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

Click For Summary
Poisson's ratio is relevant only under loading conditions, as it describes the relationship between axial strain and lateral strain in materials. When uniformly heating a free rod, there are no applied stresses, making Poisson's ratio inapplicable. Instead, the change in diameter due to temperature increase can be calculated using the formula Δd = α*ΔT*d, where α is the coefficient of linear expansion. This approach is particularly important for applications involving tight-fitting components, such as bearings. In scenarios involving both mechanical loads and high temperatures, the behavior of materials may change, affecting their Poisson's ratio.
question4
Messages
2
Reaction score
0
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.
 
Engineering news on Phys.org
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.
 
Last edited:
  • Like
Likes DeBangis21, Lnewqban and Chestermiller
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.
 
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.

 
Last edited:
This video can be helpful:
 
Thread 'Local pressures in turbocharger housing?'
This is question for fluid mechanics. Static pressure in the exhaust manifold(turbo car engine) is usually 1.2 to 2.5 times higher than the boost pressure(intake manifold pressure).Boost pressure is around 1bar guage pressure(2bar absolute). Can the local static pressure somewhere inside a turbine housing ever be lower than atmospheric pressure, is this possible? here some links where CFD is used...

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K