compressible than front tyre for a smooth ride. - - - - Unit Properties of matter Poisson’sratio, Lateralstrain Longitudinalstrain µ= ( . ) Consider a wire of length L with diameter D . Due to applied force, wire stretches and let the increase in length be l and decrease in diameter be d .
Then µ =− =−× d D l l d D Negative sign indicates the elongation is along longitudinal and the contraction along lateral dimension. Further, notice that it is the ratio between quantities of the same dimension. So, Poisson’s ratio has no unit and no dimension (dimensionless number). The Poisson’s ratio values of some of the materials are listed in Table .
. EXAMPLE . A metal cube of side . m is subjected to a shearing force of N.
The top surface is displaced through . cm with respect to the bottom. Calculate the shear modulus of elasticity of the metal. Solution Here, L = .
m, F = N, x = . cm = . m and Area A = L = . m Therefore, η R = F x N m .
. Poisson’s ratio Longitudinal Strain Longitudinal Strain Lateral Strain Lateral Strain Figure . Lateral strain versus longitudinal strain Suppose we stretch a wire, its length increases (elongation) but its diameter decreases (contraction). Similarly, when we stretch a rubber band (elongation), it becomes noticeably thinner (contraction).
That is, deformation of the material in one direction produces deformation in another direction. To quantify this, French Physicist S.D. Poisson proposed a ratio, known as Poisson’s ratio. It is defined as the ratio of relative contraction (lateral strain) to relative expansion (longitudinal strain).
It is denoted by the symbol µ. Table . Poisson’s ratio of some of the materials Material Poisson’s ratio Rubber . Gold .
- . Copper . Stainless steel . - .
Steel . - . Cast iron . - .
Concrete . - . Glass . - .
Foam . - . Cork . - - - - Unit Properties of matter