Tube-Inside-Tube Beams 7

[colar]

Has anyone ever used a smaller cross section HSS member inside another HSS member (i.e. a 2.5 x 2.5 x .125 inside a 3 x 3 x .188) to stiffen/strengthen a beam.

I am considering this becasue of size constraints on the outer envelope of a beam. I realise this is an inefficient use of material, but the size constraints leave little choice.

My main question has to do with the tranverse shear stresses between the mating surfaces of the two beams. There is lots of literature about built up beams, but in this case the two beams share the same neutral axis. Will there be transverse shear between the outside of the 2.5" and the inside of the 3" that will need to be picked up by some plug welds or end welds? Is it just a matter of needing shims between the tubes at the ends so that the normal force can be transmitted and the two tubes are engaged simutaneously?

Look forward to some advice

Colar

 

[Lcubed]

You will need a shear connection between the two tubes, preferably along the full length rather than just at the ends. You might be better off to start with the smaller tube and weld plates on the top and bottom, rather than using two nested tubes. Suggest you check the section modulus for several variations.

 

[redhead]

A simole way to provide shear coupling between the two tubes would be to provide 3/4" dia. holes along two opposite faces of the outer tube and use puddle welds for the shear transfer.

 

[dlew]

colar,

A few comments:

1) If you connect both tubes along their length as suggested by Lcubed and Redhead, the moment of inertia of the built-up section is the sum of the moments of inertia of each tube. The two tubes would share the load in proportion to their moments of inertia. There is no a large horizontal shear between the two tubes, but the connection is needed to transmit the load from one tube to the other.

2) There is a difference of 1/8" between the outside of the 2.5 HSS and the inside of the 3 HSS. If you keep a common neutral axis for both HSS and do not tie both tube together along their length, the outer tube would have to deflect 1/16" before starting to transmit the shear to the inner tube. The outer tube would carry more load than the load that it will carry if the tubes are connected.

3) If the direction of the load does not reverse, the two tubes could be installed touching each other so the shear load is transmitted by bearing.

4) The total load that the two tubes can carry, independent of how you connect the tubes, is less than the sum of the loads that each tube can carry by itself.
The load that the inner tube would carry is limited by deflection, since the deflection of both tubes has to be the same.

5) Could you use a heavier wall 3" HSS, or a rectangular HRS instead of two tubes?

 

[colar]

So do I have two opposing views between Lcubed and dlew? i.e. is there transverse shear that needs to be picked up as well?

 

[VoyageofDiscovery]

Hi Colar,

In the situation you propose you have:

1) Bending stress resisted by the new Section Modulus
2) Transverse shear stress resisted by the outer section
3) Longitudinal shear stress resisted by the shear connectors that ensure composite strain (simultaneous bending)throughout the new section when loaded transversely.

Regards

VOD

 

[par060]

This configuration occurs frequently when reinforceing curtain walls. An aluminum mullion is reinforced with steel channels. The channels are fairly snug inside the mullion so that as the mullion deflects under loading the steel channel picks up a portion of the load.
I see no reason to connect the members together...think of 2 2x10's spaning between 2 saw horses...if you walk across them they both will be supporting your weight regardless if they are fastened together or not

 

[steve1]

It seems to me that in this particular case it dosen't matter if the two tubes are tied together or not. The reason being is that if the tubes do act compositly than the calculation of the combined moment of inertia would be the sum of I + Ad squared. But since the d term would be zero (i.e. same neutral axis for both the individual and the composite sections) then this reduces to the sum of the I's, which is exactly what you would calculate if there was no composite action. Therefore it seems that you just need shims to ensure that the neutral axis' remain concurrent, and to ensure load transfer.

 

[RockEngineer]

If you use plug welds along their length to connect the two tubes the moment of inertia becomes
I=I1+A1*d1^2+I2+A2*d2^2 For the concentric tubes the d1 and d2 are zero and you get I=I1+I2

If the tubes aren't connected along the length and share the load only due to equal deflection

I=1/(1/I1+1/I2)

These are not the same. This is why when you build a built up girder you don't just lay the horizontal plate on top of the vertical plate and weld them at the end. You have to transfer the longitudinal shear to be able to use the first equation.

 

[vonlueke]

colar: Because the centroids of both cross sections reside at (or essentially at) the same elevation, the moment of inertia of the built-up section is the sum of the moments of inertia of each tube regardless of whether welded or not, provided both cross sections start deflecting at exactly the same time. Welding certainly achieves the requirement to make the inner tube start deflecting at the same time as the outer tube. But if your applied load is downward and nonreversing, then if you insert the smaller tube inside the larger tube, then insert a nontapered plate shim (which also must function as a "bearing pad&quot between the two bottom walls of the tubes at each end of your beam (in order to push the inner tube upward in contact with the upper wall of your outer tube), then you achieve the same goal (simultaneous engagement).

There will be no transverse (nor longitudinal) shear transfer between the two tubes that needs to be picked up by plug welds, etc., provided the top surface of the inner tube (especially at midspan) is initially in contact with the top wall of the outer tube.  It's therefore just a matter of needing shims underneath the the inner tube at the beam ends so that these two surfaces are initially in complete contact at midspan, so that the two tubes will be engaged simultaneously. (This is item 3 by dlew.)

The question in the above option is, how do you ensure these two surfaces are initially in contact at midspan? Just seeing the inner tube shimmed upward in contact with the top wall of the outer tube at the beam ends perhaps doesn't necessarily guarantee the top walls are completely in contact at midspan. Possible solution, for review: If the two tubes have a camber, and you orient the cambers in opposite directions, then force the inner tube inside the outer tube, keeping track of the "top," then after forcibly shimming up the bottom of the inner tube at the beam ends, you are perhaps sure the two top walls are in contact at midspan.

Assumptions: All of my above paragraphs assume your beam is horizontal and simply supported, and that the applied load(s) are downward and nonreversing.

Before considering the built-up section option(s), first strongly consider the first part of item 5 by dlew, to see if your analysis shows it has sufficient moment of inertia, as item 5 by dlew isn't limited to the above assumptions.

 

[colar]

Thanks for the info everyone.

Obviously there is a wealth of varying opinions on this subeject! Those of you who are of the opinion that a shear joint is not needed, I agree that if the assembly would work at all there needs to be contact to transfer the loads.

However, I still don't have a fundamental feel or indication that there is no shear at this interface. Certainly in the case of two planks on top of one another, there is shear that needs to be picked up. Try bending a softcover book without restraining the ends. It will be floppy and the pages will slide past each other. If the pages are restrained at the ends, it is much stiffer. This is pretty fundamental stuff. However, when the two members are on the same neutral axis we are looking at a different situation (or are we?)

As I picture it in my mind, I see slippage. On the other hand I see the logic in the "restrain only" camp. Maybe I will need to try an experiment to find out. I tried modeling it on COSMOS and it tells me that a shear joint is not needed. (However, I just started playing with COSMOS so I am not sure of the solution legitimacy).

Incidentally, I can not add plates to the outside edges to build up the section as there is not room in my application. Also, going to a thicker wall would not work either for my case, as there is a predefined component that needs to fit into one end of the large tube. I will just make the inner tube slightly shorter to meet this requrement.

Maybe this should be a student project?

Thanks again,
Colar

 

[RockEngineer]

I concede that if the following requirements are met you should be able to add the moments of inertia directly.

1. The ends are fixed so that there is no relative movement or sliding between the components.
2. The neutral axis coincides.
3. The items are fixed so they deflect as one.
Slop removed.

This has been a controvery on several occasions amoung engineers in our office when you have one tube inside another. It always gets the discussion going.

Disregard my previous post where I went from the spring constants to adding moments of inertia. I had a brain fart and didn't test out my thought before opening my mouth.

I should have read the statement on the base of the little statue that has set on my desk for the past 22 years. It says "Be sure brain is engaged before putting mouth in gear".

 

[trainguy]

You can talk structural theory all you want, but at the end of the day, the guy in the fab. shop will be phoning YOU when he can't make it work!

Just use a smaller tube, with welded bar reinforcements, and watch how easily it all goes...

 

[prex]

colar
to understand why shear pins are not required in your case just follow this way of reasoning.
Take the two planks one on top of the other of your example: the shear connection that you need in that case serves to transfer the forces that are compressive on the upper plank to the tensile forces that act on the lower plank. If the shear pins are not present, each plank will have its separate linear distribution of stresses, so no resultant is transmitted between the planks, but of course the inertia and the resistance are much smaller.
Now if you look at your tubes one inside the other (but they could well be one on the side of the other, if they are forced to have the same deflection), you'll see that each one of the tubes has its own linear and symmetric stress distribution in the height of the section: this distribution is equilibrated separately for each one of the tubes, so no shear force is required between them. prex

http://www.xcalcs.com

Online tools for structural design

 

[colar]

I think the way Prex has put it has finally managed to sink in to my own mind. That is

"as slong as they start bending at the same time (i.e. shimmed properly) they will add together and Itotal=Ibig+Ismall. This is no different than if they were tied to each other side by side."

This seems pretty obvious now.

Thanks Prex

Colar

 

[LPPE]

Thats a lot to read at the end of the day!
One question I dont understand - If the inner tube doesnt fit snug inside the outer tube, how are you going to shim the middle part of the inner tube so there is contact (so they both deflect at the same time)?? It seems to me just shimming the ends wont be adequate enough to ensure contact for the entire length. You'd have to make sure your DL deflections are darned close, if not exact, no?

 

[colar]

pylko

I will drill and weld the tube anyway. Even though the shear connection is not needed, it is probably the easiest from a mfg point of view and the best way to ensure that there is contact right from the start.

So after all that, I am going to do what I was gong to do from the start, but for a slightly different reason.

Colar

 

[FinnB]

Hi
My view is that two timber planks on top of each other do not make an overall beam equivalent to the depth of two planks, in terms of bending strength, without a shear connection.Laminar horizontal shear forces develop between the planks causing the planks to slip relative to one another.The two planks act indivually as opposed to one deep section.

In your situation the outer tube first takes up the load and as it deflects loads the inner tube.As the inner tube delects if may slip in relation to the outer tube.If this happens the outer tube may need to take up more load than it is designed for.To avoid this I would always connect the two tube to form a composite section.It is then possibe to add the moment of inertia of the two sections with safety.

Note: If you dont connect the two tubes are you limiting the maximum deflection to that of the inner tube as opposed to that of a composite section ?(This wont allow you the full capacity of the outer tube to be mobilised) Can you predict how much each tube will deflect?

The fabrication cost of connecting the two tubes is peanuts, why bother with putting one tube inside another without a connection when it is difficult(impossible I would say)to predict what will happen, especially assessing the deflection/loadsharing of the tubes.

Sorry if this got long winded
Finn

 

[adlington]

I read with interest all the various opinions as to what is occurring when load is applied to your double tube arrangement. I do feel however the first analytical decision should be to consider your size restrictions to ascertain the maximum size of your beam (depth and width) you can use. Once this is known calculate your beam forces (moments and shears) and based on these size restrictions calculate the section required, i.e. make up your top and bottom flanges with plates etc to satisfy your bending and deflection requirements.

Trying to make a tube work inside another tube compositely is very difficult and unless you can ensure the two sections work compositely (i.e. bonded along their entire length) you cannot with any absolute certainty analyse the tubes as a composite section having ‘singular’ properties.

Start with what you know and what you have to work within and make up a section to suit, i.e. tubular section with welded plates top and bottom. Are you sure you cannot facilitate a greater depth?

Do not worry about longitudinal shear between components, I’m uncertain why a lot of the advice given discusses this as an issue, bending stresses, deflection and bearing shears are what you need to check

Regards

 

[FinnB]

To adlington

Longitudinal shear stress is discussed because this is the force which must be resisted by longitudinal welds when making up a composite section.

Longitudinal shear stresses in standard sections are generally so low they can be ignored, but they are there all the same.

To colar

You probably don't need to weld the beams together all along there length. Intermittent welds as you discribe are likely to be sufficent. It not hard to work out the stresses and weld size you would need.It's back to the old equation which rules structural engineering:

Engineers theroy of bending.

Moment \ 2nd Moment of Inertia = stress \ distance from the neutral axis to the point in the cross section which is being considered.