An element 2 cm long is extended to twice of its initial length and then compressed to its initial length. The engineering strains for the extension and compression are, respectively :
- A1 and – 0·5Correct
- B2 and – 1
- C0·5 and – 1
- D0·693 and – 0·693
Solution
Concept
Engineering strain = change in length / original length of the stage
Formula
ε = ΔL / L₀
Step-by-step solution
1. Extension: 2 cm → 4 cm, ε = (4 − 2)/2 = 1. 2. Compression: 4 cm → 2 cm, ε = (2 − 4)/4 = −0.5.
Explanation
Engineering strain uses the length at the start of each stage. The true (logarithmic) strains would be ±0.693 (option d), which are not engineering strains.
Why the other options are wrong
Option (d) gives true strains ln 2. Options (b) and (c) use the wrong reference length.
Shortcut
Engineering strain: divide by starting length. True strain: ln(L/L₀).
Common mistake
Dividing the compression by 2 cm instead of 4 cm.
Correct answer: A) 1 and – 0·5