Question

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa m. It has been determined that fracture results at a stress of 365 MPa when the maximum internal crack length is 2.5 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm. 90 MPa

Answer 1

Answer:

$\sigma_2=288.39\,MPa$

Explanation:

Given:

Strain fracture toughness, $K=40\,MPa.\sqrt{m}$

maximum internal crack length in first case, $l_1=2.5\times 10^{-3}m$

stress level in the first case, $\sigma_2 =365\,MPa$

maximum internal crack length in second case, $l_2=4.0\times 10^{-3}m$

stress level in the second case, $\sigma_2 =?$

So, now we require a factor given as:

$Y=\frac{K}{\sigma.\sqrt{\frac{\pi.l}{2} } }$

$Y=\frac{40}{365\sqrt{\frac{\pi\times 2.5\times 10^{-3}}{2} } }$

$Y=1.75$

Now,

$\sigma_2=\frac{K}{Y.\sqrt{\frac{\pi.l}{2} } }$

$\sigma_2=\frac{40}{1.75\sqrt{\frac{\pi\times 4\times 10^{-3}}{2} } }$

$\sigma_2=288.39\,MPa$