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 = 1.39.

Answer 1

Answer: 0.0033 m = 3.3 mm (0.13 in.)

Explanation:

since this problem requires us to calculate the maximum internal crack length allowable for the 7075-T651

aluminum alloy is given that it is loaded to a stress level equal to one-half of its yield strength.

For this alloy, KIc 24 MPa $\sqrt{m}$(22 ksi $\sqrt{in}$); also, σ= σ_y/2 = (495 MPa)/2 = 248 MPa (36,000 psi).

Now solving for 2ac gives us;

2a_c=2/π (K_IC/γσ)^2=2/π [(24MPa √m)/(1.35)(248MPa) ]^2=0.0033m=3.3mm (0.13 in.)