Question

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 26 MPa√m . It has been determined that the fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm

Answer 1

Answer:

193901.39793 Pa

Explanation:

a = Crack length

$\gamma$ = Geometrical factor

$\sigma$ = Applied stress = 112 MPa

Plane strain fracture toughness is given by

$K_f=\gamma \sigma \sqrt{\pi a}\\\Rightarrow \gamma=\frac{K_f}{\sigma \sqrt{\pi a}}\\\Rightarrow \gamma=\frac{26}{112\times \sqrt{\pi 0.0086}}\\\Rightarrow \gamma=1.41231$

When a = 6 mm

$K_f=\gamma \sigma \sqrt{\pi a}\\\Rightarrow K_f=1.41231 112\times 10^6\sqrt{\pi 0.006}\\\Rightarrow K_f=193901.39793\ Pa$

the stress level at which fracture will occur is 193901.39793 Pa