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 fail under the given loading conditions, we can use the concept of fracture mechanics. Specifically, we can apply the stress intensity factor (K) and compare it to the material’s fracture toughness (Kc) to predict failure.
Given Data:
- Applied stress, ( \sigma = 280 \, \text{MPa} )
- Crack length, ( a = 10 \, \text{mm} )
- Fracture toughness, ( K_c = 50 \, \text{MPa} \sqrt{\text{m}} )
- Previous failure occurred at ( a = 6 \, \text{mm} ) and ( \sigma = 350 \, \text{MPa} )
Step 1: Use the Stress Intensity Factor formula
The stress intensity factor ( K ) for an edge crack can be calculated using the formula:
[
K = \sigma \sqrt{\pi a}
]
where:
- ( \sigma ) is the applied stress (in MPa),
- ( a ) is the crack length (in meters).
Step 2: Calculate the stress intensity factor at the given crack length
- For ( a = 10 \, \text{mm} = 0.01 \, \text{m} ) and ( \sigma = 280 \, \text{MPa} ):
[
K = 280 \sqrt{\pi \times 0.01} = 280 \times \sqrt{0.0314159} \approx 280 \times 0.177 \approx 49.56 \, \text{MPa} \sqrt{\text{m}}
]
Step 3: Compare the stress intensity factor with the fracture toughness
The fracture toughness ( K_c ) for aluminum is given as 50 MPa√m. From the above calculation, the stress intensity factor ( K \approx 49.56 \, \text{MPa} \sqrt{\text{m}} ), which is slightly below the fracture toughness value.
Thus, the applied stress intensity factor is almost equal to the fracture toughness value. Given this, the component is on the verge of failure. If the load were increased even slightly, it would likely cause fracture.
Step 4: Conclusion
The component is unlikely to fracture immediately at the given conditions (280 MPa and 10 mm crack), but it is very close to the fracture toughness limit, and a slight increase in load could cause failure. The fact that it failed previously at 350 MPa with a 6 mm crack suggests that the material is sensitive to crack growth and stress levels near the fracture toughness.
Image:
Here is an image illustrating the relationship between the stress intensity factor (K) and the fracture toughness (Kc). It shows how the stress intensity factor for the given crack length (10 mm) is almost equal to the material’s fracture toughness, indicating that the component is close to failure under the given loading conditions.