Formula for change in length of a shaft under torsion? 9

[Christine74]

It's easy to calculate the elastic angular deflection of a shaft under torsion (theta=TL/GJ), but is there a similar formula for the change in length of the same shaft, assuming pure torsion (no axial load)?

The length would have to change, right?

Thanks,

-Christine

 

[desertfox]

Hi Christine 74

According to Roarks formula for stress and strain a longitudinal stress and strain do exist, however it goes onto say that the longitudinal strain is a shortening, and that the longitudinal stress is a tension in the outer part and a balancing compression in the inner part.
A guy called Reiner on the basis of mathematical analysis, concluded that the longitudinal strain can be, a shortening,a lengthening,or zero, depending on physical characteristics of the material and goes on to quote various experiments carried out by a guy called Poynting.
The paragraph concludes by saying that for a solid circular bar involving pure elastic torsion the longitudinal stress and strain would not be large enough to have any engineering significance and hence I cannot find any formula's to help you.

regards desertfox

 

[GregLocock]

Great stuff.

Would an intelligently constructed FE model show the effect for a particular geometry and material? I can't see why not.

Cheers

Greg Locock

 

[drawoh]

desertfox,

Isn't this what Poisson's ratio is all about? I do not see anything in the textbooks I have lying around.

JHG

 

[JStephen]

Intuition says the length would change, based on experience with rubber bands.

However, in twisting a shaft, the shear deformation is much smaller than the rubber band example.

Consider the shear deformation at the surface. A square inscribed on the surface would deflect into a paralellogram. Now, inevitably, in these derivations, there is an assumption that deflection angles are "small", so that sin(theta) = theta. But that same assumption means cos(theta) =1. In other words, you assume away the axial deflection in the normal analysis.

 

[desertfox]

Hi All

An interesting post isn't it, when I first saw it I thought the shaft would elongate and like you drawoh I considered that poissons ratio would be the key, however I wondered how I would get an angular deflection to relate to either a lateral or lonitudinal strain as usually one can calculate the longitudinal strain and knowning poissons ratio calculate the lateral strain (ie tension and compression),at this point I decided to look in Roarks stress and strain and found this section on longitudinal stress due to torsion. Greg it would be interesting to construct an F.E. model I wonder if anyone in the Finite Element Forum as any experience of this.

Regards Desertfox

 

[prex]

For a circular section the answer is
[Δ]L=R²[θ]²/4L
where R is section radius, [θ] the total angle of torsion and L the length.
So, in contrast to what was expected by someone above, this is not a Poisson related effect.
It is a second order effect that is calculated observing that the longitudinal fibers deform under torsion in helices, so that they would lengthen if there was no change in the distance between sections.
The formula for other types od sections changes: for a narrow strip it is
[Δ]L=a²[θ]²/24L
where a is the long side of the strip.
As this is a second order effect a linear FEM calculation wouldn't be able to calculate it (this proves once more, if there was need to, that FEM will never give more than the underlying theory): I suppose that enabling large deformations it would come out. One must pay attention, however, as it is a really small effect: the unit elongation is also equal to [τ]²/4G² ([τ] being the stress at the outer fiber), hence at most of the order of 0.00001% (!) for common metals.


prex

http://www.xcalcs.com

Online tools for structural design

 

[GregLocock]

Wow. More like wow squared.

That is counterintuitive to me, to say the least. I'd have guessed shorter, and guessed Poisson's ratio was involved.

Here's my new model - imagine twisting a stack of bumpy discs that all interlock nicely. As you twist them they are forced apart.

So, since Poisson's ratio is not involved then we get the bizarre effect that some materials' density will increase as they are twisted, and some will decrease. - is that right?


Cheers

Greg Locock

 

[israelkk]

prex

Can you refer us to your references (books, article etc)?

 

[ivymike]

It is a second order effect that is calculated observing that the longitudinal fibers deform under torsion in helices, so that they would lengthen if there was no change in the distance between sections.

Delta-L in your equation above is the change in length of a fiber, as it goes from straight to helical, if the sections are kept at a uniform distance from each other, right? Doesn't this effect mean that the rod gets slightly shorter because the fibers resist elongation, and the outer ones stretch a bit while compressing the inner ones?

Thought experiment to demonstrate my thinking:
* Take two round boards, drill small holes around the circumference of each
* drill large holes in the center of each
* tie strings between the two boards such that each hole in one connects to a hole in the other, and the strings are roughly parallel
* loosely mount the connected boards on a shaft that passes through the central holes
* turn one board relative to the other
Results:
* The originally-parallel strings deform into helixes (as above)
* The boards get closer together

If the boards were restrained axially w/re the shaft, they would attempt to compress the shaft. If the boards were prevented from moving closer to each other, the strings would be forced to stretch by the amount indicated in your equation above (right?).

 

[btrueblood]

Prex,

Your formula makes intuitive sense when considering a thin tube. But when considering the case of a circular bar, the torsion distortion of the center of the bar is less (radius to that section is less), and there would therefore be less shortening in the center than at the outer perimeter. Thus, the outer fibers twist a greater amount (r*theta), and are in tension, and the inner fibers are twisting less, and are in compression (as desertfox's first post and Roark's handbook states).

But, I think you have found the "answer" to the question (we all intuitively know there's got to be something happening as the shaft twists, but no reference I've ever found gives any details). At least, your arguement makes sense to me, and gives a formula to calculate. A star for that.

Ben T

 

[prex]

My reference is O.Belluzzi - Scienza delle costruzioni (Strength of Materials), para.161.
This is The reference (4 volumes, 3000 pages, 2500 worked examples) for italian engineers.
As I think that most of you won't have access to it, I'll try to very shortly summarize the method(original length: 2 pages).

Longitudinal fibers tend to elongate because they deform into helices.
If section distances were not changing, they would stretch by
[ε]=[√](1+(r[θ])²)-1[≈](r[θ])²/2
where r is local radius between 0 and R, and [θ] is the unit angle of torsion (total angle divided by length).
with a corresponding normal stress
[σ]=E(r[θ])²/2
However those normal stresses would give a resultant axial stress, in contrast with the assumption of pure torsion. So we have to assume that the axial elongation of a long.fiber would be diminished by a shortening of the bar that we call [ε]_o so now
[σ]=E((r[θ])²/2-[ε]_o)
By imposing a null resultant
[∫]_o^R[σ]2[π]rdr=0
one has
[ε]_o=(R[θ])²/4.

Leaving off the derivation for a narrow rectangular strip: the result is in my previous post. Belluzzi also observes that in this case the angle of torsion may become relatively large, so that the axial stresses may contribute, with their radial component, in resisting torsion in a non negligible way, and the equation of torsion should be rewritten as
M_t=Gab³[θ]/3+Ea⁵b[θ]³/360.
He quotes an example where, in a strip with a ratio of 60 between large side and small side, by using this method the shear stress in torsion decreases by some 20%, but the axial stress is 60% of shear.
To be noted also that, as suggested by others (and Roark), the axial stress is tensile in the outer part of section and compressive inside.


prex

http://www.xcalcs.com

Online tools for structural design

 

[btrueblood]

Prex,

Wow, thanks (again). I printed out that last post and am going to slip the page into my copy of Roark's. Will give you another star if it will let me.

Ben T.

 

[EnglishMuffin]

To demonstrate such tiny effects in practice, I presume you would have to be very careful about specifying the exact details of the end fixities of the shaft, especially if it is short. After all, the torsion has to be applied to the shaft somehow. I'd bet Belluzzi's assumptions, whatever they were, are unlikely to be easily achievable in reality.