# HELP! Fracture Toughness Question

1. Apr 22, 2010

### Cambo!

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.

2. Apr 22, 2010

### Studiot

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.

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