Equivalent torsional constant for parallel beams

[fl11]

Hi,

I am looking for that answer for quite a while now. Do you guys know if there is any way to calculate an equivalent torsional constant from two rectangular beams?

For example, let say that I have two 2in x 5in rectangles separated with a distance of 2 in. Is there any way to calculate a single beam which would have the exact same torsionnal constant, the same way we could find a beam with equivalent area (sum of areas) and equivalent Inertia (parallel axis theorem)?

FEM program like Patran is not able to give equivalent properties for two separate surfaces and Catia V5 is able only for area and inertia.

Thank you very much for any help.

 

[israelkk]

If both beams are fixed to a plate on both sides of the beams and you try to rotate one plate while keeping the other plate fixed in addition to rotation of each beam (torsion stresses) in each beam there is a bending stress in each beam due to the distance between them. Therefore, I do not think you can simply find an equivalent beam. The larger the distance between the beams the larger the bending effect.

 

[rb1957]

so you have two beams connected in a fashion so that the applied load is reacted by both of them, but other than this there is no inter-connection between the beams. then you apply load to cause torsion.

i'd say that the primary reaction to the applied torque is differential bending ... the applied torque is reacted by a couple, bending the beams in opposite directions.

then to calculate the torsion constant, consider the twist that happens at the load point (due to one beam deflecting up and the other down). from the twist and the torque you can determine the torsion constant of the combination; but really aren't you interested in either the displacement (angular or linear) of some point ? obviously these are related to the twist.

 

[fl11]

This is more of a modeling question, but the principle stays the same as to find a way to calculate an equivalent torsional constant as it can be done with Inertia or Area.

I have attached to this post a picture showing the kind of beam I am talking of.

At one side, the beam is clamp and at the other side, a torquing load is applied. So if I would like to model this beam using only beam elements, in Patran lets say, I would have to input a Torsional constant. It would be quite easy, I agree, to model the two "parallel beams" as two beam elements having their own torsional constant. But let's say that I want to model both with only 1 beam element, what value of torsional constant should I use? I can I calculate it?

 

[rb1957]

in that case ( a pic is worth 1000 words) i'd use 2* the torsion constant of one "beam" ... assume each beam can absorb 1/2 the applied torque

 

[prex]

For a closed hollow section the torsional constant is proportional to the square of the enclosed area and inversely proportional to the developed length of the walls: so for two equal sections the torsional constant simply doubles (irrespective of distance).
The same conclusion holds for thin open sections (not connected in a way that creates a 'trapped' area), as the torsional constant is proportional to the developed length.
For the latter condition the Roark has some cases where the above conclusion may be checked; also some cases of sections with two unconnected open profiles are found in the first site below, under Beams -> Cross sections -> I beams | Channels | Angles

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[israelkk]

rb1957

When a torque is applied to the end plate of the two beams a reaction force is created where the beam is connected to the end plate in adition to torsional torque. At the same time because the beams are integral part of end plates each beam see's the same angle as the end plate. Therefore, the torque has to be larger than just the sum of two meams in torsion in order to recieve a defined angle. For example if the two beams were two concentric tubes then both tubes will see just pure torsion. However in this case where the two beams are not concentric the torque to avhieve a defined angle will be larger to overcome the bending of each beam as a cantilever.

 

[rb1957]

i quite agree that the exact value is going to be different ... this was meant as a quick, conservative, approximation for the OP's FEM for a beam property that probably isn't very significant anyways.

 

[GregLocock]

I assume you are applying the torque about y?

The contribution from the beam bending falls away as L^-3, and the shear contribution falls away as L^-1, so for a sufficiently long beam you can just add the torsional constant for each beam.

For interesting cases where L is fairly short then life gets much more complicated, in fact I end up building a detailed local FEA of the beam system to get an equivalent torsional stiffness for the single beam, quite often.



Cheers

Greg Locock

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

 

[fl11]

First, thank you all for your input, it is relly apprciated. I had more usefull comments here then in two weeks of questionning people in my working place.

In Roark formula for stress and strain, it seems a given that you can somehow add the torsional constants. For example, if you take two thick wall L section and put them together to make one thick wall C or U section, then it states that you only have to add the torsionnal constants. It also works fine with thin wall open section (I C ...) as you usually use sum(bt^3)/3.

But if you take two 2 x 5 rectangular sections, separated with a distance of 2 inches for example. The distance between the two sections does not seem, (still refering to roark), to have an effect on the final torsionnal constant. Then we can bring them close until they form a 4 x 5 recangular (a little bit like two L section forming a C section.) Then by adding the torsional constant, you would end up with a value of nearly 18-19 in^4 while actually, for a 4 x 5 rectangular section the torsional constant would be around 54 in^4. That is where I thought that may be the summation method wouldn't be systematic.

In the same time, I am really wondering if the effort really is worth while. What I am trying do to is to transform an assembly into a stick model in order to find the load path and find the reactions at the connections points between the part so I can make individual 3D models of those parts with those reaction apply as load boundary. So the stress actually doesn't really matter to me for the stick model, but I thought that the torsionnal constant would influence the rotation hence maybe the load path or at least the reactions at the connection points.

I'm sorry, this last paragraph is more suitable to an FEM forum, but it is somehow linke to the principal query.

 

[rb1957]

yeah, i agree with your last comment ... a bunch of work (soul searching) over something that isn't really that significant. you've got a beam element and that needs inertia properties ... unless there's something "funny" happening near either end, i'd put in some reasonably small prop and let it be. if there is a significant torque being applied nearby, you may want to detail model the structure (or wait, and hope, for the static tests to show it good).

btw, your earlier comments confused me a little ... of course torsion on an open section is different (much weaker) than a closed section ... but then i lost the thread about 1/2 way along ...

 

[prex]

francois, you can't compare the torsional stiffness of two separate rectangular sections with the join of them. If they are separate, there is no flow of shear stresses from one into the other, the flow of shear stresses remains confined in each one (recall the hydrodynamic analogy for torsion).
The same under a different perspective: if you torque two separate sections, you'll see them slide with respect to each other, you need to connect them in shear to let them behave like a single solid section.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[fl11]

Prex,

I totally agree with you. But as I said, only for the torsionnal constant point of view, Roark 5th edition Table 20 (Formulas for Torsional deformation and Stress) states in case 25 that for a thick U section or thick Z section, it is like having two thick L sections that you join together to form a U or a Z. So the torsional constant for the U section or Z section is twice the torsional constant for a L section.

So I thought that it might be the same for rectangular sections. But as I explained in my previous post, the sum of the torsional constant of two 2 X 5 rectangulars do not give the same value of torsional constant than one 4 X 5 rectangular section. That is where I got suspicious about doing a summation of the two constants for my case.

Anyhow, I think I am getting all your points which are very good. In a theoritical way, if you would have to represent two beams in parallel with only one single beam that would behave the same way, what would be the mathematical way to determine the torsional constant of that one beam. Or is it actually possible? Is it only by doing a local FEA as stated by GergLocock and that way, with the deformation or stress, find an equivalent K. I don't know. It just seems strange to me that a method was dertermined for inertia (parralel axis theorem) but nothing for Torsional Constant. And as was saying rb1957, in the end it might even not be that significant. Might be an idea for a master degree though... : )

 

[prex]

Sorry, francois, but seems we are not understanding each other.
There is no contradiction in your figures: the torsional constant of two unconnected equal sections is two times that of the single section. The torsional constant of a single section having an area equal to the sum of the formerly unconnected sections is a different thing and has a far higher value.
So the simple answer to your question is again: take two times.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[Tmoose]

http://weldingmag.com/blodgett/wdf_79843/

The new May 2008 issue of Welding Magazine has an article describing torsional characteristics of a ladder frame, and how to stiffen it. End conditions can change things quite a bit

Lincoln's "procedure handbook of arc welding" has a chapter with similar info.

 

[israelkk]

francoislambert11

The whole discussion talk about torsional equivalency however, stresses are an important ssue too. Because of the bending in the beams the stresses in the beams must be considered. Therefore, even if an equivalent single beam can be found that will give the same twist angle at a given torque the stresses will not be the same and the two beam may fail in bending stresses even though the equivalent beam will hold the torsional stresses.

 

[GregLocock]

Crazy idea, why not put in two parallel beam elements?



Cheers

Greg Locock

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

 

[fl11]

israelkk,

I agree with you, but the general idea behind what I am trying to do is to represent an assembly only with beams elements and what I am interested in, is to find the reaction forces at the pivot point between two parts of the assembly. Those reaction forces will then be re-apply on a solid 3D model of the part so the stress will be "real" and accurate. So what is important here, I think, is to be able to represent the global rigidity of the assemble in order to get a load path that is as close as possible to the reality.

GregLocock

Of course, using two beams would be appropriate and probably the best idea. But sometimes, you want to simplify symetric portions of a part. It is much more simplier to make one beam on the symetry line. Or sometimes, one beam ends with a lug. If I'm dividing the beam into multiple smaller beams, I might end up cutting the lug end getting two rectangular sections in "paralell". But like I said previously, I am probably scratching my head for something that isn't that important after all. : )