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I saw Prof. McMeeking not so

Fri, 2014-11-21 02:49

In reply to Speech of Acceptance of the 2014 Timoshenko Medal by Robert M. McMeeking

I saw Prof. McMeeking not so long ago and we exchanged a couple of jokes regarding the significant relevance of Timoshenko Medal speech of acceptance and how difficult it is to keep the level of previous awardees. 

I pleasantly see that he pulled out of his sleeve a fantastic (and humorous) speech. Congratulations for this very nice piece of text (and thanks to iMechanica for sharing it) and of course congratulations for the (well deserved) medal.

Congratulations to Professor McMeeking!

Thu, 2014-11-20 22:30

In reply to Speech of Acceptance of the 2014 Timoshenko Medal by Robert M. McMeeking

Dear Bob:  You have been a caring mentor and an inspiring role model.  Your work has had a profound effect.  In your hands, mechanics becomes a way to describe all things describable.  Crack-tip blunting, transformation toughening, powder sintering, composite fatigue, electromechanical coupling, living tissues, lithium-ion batteries...  Huajian asked, what have you not worked on?  You make old mechanics fresh, and youthful mechanics useful.  Thank you for showing the way.  Looking forward to watching your next adventure.


Thu, 2014-11-20 17:35

In reply to Lithiation of SiO2 in Li-Ion Batteries: In Situ Transmission Electron Microscopy Experiments and Theoretical Studies

Dear Kejie,

Great work and congrats on starting your faculty career !

Re: von Mises stress for beams

Thu, 2014-11-20 12:35

In reply to Re: von Mises stress for beams

Thanks for your help Alejandro! The first book has the example you mentioned.

Re: von Mises stress for beams

Wed, 2014-11-19 00:23

In reply to von Mises or Equivalent stress for Beam Elements

It depends on how it is oriented the Cartesian coordinate in your problem. I think there is an example for a bike pedal arm that is subjected to combined bending and torsion, in this book:

or perhaps in this one:




Re: von Mises for beam elements

Tue, 2014-11-18 18:46

In reply to Re: von Mises for beam elements

Which one exactly? Implementation wise it won't matter but I am trying to understand the theoretical background. Any references I can look at?

Dear fruends,

Tue, 2014-11-18 16:53

In reply to Bridge

Dear fruends,

I need to know how to  generate random particle inclusion in the matrix. I request kindly give me some idea.


Re: von Mises for beam elements

Tue, 2014-11-18 15:47

In reply to von Mises or Equivalent stress for Beam Elements

The torsional stress must be included in one of these terms:

\tau^2_{xy} + \tau^2_{yz} + \tau^2_{zx}



Re: von Mises for beam elements

Tue, 2014-11-18 12:08

In reply to Re: von Mises for beam elements

So you are saying the definition of von mises stress used for continuum elements is valid for beam element too. Here, axial stress is \sigma_{xx}, shear in y-direction is \tau_{xy}, shear in z-direction is \tau_{xz}. All other components are zero? How do you resolve the torsional stress in terms of the stress components indicated in the expression?

hi steven,

Tue, 2014-11-18 07:17

In reply to Cohesive (zero thickness): NODE NUMBERING MIDE NOT BE CORRECT (A

hi steven,

ı am struggling with the same problem.

i use python to build the zero thickness cohesive elements, some cohesive elements is fine but the others have the same problem as you. 

have you solved this problem? 

Re: von Mises for beam elements

Mon, 2014-11-17 19:37

In reply to von Mises or Equivalent stress for Beam Elements

If you know the analytical expressions for the axial, shear and torsional stresses at any point in the beam, then you can use

 \sigma_\text{VM} = \sqrt{{1\over 2}\left[\left(\sigma_{xx} - \sigma_{yy}\right)^2 + \left(\sigma_{yy} - \sigma_{zz}\right)^2 + \left(\sigma_{zz} - \sigma_{xx}\right)^2 \right] + 3 \left(\tau^2_{xy} + \tau^2_{yz} + \tau^2_{zx}\right) } 

Thanks for the response

Mon, 2014-11-17 18:45

In reply to von Mises or Equivalent stress for Beam Elements

Thanks for the response Alejandro. Yes, I am check buckling also along with stresses. But rather than enforcing multiple stress constraints (one for each stress - axial, bending shear, torsional shear), I want to enforce a single constraint on equialent stress. So back to original question, how do I compute this equivalent/von Mises stress?

Re: von Mises for Beam Elements

Mon, 2014-11-17 18:20

In reply to von Mises or Equivalent stress for Beam Elements

Before using von Mises stress to design a structure with beam elements, I would check the structure  is not determined by buckling due to axial and/or bending forces. I usually never use von Mises on a beam structure unless I am sure it will fail by yielding. A common mistake that I have seen is that many mechanical engineers try to use the von Mises stress to analize by yielding a structural element that clearly would fail by buckling.

Is your beam element really a beam element? There are some software that use beam element to refer to beam as well as frame elements.

The first few chapters of

Mon, 2014-11-17 15:03

Effect of single soft interlayer on the crack driving force

Mon, 2014-11-17 02:38

In reply to The Discussion

Dear Per,

I am glad that you will continue Wolfgang Brock's work.

I think, you raised very interesting questions in your comments to our paper.

Let us first consider a linear elastic material with inhomogeneity of the Young's modulus E, i.e. E exhibits a jump at the interface. Theory predicts that, for a crack ending directly at the interface, the crack driving force becomes infinite, if E decreases in the crack growth direction. The crack driving force becomes zero if E decreases in the crack growth direction.
As you write, this is unrealistic, and one should introduce a non-linear region arond the tip. How large should such a region be? A related question is, how sharp can a real bimaterial interface be?
A region not less than an atomic distance is maybe reasonable.

I am not sure whether I understand correctly your question in the case of yield stress inhomogeneity. If we take the length of the process zone l_proc as being proportional to the crack tip opening displacement (with the proportionality constant of the order of 2), then we have considered in the paper also cases where the distance L between crack tip and interface is smaller than l_proc. I do not see a big problem here. Do we overlook something?
Another problem is - and maybe that is also what you had in mind - that not only the crack driving force, measured in terms of J_tip, changes when the crack tip approaches the interface. Also the crack growth resistance R will, in general, change. You know, the crack extends, if J_tip >= R. How can we model the change in R? Here we have had, as you also proposed, the idea of applying the cohesive zone model, since this model allows us to with an intrinsic fracture resistance of the materials, prescribed by the cohesive energy. The materials left and right of the interface can then have different material properties and different characteristic cohesive zone parameters (cohesvive energy and cohesive stress).
We have started with the analyses, first results are already available, but not yet published.

Otmar Kolednik
Erich Schmid Institute of Materials Science,
Austrian Academy of Sciences,
Jahnstrasse 12, A - 8700 Leoben, Austria

Links are re-connected

Sun, 2014-11-16 09:30

In reply to Why your lecture notes are

Not sure what have happened.  I have just reconnected all the links.  Thank you so much for pointing this out.

Why your lecture notes are

Sun, 2014-11-16 08:05

In reply to Engineering Sciences 247: Fracture Mechanics

Why your lecture notes are unavailable now?

hi brother iam trying to

Sun, 2014-11-16 02:52

In reply to any question any answer in Abaqus Ansys and finite element modeling

hi brother iam trying to model rc slab with opening and strengthening that opening with CFRP sheets so i tried to select the engineering constants form the elastic window and selected the hashin damage and give it damage evloution but the CFRP still dont take more than 5 KN load till failure !!! it must take about 25 KN before failure occur please help me what should i do and to make the cfrp sheet more stronger ??by the way the parameters i used for engineering constants is 

80000   231000   80000   0.05   0.05    0.45   3700   79655   79655

and for the jashin damage is

4100   50   4100   50   100   100

and for damage evolution is

1   1000   1   1000

my cfrp E is 231000 Mpa and tensile strength is 4100 Mpa

so the upper parameters i calaculate them from excel sheet belong to my friend 

and my cfrp sheet is unidirectional 

please help . . . . 


Fri, 2014-11-14 09:11

In reply to you can directly use open

Thank you so much for your answer. But I dont have that software. Is that free to download from internet.

you can directly use open

Thu, 2014-11-13 22:35

In reply to SEM image conversion into finite element model

you can directly use open source software OOF to convert your SEM images to FE meshes and then use abaqus to carry out the FE analysis.


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