Question

Calculate the maximum internal crack length allowable for a 7075-T651 aluminum alloy component that is loaded to a stress one-half of its yield strength. Assume a yield strength 495 MPa, a plane strain fracture toughness of 24 MPa and that Y

Answer 1

Answer:

The answer is below

Explanation:

Assuming that Y = 1.35

$The\ plane\ strain\ fracture\ toughness\ K_{IC}= 24\ MPa ,stress(\sigma)=yield\ strength/2\\=495\ MPa/2=247.5\ MPa\\\\The\ maximum\ internal\ crack\ length\ allowable\ is\ calculated\ using:\\a_c=\frac{1}{\pi}(\frac{K_{IC}}{Y*\sigma} ) ^2=\frac{1}{\pi}(\frac{24\ MPA}{1.35*247.5\ MPA} )^2= 0.00164\ m\\\\a_c=1.64\ mm\\\\If\ the\ crack\ is\ internal\ then\ The\ maximum\ internal\ crack\ length\ allowable=2a_c=2*1.64\ m=3.28\ mm$