Question

If the length of a cylinder made with a material of Poisson’s ratio 0.4 is increased by 5\%, then the decrease in its diameter is:

• A. \( 0.5\% \)
• B. \( 2.0\% \)
• C. \( 1.0\% \)
• D. \( 1.5\% \)

Hint:

For materials with Poisson’s ratio, the change in the diameter is inversely related to the longitudinal strain. A positive longitudinal strain results in a negative lateral strain, indicating a decrease in diameter.

Answer 1

The Correct Option is B

Step 1: The relationship between the longitudinal strain (\( \epsilon_L \)) and lateral strain (\( \epsilon_D \)) for a material with Poisson’s ratio \( \nu \) is given by: \[ \epsilon_D = -\nu \epsilon_L \] where \( \epsilon_L \) is the strain in the length and \( \epsilon_D \) is the strain in the diameter.
Step 2: If the length of the cylinder increases by 5\%, then the strain in the length is: \[ \epsilon_L = \dfrac{\Delta L}{L} = 5\% = 0.05 \] Step 3: Using Poisson’s ratio \( \nu = 0.4 \), we can find the lateral strain in the diameter: \[ \epsilon_D = -\nu \epsilon_L = -0.4 \times 0.05 = -0.02 \] Step 4: The lateral strain is the fractional change in the diameter, so the decrease in diameter is: \[ \text{Decrease in diameter} = 0.02 \times 100\% = 2.0\% \]