HELP Fracture Toughness Question

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SUMMARY

The discussion centers on calculating the maximum allowable loading for aluminum alloy 7075-T651 used in aircraft wings, given a flaw size of 9mm and a fracture toughness of 26 MPa*m^0.5. The correct formula for determining the stress at which a crack will propagate is K_{1C} ≤ Yσ√(πa), where Y is the geometric factor (1.1) and a is the flaw size (0.009 m). The user initially misapplied the formula, leading to an incorrect calculation of psi as 21735. The correct approach involves multiplying the stress by the geometric factor rather than the flaw length.

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Engineers, materials scientists, and students studying structural integrity in aerospace applications will benefit from this discussion, particularly those focused on fracture mechanics and material selection for aircraft components.

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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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

{K_{1C}} \le \sigma \sqrt {\pi a}

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.

{K_{1C}} \le Y\sigma \sqrt {\pi a}
 

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