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Updated: 7 hours 42 min ago

moving load

10 hours 42 min ago

Is any similar UMAT for

12 hours 33 min ago

In reply to Writing User Subroutines with ABAQUS

Is any similar UMAT for linear and non-linear viscoelastic materials available? Please upload.

Can somebody upload a UMAT

12 hours 40 min ago

In reply to Sharing ABAQUS UMAT and VUMAT subroutines

Can somebody upload a UMAT for Schapery Material Model (1-D or 3-D)of non-linear viscoelastic materials?

Hi Jia and all,

Sat, 2015-01-24 20:44

In reply to Cohesive parameters input in Abaqus

Hi Jia and all,

what should be units for density. i find in abaqus documentation that tom/mm3 is used in SI(mm). Hence should density for steel (8700 kg/m3) be input as 8.7 e-9 ton/mm3 . Also I am creating a unit thickness adhesive layer. Later I edit the nodes to make it zero thickness layer. However since 1 mm thickness is magnified many times (real thickness of adhesive would be in microns probably). Hence how to input  the density of the adhesive. If in real adhesive density is 2000 kg/m3 would it also be input as  2 e-9 ton/mm3.

Regards,

Shouvik

Stress and Elastic Strain Coaxiality

Sat, 2015-01-24 10:38

In reply to Anisotropic stiffness of isotropic material

Yes, in this case stress is coaxial with elastic strain.

Mohsen

That is the question

Fri, 2015-01-23 15:48

In reply to Isotropy or Coaxiality

I think we can agree that an isotropic operator (in this case the stiffness tensor) produce a result that is coaxial with the argument, and I have always thought that the converse was also true, i.e. that if the result is coaxial with the argument (for an arbitrary argument) then the operator must be isotropic.

Apparently this is not the case, since in my example we can tell that the stress is coaxial with the (total and elastoc) strain.

This is the 'issue' I addressed in the OP, but according to the article cited by Biswajit there is nothing funny going on. I guess what seems to be a source to non-coaxiality will infact turn out to produce coaxial tensors when integrated, though this is just me guessing now.

Your interpretation of εdev is correct, and εiso  is the isotropic part that you subtracted from the total.

I agree that in your plasticity example there is deformation induced anisotropy when cosidering the total strain, but the stress is still coaxial with the elstic strain, right?

-Espen

 

Isotropy or Coaxiality

Fri, 2015-01-23 07:09

In reply to Anisotropic stiffness of isotropic material

Dear Espen,

There are a couple of points about your question that should be clarified. They can be itemized as follows:

1. Isotropy means that constitutive behavior remains identical for all material directions. Therefore stress should be independent of prinipal directions of strain (eigenvectors of strain tensor). To this end stress should be related to strain only through invariants of strain tensor. If this is the case isotropy is preserved for all material directions.

2. I would be so grateful if you could provide explicit definitions for εdev ( = ε - 1/3 tr(ε) I ?) and εiso. Then we can be more specific about elasticity tensor.

3. In previous simple example, it has been shown that for two perpendicular directions, the stress-strain relation (constitutive relation) is different and therefore we have deformation induced anisotropy.

Mohsen

Is the isotropy violated?

Thu, 2015-01-22 17:17

In reply to Isotropic Stress-Strain Relation

Thank you for your response, Mohsen.

I'm not quite sure I follow what you say here, do you mean that the relation isotropic or not?

In my post I didn't mention plasticity, but I agree that there are aspects in common in this case as one gets more or less the same stiffness tensor using J2 plasticity. Also I wrote the total stress--strain relation which could perhaps be interpreted as something to do with total strain; what I meant was the opposite of incremental/differential relation.

never the less, I don't follow the argument for why isotropy is lost when there is plastic strain only in the x direction. The principal directions will still coincide, right?

-Espen

Spot on!

Thu, 2015-01-22 16:59

In reply to Re: Anisotropic isotropic material aka deformation induced aniso

Thank you so much Biswajit, the article is spot on!

I also found the thesis by Fuller which I'm looking forward to reading :)

I'm surprised that this phenomenon is mentioned so scarsely in the litterature, but very good that Brannon (being the supervisor, I guess) does something about it. BIG fan of her :)

-Espen

Isotropic Stress-Strain Relation

Thu, 2015-01-22 14:51

In reply to Anisotropic stiffness of isotropic material

Actually, the relation between stress and elastic strain remains isotropic. Consider the following relation:

s = 2G (e - ep)

In this relation s is the deviatoric part of the stress and e and ep are respectively total deviatoric and plastic strain tensors. From this simple relation it is evident that s is coaxial (share the same principal directions) with (e - ep) and not e itself. It is possible that we might have plastic strain only in x direction (e11p ≠ 0) while there is no plastic strain in y direction (e22p = 0). According to previous relation we can write:

s11 = 2G (e11 - e11p)     and

s22 = 2G (e22)

Clearly, the stress-strain relations are different for x and y directions showing that the isotropy is violated due to the prescence of plastic strain in x direction only.

Mohsen

Re: Anisotropic isotropic material aka deformation induced aniso

Thu, 2015-01-22 13:58

In reply to Anisotropic stiffness of isotropic material

You can look at "On the effects of deformation induced anisotropy in isotropic materials" by Fuller and Brannon, 2013, IJNME 37(9), 1079-1094, DOI: 10.1002/nag.1139  for some insight into the problem.

-- Biswajit

 

I meant " by the material

Thu, 2015-01-22 10:32

In reply to State of stress in a Uniaxial Tensile Test

I meant " by the material present on the other side along x and z axis as" in the above comment on 6h line . Sorry for the typo

Dear Espen,

Thu, 2015-01-22 10:26

In reply to State of stress in a Uniaxial Tensile Test

Dear Espen,

Thanks a lot for your response. I undertsand that tensile test is an attemp to reach the ideal case of uniaxial tensile test. 

However my doubt is a bit different. From the way I understand there should be triaxial state of stress in the tensile specimen before the start of necking even at contnuum scale (not microstrucutre state). For e.g Consider a point inside a cylindrical uniaxial tensile test specimen. Now if you section it through x,y, z planes, y axis being the direction of loading, apart from the pull experienced by it along y axis due to external loading, there will be also be pull/push on this point by the material present on the other side along x and y axis as the material will be contarcting along x and z direction due to transverse strain. Wont this push/pull cause stress on this point even in x and z direction? As long as the point is inside the cylinder there should be triaxial stress. On surface however the normal stress would become zero.

Kindly let me know where am i going wrong in the understanding.

Thanks,

Danish

An idealization

Thu, 2015-01-22 05:19

In reply to State of stress in a Uniaxial Tensile Test

I believe that the uniaxial stress state is an idealization and very hard to obtain i reality.

However, if you ensure that the specimen ends are not restricted against transverse deformation is possible (e.g. by using bolts in a slightly large hole through the specimen) and align the applied force perfectly with the speciment, the (macroscopic) state of stress should be very close to uniaxial as long as the deformation is uniform in the guage area (the part of the speciment where you measure deformations).

On the other hand, the microscopic state of stress, i.e. on the scale of the microstructure, would generally not be in a state of uniaxial stress. For dual phase steels (that I see you're focusing on:) the stress state in each phase is triaxial after initial (local) plastic yield, and very much so after "macroscopic" yielding.

Even for an idealized homogeneous material the stress state is triaxial when the deformation is no longer uniform, i.e. when a neck starts to form. A standard form of "correction" to the longitudinal stress is the so-called Bridgeman correction which takes the radius of curvature of the nack into account.

- Espen

PS: all of the above relates to isotropic materials only!

Proceedings?

Wed, 2015-01-21 14:50

In reply to McMAT 2015 symposium on materials and manufacturing

Does your comment about abstracts mean that there is not a mandatory proceedings paper for this conference? The conference web site does not mention a paper, but it is not explicit.

Cohesive parameters input in Abaqus

Wed, 2015-01-21 03:53

In reply to Cohesive debonding using traction separation law

Hi Shouvik,
You are welcome. I am glad to help.
1.    E/Enn in Abaqus means E or Enn, not E divided by Enn. You should input 1e6.
2.    In section control, “analysis default” assume the thickness of the cohesive elements to be 1mm. Thus, strain equals displacement (separation) in value.  It’s the traction-separation law that we defined for the cohesive elements after all. You can also specify the initial thickness to be 1 in the section control, just to be sure. 
The Abaqus user’s guide is very helpful, you should check it out.

Regards,
Jia

applicability for J-integral at interfaces (two dissimilar mat.)

Wed, 2015-01-21 02:17

In reply to The J integral

Dear Prof. Suo,

Rice introduced the J integral concept which is equal to the energy release rate for a uniform, linear or nonlinear elastic material free of body forces and subjected to a 2d deformation field (plane strain, plane stress etc.).

 

In ANSYS and Abaqus, the implementation is always referenced to the work of Shih (1986). There has been numerous publications on the use of J-integral to estimate Gc (critical energy release rate) for cracks which are located at the interface of dissimilar materials.

I could not find the relevant information if this approach is applicable, if yes, under what  circumstances? Could you please give me some information or references about the implementation of Jintegral at the interface of two materials?

I am looking forward to hearing from you,

Yalcin

 

Hi Jia,

Wed, 2015-01-21 00:12

In reply to Cohesive parameters

Hi Jia,

 

thanks for resolving many doubts. I wasnt sure that traction was N/mm2 and E also N/mm2 and that displacements are in mm. Just a few last questions using E = G1 = G2 = 1e6 as you suggested and say k = 1e5.

1. What should be the value E/Enn and G1/Ess  input to abaqus for the adhesive. Should this be 1e6/1e5 = 10 or  1e6 or 1e5 for this set of data .

2. I see that the displacement at failure should be in mm (0.01 in this case). I am using an initial cohesive layer thickness of 10-2 mm. In abaqus the cohesive section uses the "'analysis default"' thickness option. Will abaqus assume thickness 1 mm and hence strain at failure = 0.01/1 = 0.01. Or will the strain be 0.01/10-2 = 1 at failure. Also does this change if option of cohesive section is set as ''üse nodal coordinates".

 

Thanks again for the most valuable inputs.

 

Regards,

Shouvik

I am checking the following

Tue, 2015-01-20 21:07

In reply to Reducing Stiffness of a FE model

I am checking the following contents from ABAQUS manual.
I hope this can help you even a little.

Choosing between full- and reduced-integration elements

Reduced integration uses a lower-order integration to form the element stiffness. The mass matrix and distributed loadings use full integration. Reduced integration reduces running time, especially in three dimensions. For example, element type C3D20 has 27 integration points, while C3D20R has only 8; therefore, element assembly is roughly 3.5 times more costly for C3D20 than for C3D20R.

In Abaqus/Standard you can choose between full or reduced integration for quadrilateral and hexahedral (brick) elements. In Abaqus/Explicit you can choose between full or reduced integration for hexahedral (brick) elements. Only reduced-integration first-order elements are available for quadrilateral elements in Abaqus/Explicit; the elements with reduced integration are also referred to as uniform strain or centroid strain elements with hourglass control.

Second-order reduced-integration elements in Abaqus/Standard generally yield more accurate results than the corresponding fully integrated elements. However, for first-order elements the accuracy achieved with full versus reduced integration is largely dependent on the nature of the problem.

Hourglassing

Hourglassing can be a problem with first-order, reduced-integration elements (CPS4R, CAX4R, C3D8R, etc.) in stress/displacement analyses. Since the elements have only one integration point, it is possible for them to distort in such a way that the strains calculated at the integration point are all zero, which, in turn, leads to uncontrolled distortion of the mesh. First-order, reduced-integration elements in Abaqus include hourglass control, but they should be used with reasonably fine meshes. Hourglassing can also be minimized by distributing point loads and boundary conditions over a number of adjacent nodes.

In Abaqus/Standard the second-order reduced-integration elements, with the exception of the 27-node C3D27R and C3D27RH elements, do not have the same difficulty and are recommended in all cases when the solution is expected to be smooth. The C3D27R and C3D27RH elements have three unconstrained, propagating hourglass modes when all 27 nodes are present. These elements should not be used with all 27 nodes, unless they are sufficiently constrained through boundary conditions. First-order elements are recommended when large strains or very high strain gradients are expected.

Shear and volumetric locking

Fully integrated elements in Abaqus/Standard and Abaqus/Explicit do not hourglass but may suffer from “locking” behavior: both shear and volumetric locking. Shear locking occurs in first-order, fully integrated elements (CPS4, CPE4, C3D8, etc.) that are subjected to bending. The numerical formulation of the elements gives rise to shear strains that do not really exist—the so-called parasitic shear. Therefore, these elements are too stiff in bending, in particular if the element length is of the same order of magnitude as or greater than the wall thickness. See “Performance of continuum and shell elements for linear analysis of bending problems,”  Section 2.3.5 of the Abaqus Benchmarks Manual, for further discussion of the bending behavior of solid elements.

Volumetric locking occurs in fully integrated elements when the material behavior is (almost) incompressible. Spurious pressure stresses develop at the integration points, causing an element to behave too stiffly for deformations that should cause no volume changes. If materials are almost incompressible (elastic-plastic materials for which the plastic strains are incompressible), second-order, fully integrated elements start to develop volumetric locking when the plastic strains are on the order of the elastic strains. However, the first-order, fully integrated quadrilaterals and hexahedra use selectively reduced integration (reduced integration on the volumetric terms). Therefore, these elements do not lock with almost incompressible materials. Reduced-integration, second-order elements develop volumetric locking for almost incompressible materials only after significant straining occurs. In this case, volumetric locking is often accompanied by a mode that looks like hourglassing. Frequently, this problem can be avoided by refining the mesh in regions of large plastic strain.

If volumetric locking is suspected, check the pressure stress at the integration points (printed output). If the pressure values show a checkerboard pattern, changing significantly from one integration point to the next, volumetric locking is occurring. Choosing a quilt-style contour plot in the Visualization module of Abaqus/CAE will show the effect.

 

Regards,

I hope solving high stiffness of C3D8

Tue, 2015-01-20 20:26

In reply to Reducing Stiffness of a FE model

Dear Maajid

 

Nice to meet you who have same problem with me.

Please recommend progressive advise.

Thank you all in advance.

 

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