Materials testing — ductility index by percentage elongation: A tensile test shows 30% elongation for material A and 40% elongation for material B (same gauge length and cross-section). Conclude whether B is more ductile than A.
-
ACorrect
-
BIncorrect
-
CCannot be concluded without % reduction in area
-
DOnly true if both are brittle materials
-
EOnly true at the same strain rate
Answer
Correct Answer: Correct
Explanation
Introduction / Context:Ductility quantifies the extent of plastic deformation a material can sustain before fracture. Common indices are percentage elongation and percentage reduction in area, both measured on standard tensile specimens.
Given Data / Assumptions:
- Two specimens of identical dimensions are tested to fracture.
- Percentage elongation for A = 30%, for B = 40%.
- Standard test conditions assumed identical (strain rate, temperature).
Concept / Approach:Greater percentage elongation indicates higher ductility because the specimen undergoes more permanent extension prior to fracture. Therefore, with all else equal, material B is more ductile than A.
Step-by-Step Solution:
Define ductility metric: % elongation = (Lf − L0) / L0 * 100%.Given: 30% for A, 40% for B.Compare: 40% > 30% → B exhibits greater plastic strain capacity.Conclude statement is correct.Verification / Alternative check:For many metals, higher % reduction in area accompanies higher % elongation; both indices generally correlate with ductility. Even if %RA differs, the given elongation already supports the conclusion under identical conditions.
Why Other Options Are Wrong:
- “Incorrect” contradicts the definition of the ductility index used.
- Requiring %RA or limiting to brittle materials is unnecessary; elongation alone is a valid ductility measure.
Common Pitfalls:Comparing elongations from different gauge lengths or non-standard specimens; always compare like with like.
Final Answer:Correct