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 a similar crack geometry will result in fracture (under identical tensile service stress of \( \sigma \)) is _________ mm (rounded off to one decimal place).

Hint:

When comparing fracture toughness for different materials, use the relationship \( K_{IC} = \sigma \sqrt{\pi a} \) to set up a ratio between the crack lengths for materials with different fracture toughness values, keeping the applied stress constant. This helps in determining how the material properties affect the fracture behavior.

Answer 1

The relationship between the fracture toughness \( K_{IC} \), crack length \( a \), and the applied stress \( \sigma \) is given by the equation: \[ K_{IC} = \sigma \sqrt{\pi a} \] Since the tensile stress \( \sigma \) is the same for both alloys, we can set up the following ratio: \[ \frac{K_{IC, B}}{K_{IC, A}} = \frac{\sqrt{\pi a_B}}{\sqrt{\pi a_A}} \] Simplifying: \[ \frac{75}{50} = \frac{\sqrt{a_B}}{\sqrt{a_A}} \] Squaring both sides: \[ \left( \frac{75}{50} \right)^2 = \frac{a_B}{a_A} \] Solving for \( a_B \): \[ a_B = a_A \times \left( \frac{75}{50} \right)^2 = 0.4 \times \left( \frac{75}{50} \right)^2 \approx 0.9 \, {mm} \] Thus, the crack length for alloy B is approximately 0.9 mm. Answer: 0.9 mm.