Problems on Poisson’s Ratio

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Example – 1:

  • When a brass rod of diameter 6 mm is subjected to a tension of 5 × 103 N, the diameter changes by 3.6 × 10-4 cm. Calculate the longitudinal strain and Poisson’s ratio for brass given that Y for the brass is 9 × 1010 N/m².
  • Solution:
  • Given: Diameter of rod = D = 6 mm, Radius of wire = 6/2 = 3 mm = 3 × 10-3 m, Load F = 5 × 103 N, Change in diameter = d = 3.6 × 10-4 cm = 3.6 × 10-6 m, Y for the brass is 9 × 1010 N/m².
  • To Find: Longitudinal strain = ?, Poisson’s ratio = ?,

Y = Longitudinal Stress /Longitudinal Strain

∴  Y = F / (A × Longitudinal Strain)

∴  Longitudinal Strain = F / (A × Y)

∴  Longitudinal Strain = F / (π r² × Y)



∴  Longitudinal Strain =  5 × 103 / (3.142 × (3 × 10-3)² × 9 × 1010 )

∴  Longitudinal Strain =  5 × 103 / (3.142 × 9 × 10-6 × 9 × 1010 )

∴  Longitudinal Strain =  1.96 × 10-3

Now, Lateral strain = d /D =   (3.6 × 10-6)/(6 × 10-3) = 6 × 10-4



Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Poisson’s ratio = (6 × 10-4) /(1.96 × 10-3) = 0.31

Ans: Longitudinal strain is  1.96 × 10-3 and poisson’s ratio is 0.31.

Example – 2:

  • A metal wire of length 1.5 m is loaded and an elongation of 2 mm is produced. If the diameter of the wire is 1 mm, find the change in the diameter of the wire when elongated. σ = 0.24.
  • Solution:
  • Given: Original length of wire = L = 1.5 m, Elongation in wire = 2 mm = 2 × 10-3 m, Diameter of wire = D = 1 mm, Poisson,s ratio = σ = 0.24.
  • To Find: Change in diameter = d =?,

Longitudinal strain = l/L = (2 × 10-3)/1.5 = 1.33 × 10-3

Poisson’s ratio = Lateral strain / Longitudinal strain



∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.24 × 1.33 × 10-3    3.2 × 10-4

Lateral strain = d / D

∴  d = Lateral strain × D = 3.2 × 10-4  × 1 × 10-3  =  3.2 × 10-7 m

Ans : The change in diameter is 3.2 × 10-7 m



Example – 3:

  • A metallic wire ( Y = 20 × 1010 N/m². and σ = 0.26) of length 3 m and diameter 0.1 cm is stretched  by a load of 10 kg. Calculate the decrease in diameter of the wire.
  • Solution:
  • Given: Original length of wire = L = 3 m, Diameter of wire = D = 0.1 cm = 0.1 × 10-2 m = 1 × 10-3 m, Radius of wire = r  = 0.1/2 = 0.05 cm = 0.05 × 10-2 m = 5 × 10-4 m,, Stretching load = 10 kg = 10 x 9.8 N, Young’s modulus of elasticity = Y = 20 × 1010 N/m². and Poisson’s ratio = σ = 0.26
  • To Find: Decrease in diameter = d =?,

Y = Longitudinal Stress /Longitudinal Strain

∴  Y = F / (A × Longitudinal Strain)

∴  Longitudinal Strain = F / (A × Y)

∴  Longitudinal Strain = F / (π r² × Y)

∴  Longitudinal Strain =  (10 x 9.8) / (3.142 × (5 × 10-4)² × 20 × 1010 )



∴  Longitudinal Strain =  (10 x 9.8) / (3.142 × 25 × 10-8 × 20 × 1010 )

∴  Longitudinal Strain =  6.24 × 10-4

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain



∴ Lateral strain = 0.26 × 6.24 × 10-4    1.62 × 10-4

Lateral strain = d / D

∴  d = Lateral strain × D = 1.62 × 10-4  × 1 × 10-3  =  1.62 × 10-7 m

Ans : The decrease in diameter is 1.62 × 10-7 m

Example – 4:

  • A copper wire 3m long and 1 mm² in cross-section is fixed at one end and a weight of 10 kg is attached at the free end. If Y for copper is 12.5 × 1010 N/m² and σ = 0.25 find the extension, lateral strain and the lateral compression produced in the wire.
  • Solution:
  • Given: Original length of wire = L = 3 m, Area of cross-section of wire = A = 1 mm² = 1 × 10-6 m², Stretching load = 10 kg = 10 × 9.8 N, Young’s modulus of elasticity = Y = 12.5 × 1010 N/m² . and Poisson’s ratio = σ = 0.25
  • To Find: Extension = l = ?, Lateral strain = ?, Lateral compression = ?

Y = Longitudinal Stress /Longitudinal Strain

∴  Y = F / (A × Longitudinal Strain)



∴  Longitudinal Strain = F / (A × Y)

∴  Longitudinal Strain = 10 × 9.8 / (1 × 10-6 × 12.5 × 1010)

∴  Longitudinal Strain = 10 × 9.8 / (1 × 10-6 × 12.5 × 1010)

∴  Longitudinal Strain =  7.84 × 10-4



Now, Longitudinal Strain = l/L

∴  l = Longitudinal strain × L

∴  l = 7.84 × 10-4 × 3 =2.352 × 10-3  m = 2.352 mm

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.25 × 7.84 × 10-4    1.96 × 10-4



Area of cross-section = A = 1 × 10-6

∴   π r²  = 1 × 10-6

∴   r²  = 1 × 10-6/ π = = 1 × 10-6/ 3.142

∴   r²  = 0.3183 × 10-6



∴   r  = 5.64 × 10-4  m

Diameter = D = 2r = 2 × 5.64 × 10-4  m = 11.28 × 10-4  m

Now, Lateral strain = d / D

∴  d = Lateral strain × D = 1.96 × 10-4  × 11.28 × 10-4 =  2.21 × 10-7 m

Ans :  Elongation = 2.352 mm, Lateral strain = 1.96 × 10-4, Lateral compression = 2.21 × 10-7 m.



Example – 5:

  • A wire of diameter 2 mm and length 5 m is stretched by a load of 10 kg. Find the extension produced in the wire if Y = 12 × 1010 N/m². If σ = 0.35 for the material of the wire, find the lateral contraction.
  • Solution:
  • Given: Original length of wire = L = 5 m, Diameter of wire = D = 2 mm = 2 × 10-3 m , Radius of wire = 2/2 = 1mm = 1 × 10-3 m, Stretching load = 10 kg = 10 × 9.8 N, Young’s modulus of elasticity = Y = 12 × 1010 N/m² . and Poisson’s ratio = σ = 0.35
  • To Find: Lateral contraction = ?

Y = Longitudinal Stress /Longitudinal Strain

∴  Y = F / (A × Longitudinal Strain)

∴  Longitudinal Strain = F / (A × Y)

∴  Longitudinal Strain = F / (π r² × Y)

∴  Longitudinal Strain =  (10 x 9.8) / (3.142 × (1 × 10-3)² × 12 × 1010 )



∴  Longitudinal Strain =  (10 x 9.8) / (3.142 × 1 × 10-6 × 12 × 1010 )

∴  Longitudinal Strain =  2.6 × 10-4

Now, Longitudinal Strain = l/L

∴  l = Longitudinal strain × L

∴  l = 2.6 × 10-4  × 5 = 1.3 × 10-3  m = 1.3 mm

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.35 ×2.6 × 10-4   9.1 × 10-5

Now, Lateral strain = d / D

∴  d = Lateral strain × D = 9.1 × 10-5  × 2 × 10-3 =  1.82 × 10-7 m



Ans :  Elongation = 1.3 mm, Lateral contraction = 1.82 × 10-7 m

Example – 6:

  • Find the longitudinal stress to be studied to a wire to decrease its diameter uniformly by 1%. Poisson’s ratio = 0.25, Young’s modulus = 2 × 1011N/m² .
  • Solution:
  • Given: Lateral strain = 1 % = 1 × 10-2, Young’s modulus of elasticity = Y = 2 × 1011 N/m² . and Poisson’s ratio = σ = 0.25
  • To Find: Longitudinal stress = ?

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Longitudinal strain =Lateral strain / Poisson’s ratio

∴ Longitudinal strain = 1 × 10-2 / 0.25 = 4 × 10-2

Y = Longitudinal Stress /Longitudinal Strain

∴ Longitudinal Stress = Longitudinal Strain × Y

∴ Longitudinal Stress = 4 × 10-2 × 2 × 1011  =  8 × 10N/m² .

Ans : Longitudinal stress = 8 × 10N/m²

Example – 7:

  • A copper wire 3 m long is stretched to increase its length by 0.3 cm. Find the lateral strain produced in the wire. If Poisson’s ratio for copper is 0.26.
  • Solution:
  • Given: Length of wire = L = 3m, Increase in length = l = 0.3 cm = 0.3 × 10-2 m = 3 × 10-3 m, Poisson’s ratio = σ = 0.26
  • To Find: Lateral strain = ?

Longitudinal strain = l/L = (3 × 10-3)/ 3 = 10-3

Poisson’s ratio = Lateral strain / Longitudinal strain



∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.36 × 10-3   3.6 × 10-4

Example – 8:

  • A steel wire having cross-sectional area 1 mm² is stretched by 10 N. Find the lateral strain produced in the wire. Young’s modulus for steel is 2 × 1011 N/m² and Poisson’s ratio is 0.291.
  • Solution:
  • Given: Area of cross-section = 1 mm² = 1 × 10-6  m², Stretching Load = 10 N,  Young’s modulus for steel= Y =  2 × 1011 N/m², Poisson’s ratio = σ = 0.291
  • To Find: Lateral strain = ?

Y = Longitudinal Stress / Longitudinal Strain

∴  Y = F  /(A  × Longitudinal  Strain)

∴  Longitudinal strain = F  /(A  × Y)

∴  Longitudinal strain = 10  /(1 × 10-6  × 2 × 1011) = 5 × 10-5

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.291 × 5 × 10-5   = 1.455 × 10-5

Example – 9:

  • A load 1 kg produces a certain extension in the wire of length 3 m and radius 5 × 10-4  m. How much will be the lateral strain produced in the wire? Given Y = 7.48 × 1010 N/m², σ = 0.291.
  • Solution:
  • Given: Load attached = F = 1 kg = 1 × 9.8 NLength of wire = L = 3 m, Radius of cross-section = r = 5 × 10-4  m cross-section = 1 mm² = 1 × 10-6  m², Stretching Load = 10 N,  Young’s modulus = Y =  7.48 × 1010  N/m²,  Poisson’s ratio = σ = 0.291
  • To Find: Lateral strain = ?

Y = Longitudinal Stress / Longitudinal Strain



∴  Y = F  /(A  × Longitudinal  Strain)

∴  Y = F  /(π r²  × Longitudinal  Strain)

∴  Longitudinal strain = (1 × 9.8)  /(3.142 × (5 × 10-4)²  ×   7.48 × 1010)

∴  Longitudinal strain = (1 × 9.8)  /(3.142 × 25 × 10-8 ×   7.48 × 1010)

∴  Longitudinal strain = 1.67   × 10-4

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.291 × 1.67   × 10-4   = 4.86 × 10-5

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