Internal Forces in Coaxial Bars of Different Materials

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Discussion Overview

The discussion revolves around the internal forces in coaxial circular bars made of different materials (steel and aluminum) when subjected to an external force of 30 kN. Participants explore the implications of this loading scenario, considering both the compressive loads and the material properties of the bars.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the internal forces in both bars will be 30 kN if the load is applied to both parts.
  • Another participant questions the clarity of the load application, indicating that if the load is applied only to the inner bar, the analysis may differ.
  • A later reply introduces the concept of calculating compressive deformation using the formula Δ = PL/AE, suggesting that the internal stress and contact pressure between the bars may need to be considered.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the internal forces in both bars are indeed 30 kN, as there is uncertainty about how the load is applied and its effects on the bars.

Contextual Notes

The discussion lacks clarity on the specific loading conditions and the assumptions regarding material properties and interactions between the bars.

chandran
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i have attached a doc file. it shows two coaxial circular bars of different material and a force(30kn) is applied in the top. I think the internal forces in both the bars
will be 30kn. Am i correct.

bar1 is made of steel and bar 2 of aluminium. the total length of the structure is
10metres.
 
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No .doc file attached.
 
i have attached a jpg since memory was less.
 

Attachments

  • coaxial.jpg
    coaxial.jpg
    3.5 KB · Views: 466
I can't really make out the picture. I can't tell if the load is applied to both parts or the inner one. If you are asking what the compressive load is and the 30kN is applied to both bars, then you are correct.

As another possibility, you may also nee to look at this is with the application of the load to the inner shaft, which will compress according [tex]\Delta = \frac{PL}{AE}[/tex] Since the volume remains constant that should give you an ability to calculate the new diameter of the shaft. The same holds true for the outer shaft, simply using the constants for a different material. Once you have calculated the new diameters, chances are that there is an interference between the two. That interference will produce a contact pressure and thus an internal stress between the two.
 

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