Question

If the longitudinal strain of a stretched wire is 0.2% and the Poisson's ratio of the material of the wire is 0.3, then the volume strain of the wire is

• A. \(0.12% \)
• B. \(0.08% \)
• C. \(0.14% \)
• D. \(0.26% \)

Hint:

For elasticity problems involving strain, remember these key relationships: Longitudinal strain: \(\epsilon_L = \Delta L / L\) Lateral strain: \(\epsilon_T = \Delta R / R\) (for radius/diameter) Poisson's ratio: \(\nu = -\epsilon_T / \epsilon_L\) Volume strain (for a cylinder): \(\epsilon_V = \epsilon_L + 2\epsilon_T = \epsilon_L(1 - 2\nu)\). This formula is derived from the fractional change in volume of a cylinder.

Answer 1

The Correct Option is B

Step 1: Identify the given parameters.
Longitudinal strain, \(\epsilon_L = 0.2\%\)
Poisson's ratio, \(\nu = 0.3\)
Convert the percentage longitudinal strain to a decimal:
\(\epsilon_L = \frac{0.2}{100} = 0.002\)
Step 2: Recall the definitions and relationships of strains.
Longitudinal strain (\(\epsilon_L\)): The fractional change in length, \(\epsilon_L = \frac{\Delta L}{L}\).
Lateral strain (\(\epsilon_T\)): The fractional change in radius (or diameter), \(\epsilon_T = \frac{\Delta R}{R}\).
Poisson's ratio (\(\nu\)): The ratio of lateral strain to longitudinal strain (magnitude), \(\nu = \left|\frac{\text{lateral strain}}{\text{longitudinal strain}}\right| = \left|\frac{\epsilon_T}{\epsilon_L}\right|\).
Since stretching a wire longitudinally causes it to contract laterally, \(\epsilon_T\) will have the opposite sign to \(\epsilon_L\). Thus, \(\epsilon_T = -\nu \epsilon_L\).
Volume strain (\(\epsilon_V\)): The fractional change in volume. For a cylindrical wire with volume \(V = \pi R^2 L\), the volume strain is given by: \[ \epsilon_V = \frac{\Delta V}{V} = \frac{\Delta L}{L} + 2\frac{\Delta R}{R} = \epsilon_L + 2\epsilon_T \] Step 3: Calculate the volume strain.
Substitute the expression for \(\epsilon_T\) into the volume strain formula: \[ \epsilon_V = \epsilon_L + 2(-\nu \epsilon_L) \] \[ \epsilon_V = \epsilon_L (1 - 2\nu) \] Now, substitute the given numerical values for \(\epsilon_L\) and \(\nu\): \[ \epsilon_V = 0.002 (1 - 2 \times 0.3) \] \[ \epsilon_V = 0.002 (1 - 0.6) \] \[ \epsilon_V = 0.002 (0.4) \] \[ \epsilon_V = 0.0008 \] To express the volume strain as a percentage, multiply by 100%: \[ \epsilon_V = 0.0008 \times 100\% = 0.08\% \] The final answer is \( \boxed{0.08\%} \).