Poisson’s Ratio

When a rubber band is stretched, it becomes considerably thinner, which is a familiar observation. The strain and stress that we utilize in the direction of the stretching force make up Poisson’s Ratio. In addition, it has to do with an object’s tensile strength. In the direction of the stretching force, Poisson’s ratio relates to the transverse or lateral shrinkage strain to longitudinal extension strain. The tensile deformation is considered to be positive and compressive deformation to be negative.

Poisson’s ratio is the inverse of the ratio of transverse strain to lateral or axial strain. It is the ratio of the amount of transversal expansion to the amount of axial compression for small values of these changes, and it is named after Siméon Poisson and denoted by the Greek symbol ‘nu.’

Poisson’s Ratio

Poisson’s ratio is the proportion of a material’s change in width per unit width to its change in length per unit length as a result of strain.

Furthermore, the Poisson’s Ratio consists of a negative sign, resulting in a positive ratio for normal materials. It’s also known as Poisson Ratio or Poisson Coefficient. In addition, the lower case Greek letter nu, ν, is commonly used to represent the ratio.

This is where the strain is defined in its basic form. It is the ratio of change in dimension to the original dimension. For a rectangular section of rubber having length L and breadth B, the following strains are:

• Lateral Strain (ε_t) is defined as,

ε_t = -ΔB / B

• While the longitudinal Strain (ε_l) is defined as, 

ε_l  = ΔL / L

Here, Δ is the change in dimensions.

The Poisson’s Ratio formula is as follows:

ν = –ε_t  / ε_l

where, ε_t is the lateral or transverse strain, ε_l is longitudinal or axial strain & ν is the Poisson’s ratio.

The negative sign in the formula gives the positive value of the ratio.

Poisson Effect

The Poisson effect is a phenomenon in which a material expands in directions perpendicular to the direction of compression. Poisson’s ratio is a measure of this phenomenon. When a material is stretched rather than crushed, it tends to contract in directions that are transverse to the stretching direction.

Poisson’s ratio values for different materials

With the application of force on a body, stress and strain relationship can be formed.

• It is a scalar and unit less quantity.
• It is positive for tensile deformation or anisotropic materials.
• It is negative for compressive deformation or isometric materials.

Even though the longitudinal strain is positive, the negative Poisson ratio indicates that the material will exhibit positive strain in the transverse direction.

The range of its value lies between -1.0 to +0.5. However, the value of Poisson’s ratio for most materials is between 0 and 0.5.

For plastics, the Poisson’s Ratio is in the range of 0 to 0.5. When the Poisson’s Ratio is 0, there is no reduction in diameter or, to put it another way, no lateral contraction occurs when the material is elongated, but the density decreases.

When the diameter of the material drops during the elongation process or when the material is elastomeric, a value of 0.5 implies that the volume of the material or item will remain the same or constant.

The following table shows the various Poisson’s Ratios for various materials.

Material	Poisson’s Ratio
Rubber	0.49
Gold	0.43
Clay	0.37
Copper	0.33
Aluminum	0.32
Cast Iron	0.24
Concrete	0.2
Cork	0

Poisson’s Ratio is usually positive since most common materials get narrower in the opposite or cross direction when stretched. Most materials resist changes in volume, as defined by the bulk modulus K or also known as B, more than changes in shape, as determined by the shear modulus G. The shape distortion also causes the interatomic connections to realign.

Sample Problems

Problem 1: The longitudinal strain for a wire is 0.02 and its Poisson ratio is 0.6. Find the lateral strain in the wire.

Solution:

Given:

Longitudinal strain of wire = 0.02

Poisson ratio = 0.6

The Poisson’s Ratio formula is as follows:

ν = lateral strain/longitudinal strain

Substitute the given values to find the lateral strain.

0.6 = Lateral strain / 0.02

Lateral strain = 0.012

Hence, the lateral strain in the wire is 0.012.

Problem 2: What is the maximum and minimum values of Poisson’s ratio for a metal?

Solution:

The Poisson’s Ratio formula is as follows:

ν = Lateral strain/longitudinal strain

It is always positive because if we apply force in longitudinal strain, lateral strain always decreases for metals. It lies between 0 to 0.5.

Problem 3: Is Poisson’s ratio affected by temperature?

Solution:

In general, lower temperatures reduce both horizontal and vertical strain, while higher temperatures increase both horizontal and vertical strain. As a result, the net effect on Poisson’s Ratio is negligible because both horizontal and vertical strain change by the same amount.

Problem 4: A 2.0 m long metal wire is loaded, resulting in a 4 mm elongation. Find the change in diameter of wire when elongated if the diameter of wire is 1.5 mm and the Poisson’s ratio of wire is 0.24.

Solution:

Given:

Length of wire, L is 2.0 m.

Change in length, ΔL is 4 mm = 0.004 m

Diameter of wire, D is 1.5 mm.

Poisson’s ratio, ν is 0.24.

The longitudinal strain in the wire is given as:

Longitudinal strain = ΔL/L

                                  = 0.004/2.0

                                  = 0.002

The Poisson’s Ratio formula is as follows:

ν = Lateral strain/longitudinal strain

Substitute the given values to find the lateral strain.

0.24 = lateral strain / 0.002

Lateral strain = 0.00048

The lateral strain in a wire is given as:

Lateral strain = ΔD / D

          0.00048 = ΔD / 1.5 mm

                 ΔD = 0.00072 mm

Hence, the change in diameter of the wire is 0.00072 mm.

Problem 5: What if a material’s Poisson’s ratio is zero?

Solution:

A Poisson’s ratio of 0 indicates that the material does not deform in either the lateral or axial directions in response to the application of force. Cork is an example of a material with a Poisson’s ratio of nearly 0 and no deformation under stress. Cork is applied as a seal in bottle stoppers because it expands and contracts under stress, protecting the substance inside.

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