A central through-the-thickness crack, 50 mm long, propagates in a thermoset polymer in an unstable manner under an applied stress of 5 MPa. Find the fracture toughness, K, of the polymer.

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### Problem Statement

A central through-the-thickness crack, 50 mm long, propagates in a thermoset polymer in an unstable manner under an applied stress of 5 MPa. Find the fracture toughness, \( K_{Ic} \), of the polymer.

### Solution

To determine the fracture toughness \( K_{Ic} \) of the polymer, we can use the following standard equation for fracture mechanics in materials science:

\[ K_{Ic} = Y \sigma \sqrt{\pi a} \]

Where:
- \( K_{Ic} \) is the fracture toughness.
- \( Y \) is a dimensionless geometric factor which can often be approximated as 1 for central cracks in large plates.
- \( \sigma \) is the applied stress.
- \( a \) is the half-length of the crack.

Given:
- Crack length (\( 2a \)) = 50 mm, hence \( a = 25 \) mm = 0.025 m.
- Applied stress (\( \sigma \)) = 5 MPa = 5 × 10^6 Pa.
- Assuming \( Y = 1 \).

### Calculations

\[ K_{Ic} = 1 \times 5 \times 10^6 \times \sqrt{\pi \times 0.025} \]

\[ K_{Ic} = 5 \times 10^6 \times \sqrt{0.07854} \]

\[ K_{Ic} = 5 \times 10^6 \times 0.280 \]

\[ K_{Ic} \approx 1.4 \times 10^6 \, \text{Pa} \cdot \sqrt{\text{m}} \]

### Conclusion

The fracture toughness \( K_{Ic} \) of the thermoset polymer is approximately 1.4 MPa·m\uprhalf.
Transcribed Image Text:### Problem Statement A central through-the-thickness crack, 50 mm long, propagates in a thermoset polymer in an unstable manner under an applied stress of 5 MPa. Find the fracture toughness, \( K_{Ic} \), of the polymer. ### Solution To determine the fracture toughness \( K_{Ic} \) of the polymer, we can use the following standard equation for fracture mechanics in materials science: \[ K_{Ic} = Y \sigma \sqrt{\pi a} \] Where: - \( K_{Ic} \) is the fracture toughness. - \( Y \) is a dimensionless geometric factor which can often be approximated as 1 for central cracks in large plates. - \( \sigma \) is the applied stress. - \( a \) is the half-length of the crack. Given: - Crack length (\( 2a \)) = 50 mm, hence \( a = 25 \) mm = 0.025 m. - Applied stress (\( \sigma \)) = 5 MPa = 5 × 10^6 Pa. - Assuming \( Y = 1 \). ### Calculations \[ K_{Ic} = 1 \times 5 \times 10^6 \times \sqrt{\pi \times 0.025} \] \[ K_{Ic} = 5 \times 10^6 \times \sqrt{0.07854} \] \[ K_{Ic} = 5 \times 10^6 \times 0.280 \] \[ K_{Ic} \approx 1.4 \times 10^6 \, \text{Pa} \cdot \sqrt{\text{m}} \] ### Conclusion The fracture toughness \( K_{Ic} \) of the thermoset polymer is approximately 1.4 MPa·m\uprhalf.
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