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

 

[rb1957]

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.

 

[zekeman]

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.

 

[rb1957]

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).

 

[DerbyEngineer]

Thanks for your responses, although I'm still struggling. Not sure if my first thread was very clear. The attachment should clear up what I'm talking about.

 

[Cockroach]

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

 

[RFreund]

I second zekeman's response, although I struggled with this as well:

http://www.eng-tips.com/viewthread.cfm?qid=333112

EIT

http://www.HowToEngineer.com

 

[rb1957]

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

 

[btrueblood]

"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]_r[θ] and [τ]_rz directions. Straight out of Chapter 9 or Roark's Handbook.

 

[DerbyEngineer]

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.

 

[rb1957]

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 ...

 

[btrueblood]

You had me worried there, rb.

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

 

[dhengr]

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.

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.

 

[DerbyEngineer]

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...

 

[rb1957]

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.

 

[RFreund]

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

http://www.HowToEngineer.com

 

[GregLocock]

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


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[racookpe1978]

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.)

 

[DerbyEngineer]

Thanks again. Not sure why my attachment didn't work last time. Hopefully it will now.

 

[prex]

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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