Question:easy

The units of fracture toughness (K\(_{\text{IC}}\)) are

Show Hint

Double check units to avoid simple slip-ups:
- Fracture toughness: \( \text{MPa}\cdot\text{m}^{1/2} \) (or \( \text{MPa}\sqrt{\text{m}} \)).
- Strain energy release rate (\( G_C \)): \( \text{J/m}^2 \) (or N/m).
- Yield strength / stress: MPa (or \( \text{N/mm}^2 \)).
Updated On: Jul 3, 2026
  • MPa m\(^2\)
  • MPa m\(^{1/2}\)
  • N/m\(^2\)
  • J/mol
Show Solution

The Correct Option is B

Solution and Explanation

Fracture toughness comes from the stress intensity relation \( K = Y \sigma \sqrt{\pi a} \), so its units fall straight out of the units already sitting on the right side of that formula. Stress \( \sigma \) is measured in megapascals, and crack length \( a \) is measured in metres, so the square root term contributes a factor of \( \text{m}^{1/2} \). Multiplying stress by this square root of length gives units of \( \text{MPa}\cdot\text{m}^{1/2} \), not squared metres, not plain pascals, and nothing resembling energy per mole. So fracture toughness is reported in \( \text{MPa}\cdot\text{m}^{1/2} \), option (B).
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