20, 2018, 05:39:49 AM by Yaakov »
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cri

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Re: Hill model
« Reply #4 on: July 18, 2018, 12:39:04 PM »
First of all, thank you again for your attention. I agree with you about the fact that the 'angle' or 'vect orth' commands are not necessary for zero degree (indeed I also tried to omit them), but I would like to find a more general command since I want to analyze different angle values. Thus, I think I'm doing something else wrong. I checked the post you recommended but it mainly concerns the hardening models and the command solution.

Best regards
Cri

Yaakov

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Re: Hill model
« Reply #5 on: July 18, 2018, 12:44:30 PM »
Cri,

some users have already asked some questions about Hill's Model in forum and professors also gave them the solutions.

Best,
Yaakov
« Last Edit: July 18, 2018, 01:08:39 PM by Yaakov »
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Prof. R.L. Taylor

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Re: Hill model
« Reply #6 on: July 19, 2018, 06:55:09 AM »
If the orthotropic axes are already in the coordinate directions you do not need the VECT command.
As currently coded the VECT applies to both elastic and plastic axes.


The algorithm is failing in the plane stress part.  It runs in plane strain fine.  The indicator is that things are too small for the iterative algorithm we use to enforce plane strain to work.  You might try running the problem as a 3-d sheet to simulate the plane stress.

Also the parameters for the HILL look strange, where did you get them?

cri

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Re: Hill model
« Reply #7 on: July 20, 2018, 01:07:41 AM »
Dear Prof. Taylor,
thank you for your reply. I will try to run the problem in plane strain condition and as a 3-d sheet to simulate the plane stress.
However, I set the material plastic parameters in order to obtaine, for instance, the yield value for a uniaxial test along x1-axis equal to 2 MPa, as reported in section 7.5.2 of the user-manual (Yij=Rij*sigma_y). Is there any error in the other parameters?


Thanks in advance

Prof. R.L. Taylor

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Re: Hill model
« Reply #8 on: July 20, 2018, 05:07:52 AM »
Seems you have Y0=2 and R_11 = 2 -- is that what you want?

What looked strange to me were the sheqr values.  So you have a real material to represent?

cri

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Re: Hill model
« Reply #9 on: July 20, 2018, 08:16:11 AM »
At the moment, I don't have a real material, but I want only check the material model. According to your advice, I tried to run the analysis by adopting the plane strain hypothesis, but I again doesn't work. If possible, could you provide me an input file that works with the Hill plasticity model? Thanks to this I could possibly identify where I make mistakes.

Thank you in advance

Prof. R.L. Taylor

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Re: Hill model
« Reply #10 on: July 20, 2018, 09:11:56 AM »
what version of feap are you running?  On what operating system?  Your file ran on my machine using ver8.5, the intel compiler and a Mac