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

Correct Answer: 0.4

We are given the following: Length of the rod, \( L = 1 \, {m} \),
Cross-sectional area before deformation: \( A_0 = 1 \, {cm} \times 1 \, {cm} = 1 \, {cm}^2 \),
Axial tensile force, \( F = 35 \, {kN} = 35000 \, {N} \),
Young’s modulus, \( E = 70 \, {GPa} = 70 \times 10^9 \, {Pa} \),
The cross-sectional area after deformation: \( A_f = 0.998 \, {cm} \times 0.998 \, {cm} = 0.996004 \, {cm}^2 \),
Poisson's ratio, \( \nu \), is the quantity we need to find. Step 1:
Calculate the axial strain \( \epsilon \). The axial strain \( \epsilon \) is given by the formula: \[ \epsilon = \frac{\Delta L}{L} = \frac{F}{A_0 E} \] Substitute the given 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 \( \epsilon_{{lateral}} \). The lateral strain is related to the axial strain by Poisson’s ratio \( \nu \): \[ \epsilon_{{lateral}} = - \nu \epsilon \] The change in the cross-sectional dimension is: \[ \Delta A = A_0 - A_f = 1 - 0.996004 = 0.003996 \, {cm}^2 \] Thus, the lateral strain is: \[ \epsilon_{{lateral}} = \frac{\Delta A}{A_0} = \frac{0.003996}{1} = 0.003996 \] Step 3:
Relate lateral strain and Poisson’s ratio. From the relationship: \[ \epsilon_{{lateral}} = - \nu \epsilon \] Substitute the values: \[ 0.003996 = - \nu \times 0.005 \] Solve for \( \nu \): \[ \nu = \frac{- 0.003996}{-0.005} = 0.7992 \] Thus, the Poisson’s ratio is approximately: \[ \boxed{0.3} \]