Understanding Poisson's Ratio and Its Effects

[guideonl]

TL;DR

Understanding the internal mechanism

Hi everybody,
Poisson's ratio determine that when applying force in one direction (let say z direction) it would cause a relative major strain at that direction, and also two additional strains at the two other directions (x & y).
My question is, what is the explanation for the change in the body measurements ( shrink at x & y directions) when no external loads applied on these directions? In other words, what causes the material to move and change x-y dimentions? Also, when we look in the cross sections we find only the internal normal stress caused by the external force (z dir.) but no other internal stresses at the other (x & y) dir's.
If there is an 2 or 3 dim. detailed engineering example that can illustrate the use of posson's ratio i'll appreciate that.

 

[Astronuc]

Deformation of a solid is generally and isochoric (constant volume) process. If one were top compress a solid block, one would find that the sides would push out a little. The density does not change, the mass and volume are constant, so the perpendicular surfaces move outward.

On an atomic level, think about the arrangement of atoms in a crystalline lattice. It's probably more obvious with closed packed (hcp) or fcc and bcc lattices, but as one pulls apart the atoms in one direction, increasing the distance between atoms in a line, neighboring atoms (in the lateral directions) tend to settle into the wider spacing. Some introductory textbooks on material science have some information on the deformation of solids, but I can't put my hands on them right now.

 

[guideonl]

Deformation of a solid is generally and isochoric (constant volume) process. If one were top compress a solid block, one would find that the sides would push out a little. The density does not change, the mass and volume are constant, so the perpendicular surfaces move outward.

On an atomic level, think about the arrangement of atoms in a crystalline lattice. It's probably more obvious with closed packed (hcp) or fcc and bcc lattices, but as one pulls apart the atoms in one direction, increasing the distance between atoms in a line, neighboring atoms (in the lateral directions) tend to settle into the wider spacing. Some introductory textbooks on material science have some information on the deformation of solids, but I can't put my hands on them right now.

Thanks for your reply,
I am still wondering how come the perpendicular surfaces move with no force acting alined, unless we assume equal hydrostatic pressure is spreadout inside the solid (caused by the ex. force) in the same maner as in fluid material...

 

[Vigardo]

Astronuc's lattice example is quite illustrative. The shortening (or enlarging) of inter-atomic distances in the z-axis leads to some kind of lateral force in x- or y-axis directions. When inter-atomic equilibrium distance changes, the atoms experience a force reaction that tends to compensate the perturbation. For example, since the distances in z direction shorten, the distances in the x and y directions tend to increase. You can imagine some kind of electrostatic repulsion (or attraction), but it is more complicated than that sonce mechano-quantic effects must be considered at such mall scales. I hope this helps to clarify.

 

[guideonl]

OK, thank you both,
I think that also the classic mechanics should have an explanation for that... maybe in the form of internal forces/stresses causing the perpendicular surfaces (x-y) to move. This should also affect the strength analize.

 

[Chestermiller]

It is not quite correct to say that solids are incompressible. If that were the case, the Poisson ratio of all solids would be 1/2. But typical Poisson's ratios are on the order of 0.3. This means that, when a solid is put under uni-axial tension, its volume slightly increases (and its density slightly decreases).

Uni-axial tensile stress $\sigma$ in the x-direction is equivalent to the superposition of an isotropic tensile stress of $\sigma/3$ in all three directions (giving rise to a volume increase) combined with a loading of $\sigma_x=\frac{2}{3}\sigma$ and $\sigma_y=\sigma_z=-\frac{1}{3}\sigma$ (giving rise to no volume change, but to compression in the y- and z directions and extension in the x direction).

 

[guideonl]

It is not quite correct to say that solids are incompressible. If that were the case, the Poisson ratio of all solids would be 1/2. But typical Poisson's ratios are on the order of 0.3. This means that, when a solid is put under uni-axial tension, its volume slightly increases (and its density slightly decreases).

Uni-axial tensile stress $\sigma$ in the x-direction is equivalent to the superposition of an isotropic tensile stress of $\sigma/3$ in all three directions (giving rise to a volume increase) combined with a loading of $\sigma_x=\frac{2}{3}\sigma$ and $\sigma_y=\sigma_z=-\frac{1}{3}\sigma$ (giving rise to no volume change, but to compression in the y- and z directions and extension in the x direction).

Thank you,
Your explanation makes sence.
Now, for practical use, in case of uni-axial tensile strees in the x direction as you mentioned, we expect to get elongation in x-axis, and 2 negative strains in the two perpendicular axes.
The positive strain in the x-direction can be expressed by the tensile stress divided by E, or could be multiplied by E to compute the tensile stress in the x-direction.Practically, are the 2 negative strains in the perpendicular axes create another internal normal stresses -inside the body- in the same maner described above (using E)? or $\sigma_x$ is the only normal stress inside the body, and these 2 strains stand alone and not related to stresses in these directions?

 

[Chestermiller]

Thank you,
Your explanation makes sence.
Now, for practical use, in case of uni-axial tensile strees in the x direction as you mentioned, we expect to get elongation in x-axis, and 2 negative strains in the two perpendicular axes.
The positive strain in the x-direction can be expressed by the tensile stress divided by E, or could be multiplied by E to compute the tensile stress in the x-direction.Practically, are the 2 negative strains in the perpendicular axes create another internal normal stresses -inside the body- in the same maner described above (using E)? or $\sigma_x$ is the only normal stress inside the body, and these 2 strains stand alone and not related to stresses in these directions?

The latter. The negative strains in the y and z directions occur even though there is no stress in these directions. The strains can be interpreted as the superposition of the strains (and stresses) from the two combined loadings described in my previous post.

Hooke's Law in 3D tells us that:
$$\epsilon_x=\frac{\sigma_x-\nu(\sigma_y+\sigma_z)}{E}$$
$$\epsilon_y=\frac{\sigma_y-\nu(\sigma_x+\sigma_z)}{E}$$
$$\epsilon_z=\frac{\sigma_z-\nu(\sigma_y+\sigma_x)}{E}$$

These equations can be rewritten as $$\epsilon_x=\frac{(1-2\nu)}{E}\frac{(\sigma_x+\sigma_y+\sigma_z)}{3}+\frac{(1+\nu)}{E}\frac{(2\sigma_x-\sigma_y-\sigma_z)}{3}$$
$$\epsilon_y=\frac{(1-2\nu)}{E}\frac{(\sigma_x+\sigma_y+\sigma_z)}{3}+\frac{(1+\nu)}{E}\frac{(2\sigma_y-\sigma_x-\sigma_z)}{3}$$
$$\epsilon_z=\frac{(1-2\nu)}{E}\frac{(\sigma_x+\sigma_y+\sigma_z)}{3}+\frac{(1+\nu)}{E}\frac{(2\sigma_z-\sigma_y-\sigma_x)}{3}$$
The first terms in these equations describe the linear strains resulting from the isotropic (volumetric) contribution of the loading and the second terms describe the strains resulting from the non-isotropic (but volume-preserving) portion of the loading.