Bending stiffness of circular bars

In summary: I: Second moments of area, which directly relate to the stiffnessR: Radius of big rod and outer radius of tuber: radius of small rod and inner radius of tubeRODIbigRod = pi/4*R4ROD + SHELLIRodAndTube = IsmallRod + Ishell = pi/4*r4 + pi/4*(R4-r4) = pi/4*R4Hence:IbigRod = IRodAndTubeNote: 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
  • #1
itsmi
5
0
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!
 
Engineering news on Phys.org
  • #2
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 ?
 
  • #3
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?
 
  • #4
The problem statement says that the tube is not bonded and can freely move ?
 
Last edited:
  • #5
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?
 
  • #6
Anyone else?
 
  • #7
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
 
  • #8
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?
 
  • #9
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 ?
 

What is the definition of bending stiffness?

Bending stiffness is the measure of a material's resistance to bending under an applied force. It is a property that describes how easily a material can be bent or deformed.

How is the bending stiffness of circular bars calculated?

The bending stiffness of a circular bar can be calculated using the formula EI = (π/64) x (d^4), where E is the Young's modulus of the material and I is the moment of inertia of the cross-sectional area.

What factors affect the bending stiffness of circular bars?

The bending stiffness of circular bars can be affected by the material's Young's modulus, cross-sectional area, and the length of the bar. The type of loading, such as point load or distributed load, can also impact the bending stiffness.

How does the bending stiffness of circular bars compare to other shapes?

Circular bars have a higher bending stiffness compared to other shapes, such as rectangular or triangular bars, due to their symmetrical cross-sectional area. This makes them more resistant to bending and deformation under applied forces.

Why is the bending stiffness of circular bars important in engineering and construction?

The bending stiffness of circular bars is an important factor to consider in engineering and construction because it determines the structural integrity and stability of a building or structure. It also affects the load-bearing capacity and overall performance of the materials used in construction.

Similar threads

Replies
19
Views
374
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • General Engineering
Replies
8
Views
1K
Replies
8
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
746
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
22
Views
4K
  • Mechanical Engineering
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Back
Top