Saturday Morning Question - Torsion 6

[inertia4u]

OK, lets say I have a Stanely 25' tape measure. It doesn't have to be Stanely, but just pretend it is. The tape cross-section is a little concave, essentially creating a shallow "u" section.

Now, let's say that I extend that tape measure out, while holding it in my hand, and effectively creating a cantilever of the tape since the base is still in my hand. If I extend the tape out 3 inches, and rotate the base 90 deg in my hand, the free end of the tape rotates 90 deg. However, if I carefully extend the tape out 10 feet still maintaining a cantilever (without the tape "buckling"), I can rotate the base of the tape measure 90 deg and the end of the tape doesn't rotate at all.

Question 1: What is the mechanism behind this phenomena?
Question 2: How do you free body this?





-----
Nert

 

[mudflaps]

Just tried it - Cool!

I suspect it has something to do with the downward curvature helping keep the CG of the cantilever as low as possible. My tape did twist a little at the end but nowhere near as much as the end I was holding. Good thought provoker.

old CA SE

 

[inertia4u]

I've attached a picture for additional clarification, because this can be interpreted different ways



-----
Nert

 

[haynewp]

There is a certain amount of torsion induced into the end of the tape when twisted. The opposing torsion is distributed along the length of the tape. The longer the tape, the more resistance to the induced twist at the end.

I think of the opposing torsion as one-half the width across the short face of the tape times the weight of tape per length. The more rigid the tape, the more torsion produced into the end when twisted. And therefore, a longer length of tape must be extended in order to accumulate enough to resist that torsion.

 

[EdR]

inertia4u:

While the same equations don't apply just think back to your old MM equations for torsion of a circular shaft...

angle of twist = torque*length/(j*g)

note that the angle of twist is proportional to the length...same idea applies to your tape except the equations are modified due to the cross section (and probably due to the rotational large displacement)..otherwise the concepts are the same...

As for a FBD it is just the same as any other FBD...cut a section and put the internal/external forces on the diagram....

Ed.R.

 

[haynewp]

What is resisting the twist? The FBD would have an applied torque at the end and gravity acting along the length of tape.

 

[asixth]

I wouldn't even attempt to apply structural mechanics to this problem, however, if I remember back to structural mechanics 101, I believe the torsional stiffness of a 3D beam element is G*J/l. So when the tape is cantilevered 10ft, the torsional stiffness of the tape is significantly reduced because of the increase length (by a factor of 40 compared with the 3 inch cantilever).

But for the first sitution, where the 3 inch cantilever is rotated 90 degrees, I don't believe there is any applied torsion for this problem because the system rotates as one unit (rigid body rotation).

For the second situation, gravity is applying a torsional moment on the tape (concave cross-section is not doubly symmetric),and because of the lack of torsional stiffness, the tape can rotate 90 degrees.

 

[youngstructural]

asixth:

Exactly what I was thinking as I read the thread; A star for you!

Cheers,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[civeng80]

I think asixth has got it. Same thing happens to I beams, short length just rotates. When you have a long length and try and rotate it by hand (say to prime it) its less stiff torsionally (or more flexible torsionally) so the longer the member the less stiff exactly as per the flexural stiffness equations.

 

[haynewp]

{Sighs}

 

[Tomfh]

This is an elastic stability problem. Case 1 is an unbuckled cantilever and Case 2 is essentially a cantilever that has undergone severe lateral torsional buckling.

 

[civeng80]

Remember also that your hand is playing a part in all this and providing for any unbalanced moment or reaction that needs to be applied for equilibrium of the body. In all a highly statically indeterminate problem and one that I wouldn't attempt by structural mechanics as asixth has stated.

 

[GregLocock]

Energetically, the torsional elastic energy taken to twist the tape is less than the increase in potential energy required to lift the mass of the remaining segments of tape higher, as would be required if the tape is to remain untwisted.

Proving the above might take more than a Saturday!




Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[EdR]

As I tried to say earlier this problem is simply a cantilever beam problem (with the fixed support at the opposite (LEFT end in the picture) end from the applied rotational moment...note that the rotation is zero at the support which is the definition of a fixed support) and a torsional moment applied at the free end.....there are no indeterminate reactions to the problem....the rotational moment is constant over the length of the beam....

As for the bending if it is elastic (and it is) it can be decoupled from the torsion and its just a cantilever with a self load (although the resulting displacements may be large).

It is true that the actual computations require modification to the basic torsion equations (and bending) but as I tried to indicate earlier the basic system is just a cantilever unless I'm missing something.....

Ed.R.

 

[GregLocock]

Oh, and the elastic bending energy in the two cases (twisted vs untwisted) will also be different.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[civeng80]

I just think that when the tape is cantilevering about 3m (10') it just hasn't got the torsional stiffess to rotate like the handle does, like a piece of string.

 

[prex]

Very interesting indeed. However, sorry, I'm afraid no one has got it.
The fact that the section is not doubly symmetric doesn't explain it, as, following this way of reasoning, it would be necessary for the CoG to go up during the rotation, to explain the resistance to the rotation at the free end. But if the rotation applied by hand is about the CoG, then the whole length could rotate about it without any change in potential energy and consequently with the same energy of deformation.
The explanation is more subtle and involves the (much) different bending stiffness of the tape along its two principal axes. I'm sure (though didn't try it) a similar phenomenon can be observed with a long strip of paper or a metallic flat bar or anything similar.
The point is that, when the strip is cantilevered the large face horizontal, it will experience a bending deformation much larger than when it is cantilevered the large face vertical.
This means that, when going from the first to the second configuration, as in the experiment proposed by inertia4u, the free end of the strip would have to go up to arrange for the lower bending deformation. Going up would raise the potential energy, and this explains the resistance to rotation.
With a short strip the phenomenon is not (or much less) observed, because the bending deformation is lower (goes down with the 4th power of length, or a bit less when dealing with large deformations) and the torsional stiffness (goes up with the inverse of length) is sufficient to overcome the change in potential energy.

prex

http://www.xcalcs.com

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[civeng80]

prex

The fact that the stiffness of the tape is greater in a vertical position than in a horizontal position is another way of saying that the tape has a low torsional stiffess and this decreases as the tape extends)e.g. if the tape cross section were a tube then stiffness would be the same in both hor and vert but the torsional stiffness would be very high (and could thus also rotate maybe) so really asixth and I are saying the same thing that you are in different words.

Cheers Mate

 

[csd72]

Some things are just not worth trying to put numbers to.

It just goes to prove that engineering does not have answers to everything.

 

[Dinosaur]

I think Tomfh beat me to the answer. A section bending about its weak axis is in a state of particular equilibrium. The beam can rotate at the base without buckling, but the energy to rotate the end is not there, unless you manipulate your hand a set way. Inertia plays a role in getting the beam shaped properly without buckling it in the process. To illustrate that Tomfl's idea is proper, I ask that you set up the tape in its 90 degree rotated position while supporting the "free end." Then remove the support at the "free end" and see if it doesn't buckle by LTB. I also think the non-symetric section plays a role. Otherwise, you could do the same thing with the tape inverted first. I don't think this works.

You can't pay me enough to prove this analytically. But it is a nice subject for us.