This article is rated Start-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Physics Mid‑importance This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics articles Mid This article has been rated as Mid-importance on the project's importance scale.

F should be defined somewhere. — Preceding unsigned comment added by 147.83.143.248 (talk) 16:46, 6 March 2014 (UTC)[reply]

Added hyperlink. Bbanerje (talk) 01:37, 7 March 2014 (UTC)[reply]

Wikipedians will be better served if instead of having a single article covering half a dozen stress measures wikipedia hosted individual articles dedicated to each of these topics. I would create the articles myself but at this moment my knowledge of any stress measure beyond Cauchy's is rather limited. So, is anyone up for the job? -- Mecanismo | Talk 22:45, 21 September 2010 (UTC)[reply]

I feel that the article should be added to instead of being broken up. Right now it's sketchy and needs explanations of the various stress measures that a beginner can understand after they've become familiar with the Cauchy stress. Bbanerje (talk) 04:53, 9 December 2010 (UTC)[reply]

The problem with the current setup is that the article cannot even define properly what each measure is. Perhaps it would be better if all math were moved to the respective articles, leaving here only an intuitive' explanation of the concepts, their main differences and their uses. --Jorge Stolfi (talk) 01:02, 24 February 2013 (UTC)[reply]

The Cauchy stress tensor is NOT what engineers call stress. How does one explain that without confusing everyone? Bbanerje (talk) 23:57, 24 February 2013 (UTC)[reply]

The definition of Cauchy stress says

${\displaystyle d\mathbf {f} =\mathbf {t} ~d\Gamma ={\boldsymbol {\sigma ))^{T}\cdot \mathbf {n} ~d\Gamma }$

Why ${\displaystyle {\boldsymbol {\sigma ))^{T))$ and not just ${\displaystyle {\boldsymbol {\sigma ))}$? Is ${\displaystyle {\boldsymbol {\sigma ))}$ a 3×3 matrix or a 1×6 vector?
--Jorge Stolfi (talk) 02:59, 27 February 2013 (UTC)[reply]

Here ${\displaystyle {\boldsymbol {\sigma ))}$ is a second order tensor. There is no mention of coordinate systems in the article and one can use whatever coordinate representation one likes. The use of the transpose is to keep the notation consistent with that for other stress measures which may not be symmetric and to keep the linear operator interpretation of the tensor obvious. A more commonly seen expression is ${\displaystyle \mathbf {n} \cdot {\boldsymbol {\sigma ))}$.Bbanerje (talk) 01:43, 7 March 2014 (UTC)[reply]

Perhaps there are multiple conventions, which would be unfortunate, but the Malvern text has the opposite definitions for the nominal and first Piola-Kirchoff stresses from what you are using.146.6.104.189 (talk) 17:45, 11 August 2016 (UTC)[reply]

Also the nomenclature is confusing because "stress" is elsewhere defined as a vector (the traction force divided by the area), while "stress tensor" is the linear map from normal to stress vector. Perhaps some occurrences of "stress" in this article should be changed to "stress tensor" for clarity?
--Jorge Stolfi (talk) 03:09, 27 February 2013 (UTC)[reply]

Stress is always a tensor. What you are calling the "stress vector" is usually known as the "traction vector". Bbanerje (talk) 01:46, 7 March 2014 (UTC)[reply]

Sorry but I am quite confused. I presume that ${\displaystyle F}$ is the Jacobian matrix of the displacement map, correct? Then, suppose a cube of "mathematical rubber" 1 meter on each side is stretched 100 times in the z direction, and unchanged in x and y. The Jacobian ${\displaystyle F}$ will be

${\displaystyle {\begin{bmatrix}1&0&0\\0&1&0\\0&0&100\end{bmatrix))}$

So if the z traction force ${\displaystyle f}$ in the current configuration is 1 newton vertical, the "pulled back" force ${\displaystyle f_{0))$ in the original configuration will be only 0.01 newton vertical. The original stress will be 1 Pa, the "pulled back" stress will be only 0.01 Pa. Is this correct? --Jorge Stolfi (talk) 04:52, 27 February 2013 (UTC)[reply]

Traction has units of force per unit area. You'll therefore have to make sure you are using the right area when you move between configurations. Use Nanson's formula for the conversion. Also note that the 2nd P-K stress doe snot have any obvious physical interpretation. Bbanerje (talk) 01:49, 7 March 2014 (UTC)[reply]