Question

A 1 m long rod of 1 cm × 1 cm cross section is subjected to an axial tensile force of 35 kN. The Young’s modulus of the material is 70 GPa. The cross-section of the deformed rod is 0.998 cm × 0.998 cm. The Poisson’s ratio of the material is __________ (rounded off to one decimal place).

Hint:

To find Poisson’s ratio, use the relation between axial and lateral strain. Poisson’s ratio is the negative ratio of lateral strain to axial strain.

Answer 1

We are given the following:
Length of the rod, \( L = 1 \, \text{m} \),
Cross-sectional area before deformation: \( A_0 = 1 \, \text{cm} \times 1 \, \text{cm} = 1 \, \text{cm}^2 \),
Axial tensile force, \( F = 35 \, \text{kN} = 35000 \, \text{N} \),
Young’s modulus, \( E = 70 \, \text{GPa} = 70 \times 10^9 \, \text{Pa} \),
The cross-sectional area after deformation: \( A_f = 0.998 \, \text{cm} \times 0.998 \, \text{cm} = 0.996004 \, \text{cm}^2 \),
Poisson's ratio, \( \nu \), is the quantity we need to find.

Step 1:
Calculate the axial strain \( \epsilon \):
\[ \epsilon = \frac{\Delta L}{L} = \frac{F}{A_0 E} \]
Substituting values:
\[ \epsilon = \frac{35000}{1 \times 10^{-4} \times 70 \times 10^9} = \frac{35000}{7 \times 10^6} = 5 \times 10^{-3} \]
So, the axial strain \( \epsilon = 0.005 \).

Step 2:
Calculate the lateral strain \( \epsilon_{\text{lateral}} \):
\[ \epsilon_{\text{lateral}} = \frac{\Delta A}{A_0} = \frac{1 - 0.996004}{1} = 0.003996 \]

Step 3:
Relate lateral strain and Poisson’s ratio:
\[ \epsilon_{\text{lateral}} = -\nu \epsilon \]
Substituting:
\[ 0.003996 = -\nu \times 0.005 \Rightarrow \nu = \frac{-0.003996}{-0.005} = 0.7992 \]

Final Answer:
Since Poisson's ratio typically falls between 0.25 and 0.35 for most materials, a value of \( 0.7992 \) is unphysically high. This suggests a misinterpretation: the area reduction should be used to infer lateral strain per dimension, not area change directly.

The width changes from 1 cm to 0.998 cm:
\[ \epsilon_{\text{lateral}} = \frac{0.998 - 1}{1} = -0.002 \]
So:
\[ -0.002 = -\nu \times 0.005 \Rightarrow \nu = \frac{0.002}{0.005} = 0.4 \]

Corrected Final Answer:
The Poisson’s ratio is \( \boxed{0.4} \).