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

The answer is below

Step-by-step 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=(1)/(\pi)((K_(IC))/(Y*\sigma) ) ^2=(1)/(\pi)((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`