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Concentric Tubes in Bendine 2

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toothless48

Mechanical
Oct 27, 2014
29
Hello all, I am analyzing a design that involves a stainless steel tube in bending. This tube is reinforced with a smaller "doubler tube" snugly fit inside (its OD being equal to the larger tube's ID). The tubes are not bonded, and are free to slide, ignoring friction. I have scoured the internet and my textbooks to no avail - how is the section modulus for this configuration calculated? The area moment of inertia of the cross section is the same regardless of the tubes being bonded or free to slide. I assumed, similar to composite beams, that two sliding tubes would be less stiff than one thick tube... Is this not true?

Many thanks
Mike
 
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It occurs to me that we normally neglect shear deflection in beam deflection problems, but if that was considered, it may no longer be possible to get them to deflect identically under the assumptions made. That is in addition to any end effects mentioned above.
 
the two tubes are constrained by geometry to deflect the same.

another day in paradise, or is paradise one day closer ?
 
I finally ran the numbers on this and Koot was correct all along.....as he stated, composite action is not required in this example and the stiffeness is the same for 2-tube setup versus the solid equivalent tube.......learn something new everyday wheather one wants to or not...Timoshenko is again resting in his grave, if not in the same position....
 
SAIL3 said:
Koot was correct all along

Hey, even a blind squirrel finds a nut once in a while. Thanks for doing the leg work and for reporting back to share the results.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
The above discussion has been predicated on the two cylinders being of the same material with equivalent elastic moduli. Can I assume that if they were different materials with different elastic moduli that the equivalent EI would be (EI)outer + (EI)inner? I have a real-life reinforcement problem with some wood beams that I would like to reinforce with thick wall steel pipe to counter a termite concern. Thanks!
 
It'll work so long as the respective neutral axes of each piece are at the same location.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
If there is a small gap between the inner and outer pipe in order to achieve a sliding fit, the inner pipe will have a smaller deflection than the outer pipe and will tend to feel a single load at or near midspan.

BA
 
I agree with BAretired. When placing the smaller tube we will have 3 cases.

First one, the smaller tube is a little bigger than specified and then it won't fit without getting a lot of residual stresses.
Secondly, the members perfectly size each one another and then it will behave as described.
Third, probably the most common situation will be to have a gap.

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I agree, the truthiness of the theory discussed above is sensitive to the ratio gap/deflection. An eighth inch gap in a system deflecting three inches is surely of little consequence. A half inch gap would be a different animal.

KootK said:
It'll work so long as the respective neutral axes of each piece are at the same location.

In the interest of precision, let's revise that to:

It'll work so long as the respective neutral axes of each piece are at the same location and are made to travel in unison.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
@rockitz, I'd prefer to use the "rule of mixtures" and convert both sections to the same material (E)

another day in paradise, or is paradise one day closer ?
 
Fascinating discussion. Among the products my company makes are carbon fiber composite tubes that fit within each other with 0.001" gap between them. The outer tube has an OD of 0.300" and a 0.025" wall thickness and 32" length, to give some perspective. Both the outer and inner tubes are designed to have similar bending stiffness so the inner tube has a greater wall thickness.

rb1957's intuition is correct. Without bonding there is no composite action.

The logic that because the neutral axis is the same for both tubes whether they are bonded or not proves that there must be composite action is flawed. Let's take the case of three point bending. If you take the two tubes separately and bend them the same amount they will each have a certain load in the center of the span. If you place one tube in the other and deflect them the same amount, the measured load will be the sum of the loads measured for each tube. This is non-composite action. So why is this?

The factor that has been overlooked in all of the discussion so far, are the hoop stresses in the tube. When a tube is flexed it wants to ovalize where the tension and compression sides move closer together, which results in a lower "I" in axial bending. In an I-beam the web prevents this. In a tube only the walls can do this and they behave as two "C" shaped springs. So, with a tube in a tube that are not bonded you have four thin springs. When they are bonded they act as two thicker springs, which are stiffer than the four (composite action).

A unique feature of fiber reinforced composite material is that you can tailor material properties by changing the fiber orientations. One way to make a carbon fiber tube is by pultrusion, where fibers and resin are pulled through a shaping die. All the fibers are axial in the tube wall (think fiberglass tent poles). When flexed too far, these tubes fail by splitting axially because they ovalize, and there is no fiber reinforcement in the hoop direction. They have low axial stiffness because of the low hoop stiffness.

Tubes can also be made by table rolling where two sheets of unidirectional preimpregnated fibers (prepreg) are laminated together in a 0/90 orientation and then rolled around a steel mandrel and cured in an oven. Interestingly, a tube that is made from 50% 0 and 50% 90 fiber is stiffer in axial bending than a tube made from the same amount of fiber that is all in the 0 degree direction.
 
Compositepro, you are mixing different matters. All this discussion was about the behavior of two concentric tubes in the frame of the classical elastic beam theory. If you add ovalization to that, then the subject changes.

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compositepro said:
If you take the two tubes separately and bend them the same amount they will each have a certain load in the center of the span. If you place one tube in the other and deflect them the same amount, the measured load will be the sum of the loads measured for each tube. This is non-composite action. So why is this?

Actually, in this instance, that is composite behavior. That was one of the interesting findings above. Here, IE_comp = IE_non-comp.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
You are saying that bonded or not bonded the tube in a tube will have the same bending stiffness. That is not true. Bonding the tubes together will increase stiffness.
 
I disagree strongly. That said, I don't have any persuasion arrows in my quiver other than what has already been set out above.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
If there is no gap, I agree with KootK. The moment of inertia of any shape is the sum of the moment of inertia of its parts. If there is a gap, the bending stiffness of the solid ring is slightly greater than the sum of the two separate rings. The difference is Eπ(d[sup]4[/sup]-di[sup]4[/sup])/64 where d = inner diameter of the outer ring and di = outer diameter of the inner ring.



BA
 
It's true and not true. If shear isn't transfer then the members acts individually, but while having concentrics tubes without gap the inertia is not sensitive to shear transfer, just because "y" is continuous in between the members. Acting LIKE a composite action, even when it's not a composite member.
draw2_ufcako.jpg
 
I suppose it's a matter of how one defines just what a composite member "is". Is it about the degree of shear transfer available which may vary from zero to infinity? Or is it about achieving certain section properties based on maintaining a common strain profile? For me it's the latter.

I would characterize the two tube setup as "trivially composite". That, implying:

1) Sure, it's composite but then the composite properties are no improvement over the non-composite properties. A bit like concentric sistering in wood construction.

2) Demand for shear flow shear transfer to achieve "composite" is nominally zero unless one introduces a meaningful gap or drills down deep into more granular effects.

I believe that #2 is the important insight from the perspective of OP's original question.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
 
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