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A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa/m (50 ksi/in ) is exposed to a stress of 1077 MPa (156200 psi). Assume that the parameter Y has a value of 1.06. If the largest internal crack is 1.8 mm (0.0709 in.) long, Will this specimen experience fracture? Attach File Browse My Computer Browse Content Collection

A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa/m (50 ksi/in ) is exposed to a stress of 1077 MPa (156200 psi). Assume that the parameter Y has a value of 1.06. If the largest internal crack is 1.8 mm (0.0709 in.) long, Will this specimen experience fracture? Attach File Browse My Computer Browse Content Collection

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**Question 11** A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa√m (50 ksi√in) is exposed to a stress of 1077 MPa (156200 psi). Assume that the parameter Y has a value of 1.06. If the largest internal crack is 1.8 mm (0.0709 in.) long, will this specimen experience fracture? *[Attach File]* *[Browse My Computer]* *[Browse Content Collection]* --- ### Analysis In this problem, we're asked to determine if a steel alloy specimen will experience fracture under given conditions. The key values provided are: - **Plane Strain Fracture Toughness, K_Ic**: 54.8 MPa√m - **Stress, σ**: 1077 MPa - **Parameter Y**: 1.06 - **Crack length, a**: 1.8 mm ### Key Concepts The critical condition for fracture can be assessed using the following formula: \[ K_I = Y \cdot \sigma \cdot \sqrt{(\pi \cdot a)} \] Where: - \( K_I \) is the stress intensity factor. - \( Y \) is a dimensionless parameter specific to the geometry. - \( \sigma \) is the applied stress. - \( a \) is the crack length. The specimen will experience fracture if \( K_I \) reaches or exceeds \( K_Ic \). ### Step-by-Step Calculation 1. **Convert the crack length to meters**: \( a = 1.8 \, \text{mm} = 0.0018 \, \text{m} \) 2. **Calculate \( K_I \) using the given stress and crack length**: \[ K_I = 1.06 \cdot 1077 \, \text{MPa} \cdot \sqrt{(\pi \cdot 0.0018 \, \text{m})} \] 3. **Compare \( K_I \) to \( K_Ic \)**: - If \( K_I \geq 54.8 \, \text{MPa}\sqrt{\text{m}} \), the material will fracture. ### Considerations Understanding these calculations is crucial for predicting material failures in safety-critical applications such
Transcribed Image Text:**Question 11** A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa√m (50 ksi√in) is exposed to a stress of 1077 MPa (156200 psi). Assume that the parameter Y has a value of 1.06. If the largest internal crack is 1.8 mm (0.0709 in.) long, will this specimen experience fracture? *[Attach File]* *[Browse My Computer]* *[Browse Content Collection]* --- ### Analysis In this problem, we're asked to determine if a steel alloy specimen will experience fracture under given conditions. The key values provided are: - **Plane Strain Fracture Toughness, K_Ic**: 54.8 MPa√m - **Stress, σ**: 1077 MPa - **Parameter Y**: 1.06 - **Crack length, a**: 1.8 mm ### Key Concepts The critical condition for fracture can be assessed using the following formula: \[ K_I = Y \cdot \sigma \cdot \sqrt{(\pi \cdot a)} \] Where: - \( K_I \) is the stress intensity factor. - \( Y \) is a dimensionless parameter specific to the geometry. - \( \sigma \) is the applied stress. - \( a \) is the crack length. The specimen will experience fracture if \( K_I \) reaches or exceeds \( K_Ic \). ### Step-by-Step Calculation 1. **Convert the crack length to meters**: \( a = 1.8 \, \text{mm} = 0.0018 \, \text{m} \) 2. **Calculate \( K_I \) using the given stress and crack length**: \[ K_I = 1.06 \cdot 1077 \, \text{MPa} \cdot \sqrt{(\pi \cdot 0.0018 \, \text{m})} \] 3. **Compare \( K_I \) to \( K_Ic \)**: - If \( K_I \geq 54.8 \, \text{MPa}\sqrt{\text{m}} \), the material will fracture. ### Considerations Understanding these calculations is crucial for predicting material failures in safety-critical applications such
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