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Poisson’s ratio

 Poisson’s ratio 

It is the ratio of lateral strain to longitudinal strain and it is constant with in the elastic limit of the material. 
Poisson's Ratio is generally denoted by 'μ ' or u

The ratio is named after the French mathematician and physicist Siméon Poisson.
Note:
The tensile longitudinal stress produces compressive lateral strains or the compressive longitudinal stress produces tensile lateral strains.
(+ve Longitudinal strains produce -ve lateral strains or vice versa)  
 
Let us consider a rectangular bar of cross section (bxd) and length l and it is subjected to a axial load acts in the direction of length of the bar.
The longitudinal tensile strain in the bar 
e longitudinal = δl / l
Lateral compressive strains in the directions of breadth b and  depth d of bar
elateral = δb/ b and 
elateral  =  δd /d
where δl = Change in length
δb = Change in width =b' - b
δd = Change in depth = d' - d
The poisson's Ratio
 
                      m = elateral / e longitudinal
Lateral strain ,elateral = m.elongitudinal
Note:
Perfectly incompressible isotropic material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5.
Example: Rubber has a Poisson ratio of nearly 0.5. and Cork's Poisson ratio is close to zero. Most of the metals it is in between 0.25 to 0.33

Examples:

1.A metal rod 20 mm diameter and 2 m long is subjected to a tensile force of 60 kN, it showed and elongation of 2 mm and reduction of diameter by 0.006 mm. Calculate the Poisson's ratio.
Solution:
The longitudinal strain = e longitudinal = δl / l
                                    =  2mm/ 2000 = 0.001
The Lateral Strain = elateral  =  δd /d
                              = 0.006mm/20mm =0.0003
The poisson's Ratio = m = elateral / e longitudinal
                                                = 0.0003/0.001 = 0.3 Ans.

Assignment :
1. A metal bar 50mmx50mm in section is subjected to an axial compressive load of 500 kN. If the contraction of a 200mm gauge length was found to be 0.5 mm and the increase in thickness 0.04 mm, find Poisson’s ratio.
(Ans: Poisson's ratio=0.32)
2.A steel rod 4 m long and 20mm diameter is subjected to an axial tensile load of 45 kN. Find the change in length and diameter of the rod. If
E= 200 GPa, and m = 0.25
(Ans: Change in length=2.86mm, change in diameter =0.003575mm)
3. A steel rod 3m long and 25mm diameter is subjected to an axial tensile load of 60 kN. Calculate the change in length and diameter of rod. Take E=210 Gpa and m= 0.28.
(Ans: Change in length=1.75mm, change in diameter =0.0041mm)
4.A steel bar 1.2 m long, 40mm wide and 20 mm thick is subjected to an axial tensile load of 50 kN in the direction of its length. Find the change in length and thickness of the bar. Take E=200 Gpa and Poisson's ratio =0.26.
(Ans: Change in length=0.375mm, change in thickness=1.62510-3mm)
5. A metal bar 40mm40mm section, is subjected to an axial compressive load of 480 kN. The contraction of a 200 mm gauge length is found to be 0.4 mm and the increase in thickness 0.04 mm. Find Young’s Modulus and Poisson’s ratio.
(Ans: E= 150 GPa, m =0.2)

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