##### Poissons Ratio
$\nu = - \dfrac{\mathrm{lateral\ strain}}{\mathrm{longitudinal\ strain}}$

• Describe and define Poisson’s ratio.

Consider a rod under an axial tensile load P such that the material is within the elastic limit. The normal stress on x plane is $\sigma_{xx} = \dfrac{P}{A}$ and the associated longitudinal strain in the x direction can be found out from $\epsilon_x = \dfrac{\sigma_{xx}}{E}$. As the material elongates in the x direction due to the load P, it also contracts in the other two mutually perpendicular directions, i.e., y and z directions. Hence, despite the absence of normal stresses in y and z directions, strains do exist in those directions and they are called lateral strains.
The ratio between the lateral strain and the axial/longitudinal strain for a given material is always a constant within the elastic limit and this constant is referred to as Poisson’s ratio.
It is denoted by $nu$. Since the axial and lateral strains are opposite in sign, a negative sign is introduced in the definition to make $\nu$ positive. $\nu = - \dfrac{\mathrm{lateral\ strain}}{\mathrm{longitudinal\ strain}}$
Poisson’s ratio can be as low as 0.1 for concrete and as high as 0.5 for rubber. In general, it varies from 0.25 to 0.35 and for steel it is about 0.3.
Young’s modulus = E = $\dfrac{\mathrm{tensile\ stress}}{\mathrm{tensile\ strain}}$
Young’s modulus = E = $\dfrac{\mathrm{compressive\ stress}}{\mathrm{compressive\ strain}}$
Shear modulus = G = $\dfrac{\mathrm{shear\ stress}}{\mathrm{shear\ strain}}$
Bulk modulus = K = $\dfrac{\mathrm{Volumetric\ stress}}{\mathrm{volumetric\ strain}}$
K= $\dfrac{\mathrm{pressure}}{\mathrm{volumetric\ strain}}$ = $\dfrac{p}{(\delta V/ V)}$

###### Solved Example: 41-3-01 Page 131

The value of Poisson’s ratio for steel is between:

###### Solved Example: 41-3-02 Page 131

Poisson’s ratio is defined as the ratio of:

###### Solved Example: 41-3-03 Page 131

A bar of 30 mm diameter is subjected to a pull of 60 kN. The measured extension on gauge length of 200 mm is 0.1 mm and change in diameter is 0.004 mm. Calculate Poisson’s ratio.

###### Solved Example: 41-3-04 Page 131

Poisson's ratio of a material is 0.5. Percentage change in its length is 0.04%. What is the change in percentage of diameter?

###### Solved Example: 41-3-05 Page 131

Which of the following statements is NOT true?

###### Solved Example: 41-3-06 Page 131

Which of the following describes the concept of Poisson's ratio most accurately?

###### Solved Example: 41-3-07 Page 131

What is the unit of the modulus of elasticity?

###### Solved Example: 41-3-08 Page 131

Within elastic limit, the volumetric strain is proportional to the hydrostatic stress. What is the constant that relates these two quantities called?

###### Solved Example: 41-3-09 Page 131

What is another term for modulus of rigidity?

###### Solved Example: 41-3-10 Page 131

The ratio of lateral strain to the linear strain within elastic limit is known as:

###### Solved Example: 41-3-11 Page 131

Young’s modulus is defined as the ratio of:

###### Solved Example: 41-3-12 Page 131

The materials having same elastic properties in all directions are called:

###### Solved Example: 41-3-13 Page 131

The value of modulus of elasticity for mild steel is of the order of:

###### Solved Example: 41-3-14 Page 131

A metallic rod of 500 mm length and 50 mm diameter, when subjected to a tensile force of 100 kN at the ends, experiences, an increase in its length by 0.5 mm and a reduction in its diameter by 0.015 mm. The Poisson's ratio?

Solution:

Lateral strain $=\dfrac {0.015}{50}=0.003$

Longitudinal strain $=\dfrac {0.5}{500}=0.001$

Poisson's ratio $\nu =\dfrac {0.0003}{0.001}=0.3$