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

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

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

(OP)
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

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

shear stresses are zero on all surfaces (what can the material at the surface shear against ?)

i'm not quite sure how to read "shear stress on the outer fibres peak at the mid-points, and are ZERO at the corners" ... i think you're confusing torsion on open thin walled sections with torsion on thick sections.

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

The OP is talking about the corner shear on an internal cross section of the shaft.
A good explanation is available on any good stress book.
My favorite, Den Hartog's Advanced Strength of Materials covers this quite well on chapter 1.

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

sorry, but what's the distinction between a corner and the surface ? torsion shear stresses are zero on the surface of a section (be it a corner or not).

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

RB1957 is absolutely correct. You can reference this in the reduction of equations to Mohr's Circle. Look in your statics textbook, second year engineering or so, Mechanics of Materials. Beers & Johnson is a good reference on this.

Regards,
Cockroach

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

that's not how i remember being taught ... i remember it as the "soap bubble" analogy ... zero on the surface, increasing towards the center.

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

"torsion shear stresses are zero on the surface of a section"

No. Torsion shear stress on a round bar is a maximum at the surface, in the τ and τrz directions. Straight out of Chapter 9 or Roark's Handbook.

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

(OP)
rb1957 - if my understanding of the "soap bubble" analogy is correct, it is the GRADIENT of the bubble that represents shear stress. Therefore, for a circular section, maximum shear stress (i.e. maximum slope) is at the outer surface and zero shear stress (i.e. zero gradient) is at the centre. See attachment for illustration.

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

i stand corrected (again) ...

reviewing Bruhn A6, shear stess for a round bar increases with radial distance (obviously peaking on the surface). for rectangular sections, i think the corners have zero stress 'cause the stress contours round off the corners (i'm guessing that the soap bubble analogy will show zero slope in the corners ... not as i'd've thought, but then that's what Bruhn shows so it must be so ...).

i guess the stress can't go into the corner, crank 90deg and go out again, i think it wants to flow smoothly.

so then for a rectangular section, the peak shear stress is mid-side nearer from the center ...

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

You had me worried there, rb.

"i stand corrected (again) ... " Join the club :)

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

Ah! One of the differences btwn. shear stresses associated with normal bending stress and torsional shear stress and its flow around the periphery of the bar/shaft. Rb... I knew you’d finally come to your senses. smile

The magnitude of the torsional shear stress is in direct proportion to the slope of the soap bubble, and the closeness of the contour lines, and is zero where the soap film is horizontal. The attachment is from Zekeman’s ref. by J.P. Den Hartog.

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

(OP)
Thanks everyone, I think I'm almost getting it. Although not quite...

Dhengr - from your attachment:

"The material in protruding corners has no shear stress: it is dead material. (This conclusion can be immediately verified by assuming a shear stress in the corner, by resolving that shear stress into components perpendicular to the two sides locally, and remarking that both components must be zero by virtue of Fig. 1.)"

I can't seem to do this "immediate verification". When I draw a free body diagram of a corner element (see attachment), I don't see why the shear stresses have to be zero...

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

i think they're saying that the shear stress in the corner can't have two components (along the two surfaces) ... that the shear stress at the surface wants to be along the surface (as in a round bar). if the stress is happily going along one surface and it hits the corner it like says "oh crap" and doesn't want to turn 90deg. the practical idea is that you can bevel off the corners without increasing the torsional shear stresses.

one thing i found interesting in reviewing this is that an oval section bar isn't like a round bar, that the area outside the largest involute circle (i think that's the right term to describe a cirle within the cross-section) is less stressed, that the stress peaks on this circle.

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

Derby - I don't think your attachment worked correctly.
For 'figure 1' see the thread that I previously referenced it is in the OP attachment. I'll see if I can scan page 1 as well. I'v read through it again but I think I'm getting myself confused again...

EIT
www.HowToEngineer.com

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

Another way of looking at it is that at the corner of a rectangular section you would have to have shear stresses on the two free surfaces to maintain equilibrium if there was a shear stress due to torsion in the cut plane.

Cheers

Greg Locock


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

A warning to all concerned and all readers here:

Everything above is valid - but only for the case of a "pure" torsion stress across a simple square bar being twisted at both ends (or twisted at one end and held in place at the other.)

At both ends of the bar - at the two places where the "key" or the bar is clamped by the tooling/wrench/broached hole/whatever - the resistance to torsion is maximized at the square corners. Simple example: put a wrench on a square piece of steel from the hardware store held vertically in a vise. Twist the vertical steel and look at where the "rounding" (yielding of the bar) takes place as the wrench tries to separate and the resistance is localized at the corners. (This is why square nuts are seldom used. A hex nut spreads the induced load better, and allows easier turning in tight spots.)

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

You can see this using the hydrodynamic analogy of Greenhill, if this can satisfy you. Take a frictionless fluid turning into a container with a flat bottom having the same shape as the section of the bar and with vertical walls: by comparison of the governing equations, it is found that the the velocity components of the fluid are the same as the shear stress components.
Now it is clear that a fluid has zero velocity in a protruding angle, no?

prex
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RE: Rectangular bar in torsion - why are the shear stresses at the corners zero?

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

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

"Stress Cube? "

Like a stress ball, only less round.

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

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.

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

(OP)
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.

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

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 ?

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

(OP)
Afraid so sad

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

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
www.HowToEngineer.com

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

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

EIT
www.HowToEngineer.com

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

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

? ... 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.

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