A bar of 2500 mm² cross-sectional area is subjected to an axial load of 150 kN. The extension over a gauge length of 100 mm is 0·05 mm. If the decrease in each side is 0·00625 mm, the value of Poisson’s ratio is
- A1/5
- B1/4Correct
- C1/3
- D1/2
Solution
Concept
Poisson's ratio from strains
Formula
ν = lateral strain / longitudinal strain
Step-by-step solution
1. Side of the square section = √2500 = 50 mm. 2. Lateral strain = 0.00625 / 50 = 1.25 × 10⁻⁴. 3. Longitudinal strain = 0.05 / 100 = 5 × 10⁻⁴. 4. ν = 1.25×10⁻⁴ / 5×10⁻⁴ = 0.25 = 1/4.
Explanation
Poisson's ratio is the ratio of lateral to longitudinal strain. The side length of the square bar comes from the area.
Why the other options are wrong
1/5, 1/3 and 1/2 do not equal 0.25.
Shortcut
Side = √A; ν = (Δb/b)/(ΔL/L).
Common mistake
Using the area instead of the side as the lateral dimension.
Correct answer: B) 1/4