Question

For a component fabricated from an alloy A with plane strain fracture toughness $K_{IC = 50$ MPa m$^{1/2}$, fracture was observed to take place at a crack length of 0.4 mm at a tensile service stress of $\sigma$. If the same component is instead fabricated from alloy B with $K_{IC} = 75$ MPa m$^{1/2}$, the crack length at which fracture occurs under identical service stress is .................. mm (rounded off to one decimal place).}

Hint:

Fracture toughness scales with the square root of crack length under the same applied stress. If toughness increases, a larger crack can be tolerated before fracture.

Answer 1

Step 1: Recall fracture mechanics relation.
For mode I fracture: \[ K_{IC} = Y \, \sigma \, \sqrt{\pi a} \] where $a$ = crack length, $\sigma$ = stress, $Y$ = geometry factor (assumed same in both cases). Step 2: Ratio of fracture toughness conditions.
For alloy A: \[ K_{IC,A} = Y \, \sigma \, \sqrt{\pi a_A} \] For alloy B: \[ K_{IC,B} = Y \, \sigma \, \sqrt{\pi a_B} \] Dividing: \[ \frac{K_{IC,B}}{K_{IC,A}} = \sqrt{\frac{a_B}{a_A}} \] Step 3: Solve for crack length in alloy B.
\[ a_B = a_A \left(\frac{K_{IC,B}}{K_{IC,A}}\right)^2 \] Substitute: $a_A = 0.4$ mm, $K_{IC,B} = 75$, $K_{IC,A} = 50$: \[ a_B = 0.4 \left(\frac{75}{50}\right)^2 = 0.4 \times (1.5)^2 \] \[ a_B = 0.4 \times 2.25 = 0.9 \, mm \] Final Answer: \[ \boxed{0.9 \, mm} \]