Question:
A cylindrical metal component of 10 mm diameter is subjected to uniform uniaxial tension during operation. It was observed that a force of 14 kN produces a uniform reduction of \(3 \times 10^{-3}\) mm in diameter. Assume that material behaviour is homogeneous, isotropic, and linear elastic.
If its Young's modulus is 150 GPa, the Poisson's ratio of the material is ______ (rounded off to two decimal places).
Note: Assume \(\pi = 3.14\).
A cylindrical metal component of 10 mm diameter is subjected to uniform uniaxial tension during operation. It was observed that a force of 14 kN produces a uniform reduction of \(3 \times 10^{-3}\) mm in diameter. Assume that material behaviour is homogeneous, isotropic, and linear elastic.
If its Young's modulus is 150 GPa, the Poisson's ratio of the material is ______ (rounded off to two decimal places).
Note: Assume \(\pi = 3.14\).
If its Young's modulus is 150 GPa, the Poisson's ratio of the material is ______ (rounded off to two decimal places).
Note: Assume \(\pi = 3.14\).
Show Hint
Find axial stress from force and area, divide by E to get axial strain, then compare it with the diametral strain (change in diameter over diameter).
Updated On: Aug 5, 2026
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Verified By Collegedunia
Correct Answer: 0.25
Solution and Explanation
Step 1: Understanding the Question:
Poisson's ratio connects the lateral strain, the change in diameter, to the longitudinal strain, the change along the axis of pull, for a bar under simple tension, so we need to find both strains separately.
Step 2: Key Formula:
\[ \nu = -\frac{\epsilon_{lateral}}{\epsilon_{axial}}, \quad \epsilon_{lateral} = \frac{\Delta D}{D}, \quad \epsilon_{axial} = \frac{\sigma}{E} = \frac{F}{A E} \]
Step 3: Calculating each quantity:
Cross-section area:
\[ A = \frac{\pi}{4} D^2 = \frac{3.14}{4} (10)^2 = 78.5 \text{ mm}^2 \]
Axial stress:
\[ \sigma = \frac{F}{A} = \frac{14000}{78.5} = 178.34 \text{ MPa} \]
Axial (longitudinal) strain, with \(E = 150000\) MPa:
\[ \epsilon_{axial} = \frac{178.34}{150000} = 1.1890 \times 10^{-3} \]
Lateral (diametral) strain, with the diameter shrinking by \(3 \times 10^{-3}\) mm:
\[ \epsilon_{lateral} = \frac{3 \times 10^{-3}}{10} = 3.0 \times 10^{-4} \]
Final Answer:
Taking the ratio, since the diameter shrinks while the length stretches, the signs cancel and Poisson's ratio comes out positive:
\[ \nu = \frac{3.0 \times 10^{-4}}{1.1890 \times 10^{-3}} = 0.2523 \approx 0.25 \]
\[ \boxed{\nu = 0.25} \]
Poisson's ratio connects the lateral strain, the change in diameter, to the longitudinal strain, the change along the axis of pull, for a bar under simple tension, so we need to find both strains separately.
Step 2: Key Formula:
\[ \nu = -\frac{\epsilon_{lateral}}{\epsilon_{axial}}, \quad \epsilon_{lateral} = \frac{\Delta D}{D}, \quad \epsilon_{axial} = \frac{\sigma}{E} = \frac{F}{A E} \]
Step 3: Calculating each quantity:
Cross-section area:
\[ A = \frac{\pi}{4} D^2 = \frac{3.14}{4} (10)^2 = 78.5 \text{ mm}^2 \]
Axial stress:
\[ \sigma = \frac{F}{A} = \frac{14000}{78.5} = 178.34 \text{ MPa} \]
Axial (longitudinal) strain, with \(E = 150000\) MPa:
\[ \epsilon_{axial} = \frac{178.34}{150000} = 1.1890 \times 10^{-3} \]
Lateral (diametral) strain, with the diameter shrinking by \(3 \times 10^{-3}\) mm:
\[ \epsilon_{lateral} = \frac{3 \times 10^{-3}}{10} = 3.0 \times 10^{-4} \]
Final Answer:
Taking the ratio, since the diameter shrinks while the length stretches, the signs cancel and Poisson's ratio comes out positive:
\[ \nu = \frac{3.0 \times 10^{-4}}{1.1890 \times 10^{-3}} = 0.2523 \approx 0.25 \]
\[ \boxed{\nu = 0.25} \]
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