Question:easy
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
Show Solution
Correct Answer: 0.25
Solution and Explanation
Step 1: Combining everything into a single formula first:
Instead of calculating stress and then strain in separate stages, let us combine the definitions into one expression before touching numbers:
\[ \nu = \frac{\epsilon_{lateral}}{\epsilon_{axial}} = \frac{\Delta D / D}{F / (A E)} = \frac{\Delta D \cdot A \cdot E}{D \cdot F} \]
Step 2: Filling in the geometry term:
\[ A = \frac{\pi}{4} D^2 = \frac{3.14}{4}(10)^2 = 78.5 \text{ mm}^2 \]
$ D = 10 \text{ mm} $, $ \Delta D = 3 \times 10^{-3} \text{ mm} $, $ E = 150{,}000 \text{ MPa} $, $ F = 14{,}000 \text{ N} $.
Step 3: Substituting all values into the combined formula at once:
\[ \nu = \frac{(3 \times 10^{-3}) \times 78.5 \times 150000}{10 \times 14000} \]
Working out the numerator: $ 3 \times 10^{-3} \times 78.5 = 0.2355 $, then $ 0.2355 \times 150000 = 35325 $.
Working out the denominator: $ 10 \times 14000 = 140000 $.
\[ \nu = \frac{35325}{140000} = 0.2523 \]
Final Answer:
Rounded to two decimal places, the Poisson's ratio is 0.25, the same result as computing stress and strain in separate steps.
\[ \boxed{\nu = 0.25} \]
Instead of calculating stress and then strain in separate stages, let us combine the definitions into one expression before touching numbers:
\[ \nu = \frac{\epsilon_{lateral}}{\epsilon_{axial}} = \frac{\Delta D / D}{F / (A E)} = \frac{\Delta D \cdot A \cdot E}{D \cdot F} \]
Step 2: Filling in the geometry term:
\[ A = \frac{\pi}{4} D^2 = \frac{3.14}{4}(10)^2 = 78.5 \text{ mm}^2 \]
$ D = 10 \text{ mm} $, $ \Delta D = 3 \times 10^{-3} \text{ mm} $, $ E = 150{,}000 \text{ MPa} $, $ F = 14{,}000 \text{ N} $.
Step 3: Substituting all values into the combined formula at once:
\[ \nu = \frac{(3 \times 10^{-3}) \times 78.5 \times 150000}{10 \times 14000} \]
Working out the numerator: $ 3 \times 10^{-3} \times 78.5 = 0.2355 $, then $ 0.2355 \times 150000 = 35325 $.
Working out the denominator: $ 10 \times 14000 = 140000 $.
\[ \nu = \frac{35325}{140000} = 0.2523 \]
Final Answer:
Rounded to two decimal places, the Poisson's ratio is 0.25, the same result as computing stress and strain in separate steps.
\[ \boxed{\nu = 0.25} \]
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