Consider a wire having initial diameter of 2 mm. Poisson’s ratio for material is 0.5. The longitudinal strain produced in wire is 0.1%. The final diameter of wire is:

Consider a wire having initial diameter of 2 mm. Poisson’s ratio for

1.	Consider a wire having initial diameter of 2 mm. Poisson’s ratio for material is 0.5. The longitudinal strain produced in wire is 0.1%. The final diameter of wire is:
Answer» nitial diameter, D = 2 mmPoisson's Ratio σ = 0.5Longitudinal Strain, Δl/l = 0.1 % = 0.1 /100 = 0.001To Find :FINAL DiameterCalculation:- POISSON's ratio is given asσ = LATERAL strain / Longitudinal strain ⇒ Lateral Strain = σ × Longitudinal strain⇒ ΔD / D = σ × Δl/l ⇒ ΔD = σ × Δl/l × D ⇒ ΔD = 0.5 × 0.001 × 2⇒ ΔD = 0.001 MM- Final diameter = Initiat diameter - change in diameter⇒ D' = D - ΔD⇒ D' = 2 - 0.001⇒ D' = 1.999 mm- So the final diameter will be 1.999 mm

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Consider a wire having initial diameter of 2 mm. Poisson’s ratio for material is 0.5. The longitudinal strain produced in wire is 0.1%. The final diameter of wire is:​

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Consider a wire having initial diameter of 2 mm. Poisson’s ratio for material is 0.5. The longitudinal strain produced in wire is 0.1%. The final diameter of wire is: