Torsional rigidity of beam with polygonal x-sect

[borjame]

I'm trying to get an equation which will give the torsional stiffness of beams with the following cross sections: a trapezoid and a parellelogram (the polar moments of these shapes is enough). These will actually be sections in a composite beam with the end sections being triangles to close out the beam so any info calculating torsional stiffness of a composite beam structure would be a great help too!

 

[Qshake]

Should get ahold of Elasticity by Timoshenko or Linear Elasticity by Gould. The foundation for those calculations are readily available in those texts.

 

[amokhta]

For a closed box section of any shape, enclosing only one internal cell, the torsional constant of the section is given by (4A^2)/Sum of (S/t)

where:

A = area enclosed by the cell (use mid point of wall thickness)
S = Length of each side of the box
t = thickness of each side of the box

 

[GregLocock]

Does that apply to re-entrant shapes? eg suppose I collapse a six sided tube inwards to a C section, does it still work? Cheers

Greg Locock

 

[austim]

Hi, Greg.

I believe that the hollow section formula does not care whether the section is re-entrant or not, provided that it can be considered thin. It all boils down to the moment from a constant shear flow round the section, whatever the shape.

 

[GregLocock]

HANG ON, there's a mistake K=4A^2t/Perimeter t was missing from amokhta's post.

OK, for the case I mentioned, of a hexagonal tube side length b, thickness d, crushed to a C section A= 3db, perimeter=6b, so K=4*(3db)^2/6b

=6bddd

Whereas Roark says for a split tube K=1/3Pt^3, ie 1/3*6b*8ddd=16bddd

and for a solid rectangular strip (which is what my shape degenerates to, since bends don't matter) K=1.5b*d^3*(16/3- some small correction)

or roughly 8bd^3

That's neat, three different answers for the same problem.

Now, I can see why the assumptions in the tube eqn don't apply to the rectangular strip, but I'm damned if I can see why the first two disagree, if re-entrant shapes don't matter. Without doing any maths, I'd guess that the first equation only applies if the centre of rotation is inside the shape, or that the stress flow is alwasy in the same direction around the perimeter. Of those two alternatives I prefer the second, but not by much.
Cheers

Greg Locock

 

[amokhta]

Greg Locock:

The formula that I gave is correct. K=4A^2t/Perimeter only applies to hollow circular members. The formula that I provided is general.

 

[austim]

Greg,

Happy is the engineer who really understands torsion. I would not claim to do so. However, I still believe that my earlier suggestion is correct for the stiffness of a thin walled tube.

I suspect that our first problem follows from the definition of "thin-walled". Your crushed tube ends up with the tube "diameter" (at the midplane of the wall) and the wall thickness being equal. I'm sure that is well outside the bounds of "thin-walled".


I would expect that some deep consideration of the "soap bubble analogy" would give you an accurate stiffness of your channel.

The reason for the discrepancy between the two values that you get from Roark, is that you have done something rather odd with the split-tube formula. If you are going to take the wall thickness of the tube as 2d, then the perimieter should only be 3b (otherwise you would have a gross metal area of 12bd). Using 3b gives the same result from both formulae.

BUT... I question whether you can legitimately treat the squashed tube as if the result is just the same as a piece of plate 2d thick.

There is very little doubt in my mind that the presence of a midplane 'lamination' for almost the full developed length of the channel will significantly reduce the bending strength/stiffness about the 'minor' axis compared with a legitimately solid plate. I strongly suspect that it also radically changes the torsional behaviour.

Perhaps there is a case for some our more knowledgable FEA members to have a say?

 

[GregLocock]

Sorry amokhta I missed the t in the divsior, you were perfectly correct.

And austim, you are right, that 16bd^3 should be 8 bd^3, for the split tube result.

Mondays are always like that.

So we now have the thin wall tube giving 6 bd^3, and the rectangular strip and the split tube result agreeing at 8bd^3, which I suppose is close enough. One can imagine that gluing the two contacting surfaces of the crushed tube together would have a beneficial effect on the torsional stiffness. Cheers

Greg Locock

 

[austim]

Greg,

The International Date Line gives me an unfair advantage - I get Monday over before you.

The split tube formula must give you a lower bound estimate (unless you do in fact split the tube before you crush it).

In the automotive field you may have more faith than I in glue. The only text that I have which even mentions glue is a 1946 book on the Materials of Aircraft Construction. It is always good for some interesting stuff - strength testing of canvas etc. Perhaps glues have advanced since then ?

 

[GregLocock]

I'm in Geelong, Victoria, Australia so unless you're in Fiji or somewhere sheepy there won't be much in it.

OK, what I meant was suppose you take this tube, and squash it. If the two contacting surfaces can move relative to each other then the squashed tube will be able to react less torque than if the two contacting surfaces are pinned or glued. I'm rubbing my finger and thumb together to illustrate this point!

The split tube formula applies directly in the glued case, with the perimeter is half as long and twice as thick.


Cheers

Greg Locock

 

[austim]

Greg,

OOPS. I don't know how I had reached the conclusion that you were based in the USA. In fact we are much closer than I had realised - I live in the big smoke at the other end of the Geelong Highway.