An Aluminum alloy component will undergo loading of 280MPa. From NDE measurements, we do know that the component has an internal crack of 10mm. Will the component fracture (fail)? The fracture toughness of Al has a value of 50MPa(m)1/2. We also know from previous data that the component failed when it contained a crack having a length of 6mm and under a load of 350MPa.
The correct answer and explanation is :
To determine whether the aluminum alloy component will fracture under the given loading conditions, we will need to assess the situation using fracture mechanics principles, particularly the stress intensity factor (K) and the material’s fracture toughness.
Given Data:
- Loading stress, (\sigma = 280 \, \text{MPa})
- Internal crack length, (a = 10 \, \text{mm})
- Fracture toughness, (K_c = 50 \, \text{MPa} \cdot \sqrt{\text{m}})
- The component previously failed under a load of 350 MPa with a crack length of 6 mm.
Step 1: Stress Intensity Factor (K) Calculation
The stress intensity factor (K) for a crack in an infinite plate subjected to tensile loading is given by the formula:
[
K = \sigma \sqrt{\pi a}
]
Where:
- (\sigma) is the applied stress,
- (a) is the crack length.
Step 2: Previous Failure Condition
From the previous failure condition where the crack length was 6 mm and the applied stress was 350 MPa, we can calculate the stress intensity factor at that point and compare it to the fracture toughness.
Using the formula:
[
K = 350 \times \sqrt{\pi \times 6 \times 10^{-3}} = 350 \times \sqrt{\pi \times 0.006} = 350 \times 0.1386 = 48.51 \, \text{MPa} \cdot \sqrt{\text{m}}
]
This stress intensity factor (K = 48.51 \, \text{MPa} \cdot \sqrt{\text{m}}) is very close to the material’s fracture toughness, (K_c = 50 \, \text{MPa} \cdot \sqrt{\text{m}}), indicating that the material was near failure at that condition.
Step 3: Current Condition
Now for the current case where the crack length is 10 mm and the applied stress is 280 MPa, the stress intensity factor is:
[
K = 280 \times \sqrt{\pi \times 10 \times 10^{-3}} = 280 \times \sqrt{\pi \times 0.01} = 280 \times 0.177 \approx 49.56 \, \text{MPa} \cdot \sqrt{\text{m}}
]
Step 4: Comparison and Conclusion
- The calculated stress intensity factor (K = 49.56 \, \text{MPa} \cdot \sqrt{\text{m}}) is very close to the material’s fracture toughness (K_c = 50 \, \text{MPa} \cdot \sqrt{\text{m}}), which means the component is operating at a critical state.
- Since the stress intensity factor is close to the fracture toughness, the component is highly susceptible to fracture under the given loading conditions. Therefore, the component is likely to fail.
Conclusion:
Given the calculated stress intensity factor of (49.56 \, \text{MPa} \cdot \sqrt{\text{m}}) and the fracture toughness of (50 \, \text{MPa} \cdot \sqrt{\text{m}}), the component will fracture (fail) under the applied load of 280 MPa with a 10 mm internal crack.
The image above shows the stress intensity factor ((K)) plotted against the crack length for two conditions: one with a 6 mm crack under a 350 MPa load, and the other with a 10 mm crack under a 280 MPa load. The red dashed line represents the fracture toughness ((K_c)) of the material, which is 50 MPa(m)(^\frac{1}{2}).
- The blue dot represents the condition where the component had a 6 mm crack and was loaded to 350 MPa, which was near the fracture toughness.
- The green dot represents the condition with a 10 mm crack and 280 MPa loading, showing that the stress intensity factor is very close to the fracture toughness, suggesting the component is on the verge of failure.