Shear stress distribution 1

[Everydaylearning]

Is shear stress through a flat bar evenly distributed through the cross section,or maximum at center and less at the extreme fibers? If it is maximum at the center and decreasing outwards what causes this?

 

[chicopee]

How is the loading imposed on the bar?

 

[rb1957]

I thought shear stress was higher in the body of the part than at the surface (like "soap bubble" and "sand pile" analogies) ?

another day in paradise, or is paradise one day closer ?

 

[hydtools]

The shear stress has a parabolic distribution through the crossection. It is zero at the surfaces because there is no shear at the free surface. There cannot be an instantaneous jump from stress to no stress. It is maximum at the neutral axis.

http://www.bu.edu/moss/mechanics-of-materials-bending-shear-stress/

Ted

 

[LiftDivergence]

First, that is only if you are talking about transverse shear stress VQ/Ib.

Torsional shear and direct shear loadings will both lead to different distributions.

The OP has yet to state how the "flat bar" is loaded.

Also, I note the lesson linked to is specifically a discussion of transverse shear in beams. What is the geometry of the "flat bar"? We may not not talking about a beam here - the discussion for a thin plate may be different.

We already know both Euler and Timoshenko theories for beam stresses do not hold for flat plates. We need to consider Kirchoff-Love or Reissner plate theory.

A good resource for this is Theory of Aircraft Structures by Rivello Ch 13 - Bending and Extension of Thin Flates, where he notes the edge twisting can have an effect.

Keep em' Flying
//Fight Corrosion!

 

[rb1957]

"that is only if you are talking about transverse shear stress VQ/Ib" ... this is a comment against "It is maximum at the neutral axis." and not "The shear stress ... is zero at the surfaces because there is no shear at the free surface" ?

The latter comment is true for all shear stresses, no?

Possibly this (shear stress variation over a cross-section) is what the OP meant by "maximum at center and less at the extreme fibers" ?

another day in paradise, or is paradise one day closer ?

 

[LiftDivergence]

I realize it is somewhat pedantic in this instance (most likely we are talking about transverse shear, in which case I agree), but if there is an absence of bending, for example where the shear is caused by point loads with essentially no distance between them, we have direct shear in which case the stress distribution is uniform over the shearing area. And torsional shear is definitely peak at the outer surface. All of this for a beam, in particular.

Keep em' Flying
//Fight Corrosion!

 

[JStephen]

Note that when it comes to structural steel, the design codes will limit the average stress over the web, or the total shear force on the beam, without regard to how that stress is distributed.

On that "VQ/Ib" distribution, if I remember right, that applies to "long" beams, well away from any concentrated loadings or beam ends, with small deflections, that is, the normal "beam" assumptions. Meanwhile, the maximum shearing force is oftentimes at concentrated loads or very near beam ends, which may be the motivation to work with average shear stress or total shearing rather than an ideal approximation.