How to determine angle of twist for open form section

[DavidLeech]

I am trying to find the angle of twist of a thin walled open form cold rolled steel section. The application is a roof member torsionally restrained at either end and subject to a uniformly distributed torque. I have section properties including Torsion Coefficient (J), Warping Coefficient (Cw) and the location of the shear centre obtained from software. If I know the applied torque, section moduli, and length as well as the properties listed above is there a way to calculate the angle of twist mid span. This is non uniform torsion so I don't expect the formula TL/GJ for angle of twist of circular shafts will suffice. Any help most appreciated.

DL

 

[EngineerTex]

It's going to be hard. But read this previous thread:

http://www.eng-tips.com/viewthread.cfm?qid=78595

With that, I'm still going to recommend that you use a closed section if there is any way on this earth that you could possibly do it. Really. Get a tube if you can, or get two channels, put them toe-to-toe and run a skip weld along the top and bottom.

However, if there is no way you can possibly avoid using an open section, follow the advice in the above thread and refer to "Design of Welded Structures" by Omer Blodgett. Also get the "Solutions to Design of Weldments." You can find these two books at

http://www.jflf.org

and clicking on the "Educational Materials" link, then "Books".

Buy them both. There are no better books you will ever find on the subject and they are both CHEAP!

Now back to your question. I believe that you will have to perform FEA on this, but it's FEA that you can do in a spreadsheet. For your first iteration, assume that the "torsional inflection point" is in the middle of the beam.

Now, when I use the term "torsional inflection point," I am referring the the point along the beam with the maximum torsion. I'm not sure it's in the middle since your post first says that it is subjected to a uniformly distributed torque, but then later says that this is non uniform torsion.

When I do quasi-manual FEA, I set up my spreadsheet so that I have one cell that contains the finite length along the member, i.e. if you have a 100-ft beam, you consider 50 ft (inflection point to restraint length) and break the beam down into 600 1-inch sections. You put a value of 1 in that cell.

Next, I make three columns. The first is simply a counter which starts at zero and basically numbers the elements.

The second column is an equation that relates the amount of torque applied to that finite length of section. If it is a uniform torque, it is a constant divided by the length of the finite element. If not, then you have to apply an equation to determine the value at each element.

The third column determines the amount of angular twist that that particular element applies to the length of beam between the element and the support.

The nice part about doing it this way is that you only have to fill in four cells in a spreadsheet, then copy and paste three of them down to make columns of 500 - 1000 (or more).

After you've done that, you simply calculate it a few times with progressively smaller values in the finite element cell and compare the appropriate cell down at the end of the beam (remember that the end of the beam will be in a different cell depending on the length of each element) each time you enter a new element length value. After you get a change of less than 5% from one length to the next, you can be pretty sure that you have convergence.

Now, if you are talking about a setup where things really are non-uniform, it gets a little more tricky. You still have to do the technique outlined above, but you have to do it twice, and then you have to start massaging your numbers. You can start making guesses where the torsional inflection point is about halfway down the length of the beam and then start moving it around until you find a point where the torsional deflection from one side is equal to the torsional deflection on the other. It would be a few iterations, but not really a killer.

You can also automate that one by adding one more column to your spreadsheet. I will leave it to you to determine how to do that.

All that being said, you now have to consider whether your beam will be capable of supporting the bending load after it has received this twisting. For that one, you're talking about doing a buckling analysis and that's one you're going to have to do some real FEA on.

Engineering is not the science behind building. It is the science behind not building.

 

[prex]

Roark has formulae for that, with concentrated and distributed torque, that you can rework by superposition, if needed.
You'll also need to decide whether your ends are only torsionally restrained or also restrained against warping.

prex

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[GregLocock]

By hand calc I get a rotation at the centre 1/8 that of the end of the same torque applied to the end (not distributed) of the same tube held at one end only.

That seems a reasonable number, but I'm not convinced.



Cheers

Greg Locock


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

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[DavidLeech]

Thanks all for the replies. I'm going to check Roark first and see what comes of that. As someone who knows little about FEA would it be right to assume a situation like this would be ideally suited to the FE method?. I've started researching the FE method in past few months with view to developing an understanding of it.

David Leech

 

[flash3780]

Thin, open sections can be modeled as a thin rectangle:

http://www.public.iastate.edu/~e_m.424/Torsion of thin sections.pdf

Good luck.

 

[GregLocock]

OO er.

I think it would be more accurate to say that /some/ open sections can be modelled as a thin rectangle. Consider the classic case of a tube with a slit down one side.

Yes FEA is the easiest way of solving this sort of problem, but bear in mind that linear FEA won't tell you anything about second order effects that occur as the section twists. For that you need geometrically non linear FEA which is a big further step along the learning curve.

The dog has eaten my copy of Roark, or else it is at home, so I can't check that number. Do you have a rough feeling for how much twist you are expecting? If it is only a few degrees then I wouldn't worry about second order effects (famous last words).


Cheers

Greg Locock


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

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

 

[DavidLeech]

I have tried 2 calculation methods from Roark, 1 for ends which are restrained in both torsion and warping and the other for ends free to warp but torsionally restrained. The answers I get suggest twist at the centre will be approx 3 degrees and 9 degrees depending on the case. The ends are restrained by bolted connections so I am working on the assumption that these allow warping. My gut instinct tells me that the actual angle of twist for the loads that I am using would be greater than 3 but less than 9 i.e somewhere in between. I've asked someone else to review the problem to see if they get an answer something similiar.