Talk:Infinitesimal strain theory

The last section on engineering uses epsilon instead of varepsilon. Is there any reason for the inconsistency? If not, I would prefer varepsilon Berland 16:03, 18 January 2007 (UTC)[reply]

Earthquake

Hello, How would I calculate the force on an object created by a 6.0 earthquake. (The forced movement)

I'd look at Richter magnitude scale. You can't say how much force or stress an object sees without a particular problem statement. It's like asking how hot a 100W lightbulb is—it just isn't a meaningful question. —Ben FrantzDale 14:43, 15 February 2007 (UTC)[reply]

The magnitude of an earthquake is only an indication of the energy released at point of rupture. At the site you are concerned with, the force on an objects mass due to the earthquake will be a function of the earthquake strong motion (acceleration time history) at the site (F = m*a). This requires either a strong motion record that was recorded at the site during the earthquake, or if it is an estimate of a possible future event from a particular fault source, in which case you know (or estimate from historical records) the magnitude and distance, and with an attenuation relationship (describes the reduction in energy as the wave travels through the earth from fault to your site), you can determine the bedrock ground motion at your site. You then consider how the ground motion is amplified (or perhaps dampened) by the soil layers on your site (this is called site response analysis). The resulting accelerations at different modes of oscillation (frequency or period) can then be used to estimate the acceleration in the earthquake that will excite your particular object. This depends on the natural frequency of the object you are considering. If, say a tall building, high frequency waves will not have much effect, but low frequency (long period) motion will cause the building to sway about its natural frequency. This is the acceleration from an earthquake that you are most concerned with as a designer.GeoEng 07:51, 6 April 2007 (UTC)[reply]

Rewrite

HI. I think there's another error in the article. (3/7/2012) In "the Axial Vector" section ~ which sounds really weird. Wtf is an axial vector...? your omega = del X u when written in tensor notation is missing the derivative term. Books on fluid mechanics like to call this the rotation rate vector, vorticity, etc. not "axial vector". You actually switch in this segment between calling it rotation vector and axial vector. please call it one thing. subject difficult enough without odd variable names.

Hi. I think there is an error in this article. (today: 5/11/05) It's written: 'If Strain is greater than 1, the body has been lengthened; if it is smaller than 1, it has been compressed.' In fact it is: if strain > 0, the body has been lengthened and if strain < 0... according to the definition of the strain written in this same article! Am I right?

Thanks; I've rewritten parts of the article. Please review it and let me know what you think. Tom Harrison ^(talk) 04:52, 6 November 2005 (UTC)[reply]

I just did loads of stuff to this article. I would say its fairly clean, not quite featured article standard, but clean enough to remove the cleanup notice. I took out a couple of bits about earthquakes, I thought that was a bit too detailed (do we really want to know about tectonic folding in a strain article?), and changed the Small values of strain section to Engineering strain vs. true strain. I'm not done with this article yet. --Nathan 07:22, 2 December 2005 (UTC)[reply]

To do

Torsional strain
Strain hardening -- & strain softening behaviour! GeoEng 07:53, 6 April 2007 (UTC)[reply]
Necking and poisson's ratio
Engineering strain vs. True strain
- done. I'll get working on the rest soon. --Nathan 07:22, 2 December 2005 (UTC)[reply]
Mohrs Circle of Strain (yes it can be applied to strain as well!) GeoEng 07:53, 6 April 2007 (UTC)[reply]

Tom Harrison ^(talk) 14:25, 6 November 2005 (UTC)[reply]

Lagrangian strain (non linear component)Katanada (talk) 17:27, 23 January 2008 (UTC)[reply]
Eulerian strain Katanada (talk) 17:27, 23 January 2008 (UTC)[reply]
Deformation gradient tensor --Katanada (talk) 17:30, 23 January 2008 (UTC)[reply]

Strain & deformation

Hi everyone! I've got a question concerning the terms "strain" and "deformation". The article says strain characterizes deformation, while, e.g., Britannica makes no difference between these words. It says "...deformation, or strain,...". So which is correct? Are there any authorities supporting the view expressed in the article? Viktor 12:02, 15 February 2007 (UTC)

Strain is a technical term with a precise technical dfinition, so in general Britannica may be right, but in technical discussion strain means something particular whereas "deformation" isn't well-defined (as far as I know). —Ben FrantzDale 14:43, 15 February 2007 (UTC)[reply]

Deformation is the accumulation of strains. I would consider units of deformation to be mm, m, etc., whereas strain is normalised over the dimension being considered. GeoEng 07:59, 6 April 2007 (UTC)[reply]

Hi, i'm trying to find the strain in an aliminium pipe, how do i find the change in length, so that i can find the strain? Nathan —Preceding unsigned comment added by 141.241.129.75 (talk) 18:02, 20 October 2007 (UTC)[reply]

You probably want to use a strain gauge, since strain in a metal pipe is almost certainly small. —Ben FrantzDale (talk) 12:41, 18 August 2010 (UTC)[reply]

hi !! i guess i know the difference.... strain gives the deformation in comparison with the original length. for example. if strains is 0.03, it means the body has deformed 3 percentage of its length. where as deformation is just a value with out comparing with its length. if deformation is 3mm.. it means the body deformed 3mm.... —Preceding unsigned comment added by 125.21.211.5 (talk) 16:41, 16 August 2010 (UTC)[reply]

I have never heard "deformation" used with technical precision within solid mechanics as a quantitative measure. I have only heard the word modified to describe how something deformed as in "elastic deformation", "plastic deformation", etc.. Wikipedia does have Deformation (mechanics) which is linked from Strain, a disambiguation page. However, that page goes on to talk about strain. It appears to only use "deformation" with technical specificity in the mathier sections. —Ben FrantzDale (talk) 12:41, 18 August 2010 (UTC)[reply]

Merging St. Venant's Compatibility equations article

The St. Venant's compatibility equations article should be merge into the strain article as the former can be seen as a subtopic of strains. Comments? Sanpaz (talk) 16:06, 12 February 2008 (UTC)[reply]

No. It is only the case of rank 2, dimension 3 that is applied to strain. The result is for a general dimension and rank.Billlion (talk) 08:34, 29 March 2008 (UTC)[reply]

Rename?

I think that this article would be more accessible if it were simply named "Infinitesimal strain", which starts with a simple explanation of infinitesimal strain and then goes on to spend most of its time on infinitesimal strain theory. Opinions? Awickert (talk) 19:49, 19 January 2009 (UTC)[reply]

I disagree (need to think about it more). Please see Talk:Deformation_(mechanics)#Merging_articles Sanpaz (talk) 00:01, 20 January 2009 (UTC)[reply]

replied there. Awickert (talk) 00:20, 20 January 2009 (UTC)[reply]

Notation

I really dislike the notation used here, as well as in the finite strain theory article, for the displacement gradient tensor. Writing the operand before the gradient operator seems very unusual to me in the following equation:

\mathbf {E} \approx \mathbf {e} \approx {\boldsymbol {\varepsilon }}={\frac {1}{2}}\left(\mathbf {u} \nabla ^{T}+\mathbf {u} \nabla \right)\,\!

I think it should be written as:

\mathbf {E} \approx \mathbf {e} \approx {\boldsymbol {\varepsilon }}={\frac {1}{2}}\left(\mathbf {\nabla } \mathbf {u} +(\mathbf {\nabla } \mathbf {u} )^{T}\right)\,\!

Where did this notation come from? There are no references in this article which is also problematic. The continuum mechanics article uses the standard notation, and there are even instances in this article where it says $\|\nabla \mathbf {u} \|\ll 1\,\!$ . Sadd 2005 also uses the form I wrote above. Perceptual Chaos (talk) 15:00, 6 July 2009 (UTC)[reply]

Not sure whether this is still an issue, as I can't find the notation anywhere in this article. But yes, I agree that the notation of your first equation is irritating. I have seen it occasionally and (as far as I understood) it comes from the definition of the displacement gradient tensor: sometimes it is defined as

\nabla \times \mathbf {u}

, and sometimes as the transposition of that. The latter case is handwavingly abbreviated by

\mathbf {u} \times \nabla

(which is wrong if you insist of differential operators only acting on quantities to their right) and, I guess, this is where the notation in question comes from.Das O2 (talk) 08:37, 3 May 2020 (UTC)[reply]

What is F?

A vector field F makes an unannounced appearance is Sec 1 after "Furthermore". I have no idea what this bit is about. Can someone clarify? Billlion (talk) 09:18, 20 January 2010 (UTC)[reply]

Definition of infinitesimal strain theory

There is a mistake in the first paragraph.

Displacement field u cannot possibly be compared with 1 (one) because it carries dimensions (actually length).

You may be willing to compare the displacement field u with x or X

217.216.118.109 (talk) 23:31, 30 October 2010 (UTC)[reply]

You're right. The correct linearization argument is that grad u << 1 which implies that both eps << 1 and the spin omega << 1. The displacement doesn't have to be physically small, e.g., mm instead of m. Bbanerje (talk) 22:42, 31 October 2010 (UTC)[reply]

Eulerian vs Lagrangian strain

Hi everybody, I just noticed what might be an important error in the definition of the strain tensor. Following Chaikin and Lubensky's "Principles of condensed matter physics", there should be a minus sign in front of the nonlinear term of : $\mathbf {e} ={\frac {1}{2}}\left(\nabla _{\mathbf {x} }\mathbf {u} +(\nabla _{\mathbf {x} }\mathbf {u} )^{T}+\nabla _{\mathbf {x} }\mathbf {u} (\nabla _{\mathbf {x} }\mathbf {u} )^{T}\right)\approx {\frac {1}{2}}\left(\nabla _{\mathbf {x} }\mathbf {u} +(\nabla _{\mathbf {x} }\mathbf {u} )^{T}\right)\,\!$ (otherwise it coincides with Lagrangian expression anyway). I corrected this. If anyone knowledgeable at this subject finds it wrong, please correct back. (I'm a graduate student in theoretical physics). Cheers PR — Preceding unsigned comment added by 88.219.55.219 (talk) 09:17, 25 October 2013 (UTC)[reply]

A link to Eulerian and Lagrangian strain tensor could be useful. Even only if to know what E and u are. 109.164.216.10 (talk) 10:14, 14 February 2014 (UTC)[reply]

Fix red text

Can somebody please fix the red "Failed to parse(unknown function '\begin')" text? Or am I the only one seeing it? ----ANDROBETA 08:26, 15 February 2014 (UTC)[reply]

Should be fixed now, thanks for mentioning the problem. --Mark viking (talk) 11:59, 15 February 2014 (UTC)[reply]

References can be inserted

Hi everybody, it can be of some interest to refer of some published text-book in which these information are reported and (maybe) expanded.

A very similar article to the present one can be found here: www.iue.tuwien.ac.at/phd/singulani/disssu4.html. In such a text, there where two references:

[37] G.-Q. Zhang, W. van Driel, and X. Fan, Mechanics of Microelectronics. Springer, 2006, vol. 141, ch. 4.

[39] A. F. Bower, Applied Mechanics of Solids. CRC Press, 2011.

Maybe there are suitable also for the actual wikipedia article? Thank you, have a nice day. --Car.cin (talk) 15:19, 6 August 2018 (UTC)[reply]

Condition not clear

Hi all! In the first section, named Infinitesimal strain tensor, two conditions are required to perform the geometric linearization:

the displacement (vector) is small compared to unity: $\|\mathbf {u} \|\ll 1$ ; and
the displacement gradient is small compared to unity: $\|\nabla \mathbf {u} \|\ll 1$ .

It is unclear why the first condition is required. In the reference textbook given at the end of the article (namely Slaughter, W. S., 2002, The Linearised Theory of Elasticity, Birkhauser) only the second condition is reported to be necessary and sufficient for the linearization.

I suggest remouving the condition (1) unless someone clarifies why it is necessary.--Car.cin (talk) 07:53, 14 August 2018 (UTC)[reply]

Agreed & removed. Both in the analytical and in the geometrical derivation, one can see that the magnitude of the displacements doesn't matter. Instead, only the gradients mater. See also Talk:Infinitesimal_strain_theory#Definition_of_infinitesimal_strain_theory. Das O2 (talk) 21:08, 27 April 2020 (UTC)[reply]

The vector u isn't defined

The vector u isn't defined, rendering the entire mathematical content of the article meaningless. 62.30.82.22 (talk) 17:59, 12 October 2024 (UTC)[reply]