Hey student’s we know that, Hooke’s law according to the stress is directly proportional to strain. So that when load is attached to free end of wire, length of wire increases, which produces longitudinal strain in the wire along the direction of applied force. Sir Simon Poisson observed that along with the increase in length of wire towards the direction of force, there is decrease in diameter of wire is also takes place which produces the strain in perpendicular direction to applied force.
Let’s define the concept and derive its formula….!
Consider the uniform wire is suspended from rigid support and subjected to external load of magnitude, F= mg.
Let L and D be the original length and diameter of the wire. When it is loaded, let ∆l be the change in its length and ∆d be the corresponding change in its diameter.
Then the strain in the direction of applied force i.e. longitudinal strain is given by,
∴ Longitudinal strain = Change in length/Original length
∴ Longitudinal strain = ∆l/L
Along with this, the decrease in diameter gives the lateral strain given as,
∴ Lateral strain = Change in length/Original length
∴ Lateral strain = ∆d/D
It was observed that the lateral strain is directly proportional to longitudinal strain.
∴Lateral strain α Longitudinal strain
∴Lateral strain =σ Longitudinal strain
Where σ is constant known as Poisson’s ratio
The ratio of the lateral strain to the longitudinal strain is called Poisson’s ratio (σ).
Some important points to note……!
- For homogeneous isotropic material -1 ≤ σ ≤ 0.5.
- In actual practice σ is always positive & for metal σ = 0.3 0.2 ≤ σ ≤ 0.4 for rubber 0.5.
- Poisson’s ratio is pure number hence it is unit less, dimensionless quantity.
Let’s go in more detail with some numerical….!
Ex:1) Find the value of lateral strain if Poisson’s ratio for the metal wire of length 2 m undergoes elongation of 2 mm is 0.23.
Solution:
Here, L = 2 , ∆l = 2 mm = 2× 10-3 m, σ = 0.23
We have,