Rectangular bar in torsion - why are the shear stresses at the corners zero? 4

[DerbyEngineer]

Hi,

I'm trying to brush up on torsion of non-circular sections. Taking a bar of rectangular cross-section as a simple example, my textbooks and various websites all say that the shear stress on the outer fibres peak at the mid-points, and are ZERO at the corners. What I don't understand is WHY they must be zero at the corners. I feel like I'm missing something quite fundamental here, but can't figure it out.

Any help appreciated. Thanks in advance!

Martin

 

[RFreund]

Prex - Interesting.

Attached is page 1-3 one of Hartog.

I think his explanation is as follows:
Plane section remain plane. For this to happen the shear stress must be along a set of concentric circles as is seen in a circular shaft(I'm not exactly sure why but it seems to make sense.) This means that the shear stress is tangent to these concentric circles. However if the shape is rectangular we see that at the boundary the shear stress which is tangent to the concentric circle is NOT tangent to the boundary. That means we have components - One component is tangent to the boundary the other is normal to the boundary (point inward) see fig 1 in Hartog. This stress that is normal to the boundary needs another stress to maintain moment equilibrium. Well as rb stated this cannot happen as it is a 'free surface' ("nothing to shear against"). - OK so now we know that the shear MUST be tangent to the boundary. When we get to the corner however there are (2) free surfaces and thus nothing to keep the 'stress cube' in equilibrium. See next post with attachment. (How do you do multiple attachments?)

EIT
www.HowToEngineer.com

 

[RFreund]

Stress Cube?

EIT
www.HowToEngineer.com

 

[btrueblood]

"Stress Cube? "

Like a stress ball, only less round.

 

[RolMec]

If more complicated sections come into view: This (torsional stress = 0 in corners) is valid only for simple rectangular or thin walled closed cross sections. As soon as the cross section is a combined thin wall open section as C, L, U, T etc profiles, warping torsion (Wlasow-Torsion) must be considered. With warping, corners are under torsional stress, pls. consider when superpositioning stresses.

R.

 

[DerbyEngineer]

Thanks all, but I still don't understand despite all the diagrams, scanned textbooks, explanations etc.

rb1957 - I agree, it makes intuitive sense that shear stress in the corners has to be zero to prevent it having to suddenly turn 90 degrees. However, when I draw a stress cube I fail to see (in equilibrium terms) why it is not possible (RFreund - sorry, I couldn't understand your stress cube).

See attachment for my stress cube.

Any clarification on what I'm missing appreciated. Thanks.

 

[rb1957]

draw an element in the middle of the upper side (outer surface to the left, yes)

now translate this element into the corner (like you did on the rh side)

clear as mud ?

 

[DerbyEngineer]

Afraid so

 

[RFreund]

Sorry about the sloppy cube, I use this digimemo program which makes my poor hand writing even worse...anyway,
I believe that both stress cubes are wrong as there can be no shear stress on the outside (free) surface. As RB mentioned there is nothing to shear against or nothing to 'slide' against, there is no material to resist this 'want to slide' (I hope I described that correctly). The right side surface of the blue stress cube has no stress on it. The top and right side of the red cube has no stress on it.
However there is a shear stress on the 'front face' of the blue cube.

RolMec may need to clarify as I thought for the situations he describes that you would have warping stresses but that they would occur normal to the section (not shear stresses), but I may be misunderstanding what he is saying.


EIT

http://www.HowToEngineer.com

 

[RolMec]

Hi RFreund,

see attached sketch, it's a rough (not textbook) illustration of the idea.
Regards
R.

 

[RFreund]

Ah, yes, I see now what you are saying. Thanks!

EIT

http://www.HowToEngineer.com

 

[rb1957]

"DerbyEngineer (Mechanical) 30 Jan 13 17:42
Afraid so " ...

? ... the point was you have shown that a state of shear stress can exist in the corner by considering one side; however, if you consider the other side you get the opposite shear in the corner (no?) and therefore there is no shear.