Does Poisson's ratio apply when we have no loadings?

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Poisson's ratio is fundamentally related to stress and strain in materials, quantifying lateral expansion or contraction when a stress is applied. In the case of a free rod subjected to uniform heating, the change in diameter can be accurately calculated using the formula Δd = αΔΤd, where α is the coefficient of linear thermal expansion and ΔΤ is the change in temperature. This approach is valid under the condition that the linear expansion remains small (i.e., ##\frac{\delta }{L} \ll 1##). The discussion emphasizes that while Poisson's ratio is significant under mechanical loading, it does not apply in scenarios without stress, such as thermal expansion.

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Does Poisson's ratio apply when we have no loadings ? For instance if we have a free rod and we increase its temperature, in order to find the change of its diameter should i say :
Δd=-v*ε_x*d, where d is the length of the diameter or Δd= α*ΔΤ*d ?
Thanks in advance.
 
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question4 said:
Does Poisson's ratio apply when we have no loadings ? For instance if we have a free rod and we increase its temperature, in order to find the change of its diameter should i say :
Δd=-v*ε_x*d, where d is the length of the diameter or Δd= α*ΔΤ*d ?
Thanks in advance.
Poisson's Ratio is stress related. If you apply a stress in a given direction causing a strain, it quantifies what happens in lateral directions in terms of expansion/contraction for a given material.

Uniformly heating (i.e. changing the temp of) a free rod is stress free.
 
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erobz said:
Poisson's Ratio is stress related. If you apply a stress in a given direction causing a strain, it quantifies what happens in lateral directions in terms of expansion/contraction for a given material.

Uniformly heating (changing the temp) a free rod is stress free.
So in order to find the change of the diameter is it enough to say that : Δd= α*ΔΤ*d ?
 
question4 said:
So in order to find the change of the diameter is it enough to say that : Δd= α*ΔΤ*d ?
Well, I believe that formula is for ##\frac{\delta }{L} \ll 1##, but basically...yes.
 
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Welcome, @question4 ! :cool:

In practice, the linear expansion of metals is the most calculated due to its negative consequences.
Diameters of solid metal bars also grow with temperature, but that is mainly important for rings that slide tightly into cavities (like a bearing in its housing).

The diametral expansion of those rings are calculated like an unfolded section of metal expanding linearly; therefore, a coefficient of linear expansion is mostly used.

For fluids, a coefficient of volumetric expansion is used instead.

Please, see:
https://pressbooks.bccampus.ca/collegephysics/chapter/thermal-expansion-of-solids-and-liquids/

https://www.engineeringtoolbox.com/volumetric-temperature-expansion-d_315.html

https://www.engineeringtoolbox.com/thin-circular-ring-radius-temperature-change-d_1612.html

https://www.engineeringtoolbox.com/linear-thermal-expansion-d_1379.html

Now, when combining mechanical loads and high temperatures:

Copied from
https://en.wikipedia.org/wiki/Poisson's_ratio

"Most steels and rigid polymers when used within their design limits (before yield) exhibit values of about 0.3, increasing to 0.5 for post-yield deformation which occurs largely at constant volume."

The forging process shown in this video seem to demonstrate that any ratio (determined experimentally for metal in normal conditions) would change depending on sufficiently high applied loads and/or temperatures to the molecular bonds.

 
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This video can be helpful:
 

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