Bending stiffness of circular bars

AI Thread Summary
The discussion centers on the bending stiffness of two configurations: a solid circular rod and a circular rod surrounded by a non-bonded tube. Initial calculations suggest that both configurations have equal second moments of area, leading to the conclusion that their bending stiffnesses are the same. However, the presence of the non-bonded tube raises questions about relative movement and effective stiffness. Participants emphasize that the neutral axes of both components coincide, indicating no relative motion, which supports the initial conclusion. Ultimately, the consensus is that the bending stiffness of both configurations is indeed equal.
itsmi
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Which of the two is stiffer in bending?
1. A circular rod of diameter D and length L.
2. A circular rod of diameter d and length L, surrounded by a tube of inner diameter d, outer diameter D, and length L. The tube is not bonded and can freely move.

It seems like an easy solution in that we only need to calculate the second moments of area, which yields the same stiffness for both cases. However, I've had a hard time convincing people of this and would like to confirm that the bending stiffnesses are, indeed, equal. Thank you!
 
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The effective stiffness will not be the same for the two cases .

Note these words in the problem statement " The tube is not bonded and can freely move "

What does that tell you ?
 
Thank you for the quick response.

I'm aware that this is the tricky part. But the question is: Will it move (relatively to the rod)? I don't think it will. How would you approach this problem mathematically, rather than using your intuition?
 
The problem statement says that the tube is not bonded and can freely move ?
 
Last edited:
Ok, so here's my procedure
  • I: Second moments of area, which directly relate to the stiffness
  • R: Radius of big rod and outer radius of tube
  • r: radius of small rod and inner radius of tube
ROD
IbigRod = pi/4*R4

ROD + SHELL
IRodAndTube = IsmallRod + Ishell = pi/4*r4 + pi/4*(R4-r4) = pi/4*R4

Hence:
IbigRod = IRodAndTube

Note: The neutral axis remains at the same position in both cases, not like in the typical example where you place many thin plates on top of each other and compare it to one thick plate of the same total thickness.

Where do you think the mistake is and which theory/law makes you think that?
 
Anyone else?
 
Just try solving for the stiffness of a simple cantilever with point load at the end .

If you do the analysis properly you will get different answers for the two beams
 
You keep saying it is different, but you still haven't given any explanation or good reason, despite me asking multiple times. As I said, I did the calculation and get the same results. It doesn't matter if it's a three point bending or cantilever beam with a point load.

Is there anyone else who can give me an explanation for whichever solution is correct?
 
itsmi: Your answer is correct. The second moments of area of the two cases are equal. The reason is as explained by you. I.e., as you alluded to, the neutral axes of the individual components are coincident. Therefore, there is no relative motion between the telescoped parts.
 
  • #10
Consider a simply supported beam with point load at mid span . With the point load acting on the outside of the tube what load acts on the central bar ?
 
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