Poisson's ratio

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Figure 1: Rectangular specimen subject to compression, with Poisson's ratio circa 0.5

Poisson's ratio (ν), named after Siméon Poisson, is the ratio, when a sample object is stretched, of the contraction or transverse strain (perpendicular to the applied load), to the extension or axial strain (in the direction of the applied load).

When a sample cube of a material is stretched in one direction, it tends to contract (or occasionally, expand) in the other two directions perpendicular to the direction of stretch. Conversely, when a sample of material is compressed in one direction, it tends to expand (or rarely, contract) in the other two directions. This phenomenon is called the Poisson effect. Poisson's ratio ν (nu) is a measure of the Poisson effect.

The Poisson's ratio of a stable, isotropic, linear elastic material cannot be less than −1.0 nor greater than 0.5 due to the requirement that the elastic modulus, the shear modulus and bulk modulus have positive values ^[1]. Most materials have Poisson's ratio values ranging between 0.0 and 0.5. A perfectly incompressible material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5. 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.) Rubber has a Poisson ratio of nearly 0.5. Cork's Poisson ratio is close to 0: showing very little lateral expansion when compressed. Some materials, mostly polymer foams, have a negative Poisson's ratio; if these auxetic materials are stretched in one direction, they become thicker in perpendicular directions.

Assuming that the material is compressed along the axial direction:

$\nu = -\frac{\varepsilon_\mathrm{trans}}{\varepsilon_\mathrm{axial}} = -\frac{\varepsilon_\mathrm{x}}{\varepsilon_\mathrm{y}}$

where

ν

is the resulting Poisson's ratio,

$\varepsilon_\mathrm{trans}$ is transverse strain (negative for axial tension, positive for axial compression)

$\varepsilon_\mathrm{axial}$ is axial strain (positive for axial tension, negative for axial compression).

1 Cause of Poisson’s effect
2 Volumetric change
3 Width change
4 Isotropic materials
5 Orthotropic materials
6 Transversely isotropic materials
7 Poisson's ratio values for different materials
- 7.1 Negative Poisson's ratio materials
8 Applications of Poisson's effect
9 See also
10 References
11 External links

[edit] Cause of Poisson’s effect

On the molecular level, Poisson’s effect is caused by slight movements between molecules and the stretching of molecular bonds within the material lattice to accommodate the stress. When the bonds elongate in the stress direction, they shorten in the other directions. This behavior multiplied millions of times throughout the material lattice is what drives the phenomenon.

[edit] Volumetric change

The relative change of volume ΔV/V due to the stretch of the material can be calculated using a simplified formula (only for small deformations):

$\frac {\Delta V} {V} = (1-2\nu)\frac {\Delta L} {L}$

where

V

is material volume

Δ V

is material volume change

L

is original length, before stretch

Δ L

is the change of length:

Δ L = L new - L old

Derivation

A cube, made of an isotropic material, has initial volume $V$ and dimensions $L$ . An axial strain gives new dimensions to the cube: $L'$ laterally and $L t$ transversely.

Poisson's ratio gives a relationship between these new dimensions:

$\nu\dfrac{L'-L}{L}=-\dfrac{L_t-L}{L} \Rightarrow L_t=L-\nu\Delta L$

The new volume of the cube:

$V_{new}=L' L_t^2$

$V_{new}=L' (L-\nu\Delta L)^2 \Leftrightarrow V_{new}=L' L^2-2\nu\Delta L L L'+L' \nu^2\Delta L^2$

Deducting and subsequently dividing by the initial volume gives:

$\dfrac{\Delta V}{V} = \dfrac{L'}{L} - 2\nu\dfrac{\Delta L L'}{L^2} + \nu^2\dfrac{\Delta L^2 L'}{L^3}-1$

$\dfrac{\Delta V}{V} = \dfrac{\Delta L}{L} - 2\nu\dfrac{\Delta L L'}{L^2} + \nu^2\dfrac{\Delta L^2 L'}{L^3}$

$\dfrac{\Delta V}{V} = \dfrac{\Delta L}{L}(1 - 2\nu\dfrac{L'}{L} + \nu^2\dfrac{\Delta L L'}{L^2})$

Small deformation:

$\Delta L \rightarrow 0$

$\dfrac{\Delta V}{V} = (1 - 2\nu)\dfrac{\Delta L}{L}$

[edit] Width change

Figure 2: Comparison between the two formulas, one for small deformations, another for large deformations

If a rod with diameter (or width, or thickness) d and length L is subject to tension so that its length will change by ΔL then its diameter d will change by:

$\Delta d = - d \cdot \nu {{\Delta L} \over L}$

The above formula is true only in the case of small deformations; if deformations are large then the following (more precise) formula can be used:

$\Delta d = - d \cdot \left( 1 - {\left( 1 + {{\Delta L} \over L} \right)}^{-\nu} \right)$

where

d

is original diameter

Δ d

is rod diameter change

ν

is Poisson's ratio

L

is original length, before stretch

Δ L

is the change of length.

The value is negative because the diameter will decrease with increasing length.

[edit] Isotropic materials

For an isotropic material, the deformation of a material in the direction of one axis will produce a deformation of the material along the other axes in three dimensions. Thus it is possible to generalize Hooke's Law into three dimensions:

$\varepsilon_x = \frac {1}{E} \left [ \sigma_x - \nu \left ( \sigma_y + \sigma_z \right ) \right ]$

$\varepsilon_y = \frac {1}{E} \left [ \sigma_y - \nu \left ( \sigma_x + \sigma_z \right ) \right ]$

$\varepsilon_z = \frac {1}{E} \left [ \sigma_z - \nu \left ( \sigma_x + \sigma_y \right ) \right ]$

where

$\varepsilon_x$ , $\varepsilon_y$ and $\varepsilon_z$ are strain in the direction of

x

,

y

and

z

axis

σ x

,

σ y

and

σ z

are stress in the direction of

x

,

y

and

z

axis

E

is Young's modulus (the same in all directions:

x

,

y

and

z

for isotropic materials)

ν

is Poisson's ratio (the same in all directions:

x

,

y

and

z

for isotropic materials)

[edit] Orthotropic materials

For orthotropic materials such as wood, Hooke's law can be expressed in matrix form as^[2]

$\begin{bmatrix} \epsilon_{{\rm xx}} \\ \epsilon_{\rm yy} \\ \epsilon_{\rm zz} \\ 2\epsilon_{\rm yz} \\ 2\epsilon_{\rm zx} \\ 2\epsilon_{\rm xy} \end{bmatrix} = \begin{bmatrix} \tfrac{1}{E_{\rm x}} & - \tfrac{\nu_{\rm xy}}{E_{\rm x}} & - \tfrac{\nu_{\rm xz}}{E_{\rm x}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm yx}}{E_{\rm y}} & \tfrac{1}{E_{\rm y}} & - \tfrac{\nu_{\rm yz}}{E_{\rm y}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm zx}}{E_{\rm z}} & - \tfrac{\nu_{\rm zy}}{E_{\rm z}} & \tfrac{1}{E_{\rm z}} & 0 & 0 & 0 \\ 0 & 0 & 0 & \tfrac{1}{G_{\rm yz}} & 0 & 0 \\ 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm zx}} & 0 \\ 0 & 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm xy}} \\ \end{bmatrix} \begin{bmatrix} \sigma_{\rm xx} \\ \sigma_{\rm yy} \\ \sigma_{\rm zz} \\ \sigma_{\rm yz} \\ \sigma_{\rm zx} \\ \sigma_{\rm xy} \end{bmatrix}$

where

${E}_{\rm i}\,$ is the Young's modulus along axis

i

$G_{\rm ij}\,$ is the shear modulus in direction

j

on the plane whose normal is in direction

i

$\nu_{\rm ij}\,$ is the Poisson's ratio that corresponds to a contraction in direction

j

when an extension is applied in direction

i

.

The Poisson's ratio of an orthotropic material is different in each direction (x, y and z). However, the symmetry of the stress and strain tensors implies that not all the six Poisson's ratios in the equation are independent. There are only nine independent material properties; three elastic moduli, three shear moduli, and three Poisson's ratios. The remaining three Poisson's ratios can be obtained from the relations

$\frac{\nu_{\rm yx}}{E_{\rm y}} = \frac{\nu_{\rm xy}}{E_{\rm x}}~, \qquad \frac{\nu_{\rm zx}}{E_{\rm z}} = \frac{\nu_{\rm xz}}{E_{\rm x}}~, \qquad \frac{\nu_{\rm yz}}{E_{\rm y}} = \frac{\nu_{\rm zy}}{E_{\rm z}}$

From the above relations we can see that if $E x > E y$ then $ν xy > ν y x$ . The larger Poisson's ratio (in this case $ν xy$ ) is called the major Poisson's ratio while the smaller one (in this case $ν yx$ ) is called the minor Poisson's ratio. We can find similar relations between the other Poisson's ratios.

[edit] Transversely isotropic materials

Transversely isotropic materials have a plane of symmetry in which the elastic properties are isotropic. If we assume that this plane of symmetry is $y - z$ , then Hookes's law takes the form^[3]

$\begin{bmatrix} \epsilon_{{\rm xx}} \\ \epsilon_{\rm yy} \\ \epsilon_{\rm zz} \\ 2\epsilon_{\rm yz} \\ 2\epsilon_{\rm zx} \\ 2\epsilon_{\rm xy} \end{bmatrix} = \begin{bmatrix} \tfrac{1}{E_{\rm x}} & - \tfrac{\nu_{\rm xy}}{E_{\rm x}} & - \tfrac{\nu_{\rm xy}}{E_{\rm x}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm yx}}{E_{\rm y}} & \tfrac{1}{E_{\rm y}} & - \tfrac{\nu_{\rm yz}}{E_{\rm y}} & 0 & 0 & 0 \\ -\tfrac{\nu_{\rm yx}}{E_{\rm y}} & - \tfrac{\nu_{\rm zy}}{E_{\rm y}} & \tfrac{1}{E_{\rm y}} & 0 & 0 & 0 \\ 0 & 0 & 0 & \tfrac{1}{G_{\rm yz}} & 0 & 0 \\ 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm yx}} & 0 \\ 0 & 0 & 0 & 0 & 0 & \tfrac{1}{G_{\rm xy}} \\ \end{bmatrix} \begin{bmatrix} \sigma_{\rm xx} \\ \sigma_{\rm yy} \\ \sigma_{\rm zz} \\ \sigma_{\rm yz} \\ \sigma_{\rm zx} \\ \sigma_{\rm xy} \end{bmatrix}$

where we have used the plane of symmetry $y - z$ to reduce the number of constants, i.e., $E_y = E_z,~ \nu_{xy} = \nu_{xz},~ \nu_{yx} = \nu_{zx}$ .

The symmetry of the stress and strain tensors implies that

$\cfrac{\nu_{\rm xy}}{E_{\rm x}} = \cfrac{\nu_{\rm yx}}{E_{\rm y}} ~,~~ \nu_{\rm yz} = \nu_{\rm zy} ~.$

This leaves us with six independent constants $E x, E y, G xy, G yz,ν xy,ν yz$ . However, transverse isotropy gives rise to a further constraint bewtween $G yz$ and $E y,ν yz$ which is

$G_{\rm yz} = \cfrac{E_{\rm y}}{2(1+\nu_{\rm yz})} ~.$

Therefore, there are five independent elastic material properties two of which are Poisson's ratios. For the assumed plane of symmetry, the larger of $ν xy$ and $ν yx$ is the major Poisson's ratio. The other major and minor Poisson's ratios are equal.

[edit] Poisson's ratio values for different materials

Influences of selected glass component additions on Poisson's ratio of a specific base glass.^[4]

material	poisson's ratio
rubber	~ 0.50
gold	0.42
saturated clay	0.40-0.50
magnesium	0.35
titanium	0.34
copper	0.33
aluminium-alloy	0.33
clay	0.30-0.45
stainless steel	0.30-0.31
steel	0.27-0.30
cast iron	0.21-0.26
sand	0.20-0.45
concrete	0.20
glass	0.18-0.3
foam	0.10 to 0.40
cork	~ 0.00
auxetics	negative

material	plane of symmetry	$ν xy$	$ν yx$	$ν yz$	$ν zy$	$ν zx$	$ν xz$
Nomex honeycomb core	$x - y$ , $x$ =ribbon direction	0.49	0.69	0.01	2.75	3.88	0.01
glass fiber-epoxy resin	$x - y$	0.29	0.29	0.32	0.06	0.06	0.32

[edit] Negative Poisson's ratio materials

Some materials known as auxetic materials display a negative Poisson’s ratio. When subjected to positive strain in a longitudinal axis, the transverse strain in the material will actually be positive (i.e. it would increase the cross sectional area). For these materials, it is usually due to uniquely oriented, hinged molecular bonds. In order for these bonds to stretch in the longitudinal direction, the hinges must ‘open’ in the transverse direction, effectively exhibiting a positive strain.^[5]

[edit] Applications of Poisson's effect

One area in which Poisson's effect has a considerable influence is in pressurized pipe flow. When the air or liquid inside a pipe is highly pressurized it exerts a uniform force on the inside of the pipe, resulting in a radial stress within the pipe material. Due to Poisson's effect, this radial stress will cause the pipe to slightly increase in diameter and decrease in length. The decrease in length, in particular, can have a noticeable effect upon the pipe joints, as the effect will accumulate for each section of pipe joined in series. A restrained joint may be pulled apart or otherwise prone to failure.^[6]

Another area of application for Poisson's effect is in the realm of structural geology. Rocks, just as most materials, are subject to Poisson's effect while under stress and strain. In a geological timescale, excessive erosion or sedimentation of Earth's crust can either create or remove large vertical stresses upon the underlying rock. This rock will expand or contract in the vertical direction as a direct result of the applied stress, and it will also deform in the horizontal direction as a result of Poisson's effect. This change in strain in the horizontal direction can affect or form joints and dormant stresses in the rock.^[7]

[edit] See also

[edit] References

^ H. GERCEK; “Poisson's ratio values for rocks”; International Journal of Rock Mechanics and Mining Sciences; Elsevier; January 2007; 44 (1): pp. 1–13.
^ Boresi, A. P, Schmidt, R. J. and Sidebottom, O. M., 1993, Advanced Mechanics of Materials, Wiley.
^ Tan, S. C., 1994, Stress Concentrations in Laminated Composites, Technomic Publishing Company, Lancaster, PA.
^ Poisson's ratio calculation of glasses
^ Negative Poisson's ratio
^ http://www.cpchem.com/hb/getdocanon.asp?doc=135&lib=CPC-Portal
^ http://www.geosc.psu.edu/~engelder/geosc465/lect18.rtf

[edit] External links

Conversion formulas
Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these, thus given any two, any other of the elastic moduli can be calculated according to these formulas.
	$(\lambda,\,G)$	$(E,\,G)$	$(K,\,\lambda)$	$(K,\,G)$	$(\lambda,\,\nu)$	$(G,\,\nu)$	$(E,\,\nu)$	$(K,\, \nu)$	$(K,\,E)$	$(M,\,G)$
$K=\,$	$\lambda+ \frac{2G}{3}$	$\frac{EG}{3(3G-E)}$			$\lambda\frac{1+\nu}{3\nu}$	$\frac{2G(1+\nu)}{3(1-2\nu)}$	$\frac{E}{3(1-2\nu)}$			$M - \frac{4G}{3}$
$E=\,$	$G\frac{3\lambda + 2G}{\lambda + G}$		$9K\frac{K-\lambda}{3K-\lambda}$	$\frac{9KG}{3K+G}$	$\frac{\lambda(1+\nu)(1-2\nu)}{\nu}$	$2G(1+\nu)\,$		$3K(1-2\nu)\,$		$G\frac{3M-4G}{M-G}$
$\lambda=\,$		$G\frac{E-2G}{3G-E}$		$K-\frac{2G}{3}$		$\frac{2 G \nu}{1-2\nu}$	$\frac{E\nu}{(1+\nu)(1-2\nu)}$	$\frac{3K\nu}{1+\nu}$	$\frac{3K(3K-E)}{9K-E}$	$M - 2G\,$
$G=\,$			$3\frac{K-\lambda}{2}$		$\lambda\frac{1-2\nu}{2\nu}$		$\frac{E}{2(1+\nu)}$	$3K\frac{1-2\nu}{2(1+\nu)}$	$\frac{3KE}{9K-E}$
$\nu=\,$	$\frac{\lambda}{2(\lambda + G)}$	$\frac{E}{2G}-1$	$\frac{\lambda}{3K-\lambda}$	$\frac{3K-2G}{2(3K+G)}$					$\frac{3K-E}{6K}$	$\frac{M - 2G}{2M - 2G}$
$M=\,$	$\lambda+2G\,$	$G\frac{4G-E}{3G-E}$	$3K-2\lambda\,$	$K+\frac{4G}{3}$	$\lambda \frac{1-\nu}{\nu}$	$G\frac{2-2\nu}{1-2\nu}$	$E\frac{1-\nu}{(1+\nu)(1-2\nu)}$	$3K\frac{1-\nu}{1+\nu}$	$3K\frac{3K+E}{9K-E}$