ube-inside-tube beams

huangyhg

tube-inside-tube beams
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
find a job or post a job opening
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.
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.
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?  
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?
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
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.  
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.
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.
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") 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.
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