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What is the radial expansion $\delta$ of a thin elastic disk homogeneous in all directions of thickness t under a constant uniform tension T pressing on both sides, like in the picture? e.g. The tension could be due to a uniform pressure. I suppose that it can be computed from the strain $\epsilon = \frac{2 T}{t E}(1-\nu)$ obtaining $\delta = \epsilon r$. Is this correct?

Any good suggestion for a reference book? I need to compute the same for a truncated sphere (a sphere with a hole and one with the hole covered by a flat end).

enter image description here

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  • $\begingroup$ By tension do you mean compression? Does the disc have the same characteristics in all dimensions? $\endgroup$
    – Solar Mike
    Jan 31, 2022 at 17:45
  • $\begingroup$ Hi Solar Mike, yes homogeneous uniform compression on both sides, perpendicular to the surfaces. Same characteristics of the disk in any direction. $\endgroup$
    – adras81
    Feb 1, 2022 at 10:45
  • $\begingroup$ Then you should edit paragraph 1. $\endgroup$
    – Solar Mike
    Feb 1, 2022 at 10:55

1 Answer 1

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The arrows show compresion. But it dosent matter much. only signs change.

If we consider the disk as a section of a bar its axial deformation is,

$$\delta= \frac{PL}{AE}$$

$$\epsilon= \delta/L$$

  • P = stress

  • A = area of the disk.

  • $\nu=$ poison's ratio

and its orthagonal chenge is $$r_{final}=r\_{initial}*\nu*\epsilon$$

if the stress on the disk is tension the diameter will decrease, if it is compression it will increase.

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  • $\begingroup$ Thanks kamran! After some thinking, it makes sense. $\endgroup$
    – adras81
    Feb 1, 2022 at 11:37
  • $\begingroup$ @adras81, if it answers your question please check it as accepted so others can use it. thank you. $\endgroup$
    – kamran
    Feb 1, 2022 at 15:39

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