Poissons Ratio: Overview, Questions, Preparation

In the study of solid mechanics, Poisson's ratio is defined as Poisson's distribution measure. It is the ratio of change in width per unit of a given material, to the change in its length per unit length. 

Like the Poisson distribution, its ratio is also named after Simeon Poisson, who used molecular theory to explain the phenomena. It is represented using the Greek letter, 'ν'. You can also express Poisson's ratio as the amount of transversal expansion to axial compression for small values of these changes. 

Poisson's ratio is used to express the correlation between axial and lateral strain. Since it is used to measure minor changes in the former, you should always use the right equipment. The properties of Poisson's ratio are as follows - 

•  Tensile deformation is considered to be positive 
• The compressive deformation in the Poisson's ratio is considered as negative

To understand the formula in-depth, we will take the example of a rubber piece in the cuboid's usual shape. In such a scenario, the rubber will compress in the middle. If you take original length & breadth as 'L' and 'B' and pull it longitudinally, it compresses laterally. Summarising it, we can say that length increases by amount dL & breadth too increases by dB. The formula for the same is - 

ε = ΔL/L

The Poisson Effect demonstrates that an object tends to expand perpendicularly to the direction of compression. Conversely, if the object is being stretched rather than being compressed, it contracts in the directions transversal to the direction from where it's stretched. 

Poisson's ratio differs from material to material & depends on how it reacts to the lateral strain. The values for different materials are mentioned below - 

• Rubber ν = 0.499
• Cork ν = 0 
• Gold ν = 0.44

• Copper ν = 0.33

• Clay ν = 0.45
• Foam ν = 0.10 – 0.40
• Concrete ν = 0.20
• Magnesium ν = 0.35

Poisson's ratio is part of the 'Mechanical Properties of Solids' chapter in Class 11th. This chapter contributes 20 marks to the CBSE Class 11th Physics Paper when it comes to overall weightage. 

It is, therefore, quite essential to prepare and understand the concepts taught in this chapter in-depth.

Solved Examples 

1. A steel bar has a length of 100 mm & measures 50 mm in terms of width. On applying 50 newton force, its length increases to 102 mm, what’s the change in width? 

Ans. Considering that steel bar’s width reduced by dW; we know - 

                               ν = 0.3 

                             Longitudinal strain = (102-100)/100 = 2/100 = 0.02 

                                  As per formula - 

1. 3 = (dW/50)/0.02 

                                      (dW/50) = 0.3x0.02 = 0.006 

                                     dW = 0.3 

                            Hence, the metal bar’s width decreases by 0.3 mm. 

• Calculate Poisson’s ratio when lateral strain is 90 & axial strain is 5. 

Ans. v = lateral strain/axial strain 

εl = 90

εa  = 5 

v = 90/5 = 18 

Hence, Poisson ratio = 18 

• Calculate lateral strain when axial strain is 40 & Poisson’s ratio is given as 8. 

Ans. v = εl/εa

8 = x/40 

X = 40x8 = 320

A: In general, cold temperatures lead to a decrease in both strains & hot temperatures, leading to an increase in them. Hence, the net effect on Poisson's ratios is relatively smaller.

Q: Does Poisson’s ratio remain constant?

A: Yes, Poisson’s ratio remains constant within elastic limits for materials. 

Q: What is the Poisson ratio for magnesium?

A: Poisson’s ratio for magnesium is ν = 0.35

Q: What is the formula of Poisson ratio?

A: The formula is - ν = Transverse Strain/Longitudinal Strain

Q: What is the unit of Poisson’s ratio?

A: Poisson’s ratio is a unitless scalar quantity.