Question

A metal specimen elongates 0.2 mm under a tensile load. Original gauge length = 50 mm. Find engineering strain.

• A. 0.004
• B. 0.02
• C. 0.002
• D. 0.04

Hint:

Engineering strain is a measure of deformation and is dimensionless. It can be easily calculated by dividing the elongation by the original length.

Answer 1

The Correct Option is A

Engineering strain (\( \epsilon \)) is defined as the change in length (\( \Delta L \)) divided by the original length (\( L_0 \)):
\[ \epsilon = \frac{\Delta L}{L_0} \] Given that \( \Delta L = 0.2 \, {mm} \) and \( L_0 = 50 \, {mm} \), we can substitute the values into the equation: \[ \epsilon = \frac{0.2}{50} = 0.004 \] Thus, the engineering strain is \( 0.004 \).