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.
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.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44dce5c9-bcec-4e4b-a98e-07662a0f3f15%2F45d59ea7-5c15-4f13-8165-03b61af47801%2Fdffxl43.png&w=3840&q=75)
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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