Talk:Shear stress

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Old comments[edit]

Shouldn't this be connected to faults and earthquakes? Wetman 20:35, 1 Sep 2004 (UTC)

what is Q?[edit]

Richard Giuly 02:43, 27 January 2007 (UTC)[reply]

Q should be the static momentum
What static momentum? The whole equation seems plain wrong. In the area of engineering I deal with, shear stress is equal to shear force over shear area. Any reference to the bending inertia, 'I', is only relevant to longitudinal shear, not transverse shear. -- Kvetner 13:26, 9 March 2007 (UTC)[reply]
From my point of view this equation is wrong (not only this, but dimmensions are wrong). It seems to be the expression for the torsion, Kvetner, you're right, for shear stress over a beam, only moment and area matters. Somebody correct this or I'll do it myself when I learn how to edit pages. thx —Preceding unsigned comment added by Trufetes (talkcontribs) 17:33, 14 December 2007 (UTC)[reply]
I'm currently taking a class that involves this equation, and that is indeed the exact equation out of my textbook. As I understand it, F/A gives the average shear stress on a cross-section. However, shear stress is not actually constant across the whole cross-section, so this formula can be used to find the shear stress at a particular point on the cross-section of the beam. Hope this helps clarify. -- (talk) 21:12, 10 September 2008 (UTC)[reply]

the equation there is for sheer stress in a plane if that's any help

This is an interesting forum for scientific discussion[edit]

This forum is trully amazing. It provides a place where people can discuss things, even trivial like shear stress.... —Preceding unsigned comment added by (talk) 23:08, 7 July 2008 (UTC)[reply]

Important limitations of this formula to be considered.[edit]

This equation is derived from the flexure formula, and therefore is subject to some of the same assumptions made in its derivation. Namely this formula is only valid for beams made of a homogeneous linear elastic material undergoing only small deflections. Also the edges of the cross section must be parallel to the y-axis, and the shear stresses must be uniform across the cross section width. No triangles or semicircular shapes. Finally the accuracy of this model varies in a rectangular cross section with the height-to-width ratio. The most accurate approximation comes from very narrow beams (height much larger than width). —Preceding unsigned comment added by (talk) 23:08, 20 October 2008 (UTC)[reply]

Beam stress formula name[edit]

"This formula is also known as the Jourawski formula." Jourawho? Never heard of the guy; can't find much to support this (that might not have derived from here). It's possible that this is a bastardisation of a whole variety of Russian/slavic names NcLean (talk) 20:22, 12 November 2008 (UTC)[reply]

I added a ref to it. Wizard191 (talk) 02:57, 13 November 2008 (UTC)[reply]

"Hanging something" analogy[edit]

Apart from my feeling that analogies shouldn't be part of the general discription of a phenomenon in the first paragraph, the whole 'hanging something from a wall/ceiling analogy' seems counterintuitive to me. If you hang something on a wall, you usually drive one or multiple screws in there. The resultant forces on the wall are not the same as shear forces. Shear stress is more analogous to rubbing something along a wall, right? For now, I erased these sentences. If someone has a good argument to re-introduce them (either in the first paragraph, or in the body of the article, please do so. (talk) 13:03, 15 May 2009 (UTC)[reply]

I agree that it's a bad example and therefore should be removed. But just as a note, you should put something in your edit summary, because otherwise it just looks like vandalism. Wizard191 (talk) 14:55, 15 May 2009 (UTC)[reply]

Somewhat Major Edit by Jgreeter (talk) 04:14, 16 May 2011 (UTC)[edit]

The article previously basically said "shear stress is parallel to the cross section on which it acts." Though the writer probably had the concept correct when writing the article, this is mathematically and physically woefully inaccurate.

First, and most importantly, the force creating the shear stress is actually PERPENDICULAR TO THE SURFACE NORMAL of the cross-section on which it acts. This is because areas aren't defined by a coplanar vector but by a surface normal vector. This is basically the standard for defining area, probably because, in Euclidean geometry, a vector can point in infinitely many directions and still be coplanar, but there are only two directions perpendicular to a flat plane.

Second, stress is a tensor, not a vector, thus we can't say it's "perpendicular," "parallel," or "tangent" to some face. However, the components that make up stress, Forces and Areas (stress = force / area) are both scalars associated with unit vectors. Thus, it is far more correct to use the concepts to describe the differences between "shear" and "normal" components of stress.

So I rewrote the first paragraph to say the following:

"Shear stress is defined as the component of stress coplanar with a given material cross section."


"The net force vector which creates this stress is perpendicular to the surface normal vector of the cross section. Normal stress arises from a net force parallel or antiparallel to the surface normal of a material cross section."

The "normal stress" article might need fixing too; I didn't check.

Jgreeter (talk) 04:14, 16 May 2011 (UTC) Researcher in the field of biomechanics[reply]

Deleted section on micropillar shear stress sensors[edit]

Somebody was trying to plug in personal research on hypothetical sensors. I deleted it in good faith. -- (talk) 20:57, 3 July 2012 (UTC)[reply]

Shear Stress Diagram[edit]

Is the statement "...A shear stress, \tau\, is applied..." in the text below the diagram correct? It should read "...A shearing force, F, is applied..." or "normal force vector" or similar. A stress is not applied to anything, rather stress the consequence of some applied force. The tau symbol in the diagram equally needs replacing. — Preceding unsigned comment added by (talk) 12:51, 25 October 2013 (UTC)[reply]

Can you explain how shearing works in scissors or garden shears or tin-snips to cut metal, where the material is held on both sides? What is the scale of an element of the material being sheared? Does the failure of the material occur at the top. bottom, or middle of an element which is distorted from a rectangle to become an approximate parallelogram? I once had a Russian book about lathe tools which started on finite element analysis of the material being cut. Does anyone in the English speaking world understand these issues? (talk) 12:56, 6 February 2014 (UTC)[reply]

intro not at right level rewrite[edit]

The intro is NOT at the level of a general encylopedia'; it needs to bet re written at a much lower level This is not a place to show off you college undergrad physics or pchem; this is a place to be simple for you mom or dad

Assessment comment[edit]

The comment(s) below were originally left at Talk:Shear stress/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

The formula is correct for transverse sheer stress although I think the explanation is incorrect. "I" represents the moment of inertia of the entire cross-section. "Q" is A*Y, where A is the cross-sectional area measured from the plane you want to calculate TVS to the surface, Y is the centroid of this cross sectional area measured from the neutral axis. V is the load, and T is the thickness of the cross section.

Last edited at 03:19, 13 April 2007 (UTC). Substituted at 05:55, 30 April 2016 (UTC)