HELP Fracture Toughness Question

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In summary: Since this is a problem based on airframes I would have thought that the stress was defined as a function of the wing loading itself.In summary, the question is how to determine the maximum allowable loading for an aircraft wing made of aluminum alloy (7075-T651) to avoid catastrophic failure. The maximum allowable loading can be calculated using the fracture toughness of aluminum (26MPa*m^0.5), assuming a geometrical parameter of Y = 1.1 and using the formula {K_{1C}} \le Y\sigma \sqrt {\pi a}.
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Cambo!
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The Question I'm Trying to Solve Is:

Aluminum alloy (7075-T651) is used for aircraft wing. The largest flaw size monitored in the wing was 9mm. What is the maximum allowable loading so that any catastrophic failure can be avoided? The fracture toughness of aluminum is 26MPa*m^0.5. Assume the geometrical parameter, Y = 1.1.

So far this is what I've tried to do.

K = psi(pi.a.B)^0.5
26 = psi(pi.0.009.1.1)^.05
rearrange to find psi.

psi = 21735

I assume that's wrong, I don't think I'm even using the correct equation. If someone could point me in the right direction or even give me the correct equation, it would be greatly appreciated. :smile:
 
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  • #2
Well I take it this is some form of homework so I won't complete it for you.

You are on the right lines, but to make sure of the formula, the condition for a crack to propagate is that the elastic tensile stress must be such that

[tex]{K_{1C}} \le \sigma \sqrt {\pi a} [/tex]

Now I think that the geometric factor quoted is a stress intensity factor which means that it increases the stress locally. You multiply the stress found by elastic theory by this factor (1.1 in your case), not the crack length as you have done.
So the formula becomes.

[tex]{K_{1C}} \le Y\sigma \sqrt {\pi a} [/tex]
 

1. What is fracture toughness?

Fracture toughness is a material property that measures its ability to resist fracture when a crack or flaw is present. It is an important factor in determining the strength and durability of a material.

2. How is fracture toughness measured?

Fracture toughness is typically measured through experiments that involve creating a crack in a material and applying a force to see how much stress it can withstand before breaking. The results are then compared to established standards and calculations to determine the material's fracture toughness.

3. What factors can affect fracture toughness?

Fracture toughness can be affected by various factors such as material composition, microstructure, temperature, and loading conditions. Different materials may have different levels of fracture toughness due to their unique properties and structures.

4. Why is fracture toughness important in material science?

Fracture toughness is an important property in material science because it helps predict the behavior of materials under stress and can be used to design and select materials for specific applications. It also plays a crucial role in ensuring the safety and reliability of structures and components.

5. Can fracture toughness be improved?

Yes, fracture toughness can be improved through various methods such as altering the material's composition, heat treatment, and surface modifications. These techniques can increase the material's resistance to fracture and improve its overall strength and durability.

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