Question

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that 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 3.0 mm.

Answer 1

Answer: 164.2253 MPa

Step-by-step explanation:

First we find the half internal crack which is = length of surface crack /2

so α = 8.6 /2 = 4.3mm ( 4.3×10⁻³m )

Now we find the dimensionless parameter using the critical stress crack propagation equation

∝ = K / Y√πα

stress level ∝ = 112Mpa

fracture toughness K = 26Mpa

dimensionless parameter Y = ?

SO working the formula

Y = K / ∝√πα

Y = 26 / 112 (√π × 4.3× 10⁻³)

Y = 1.9973

We are asked to find stress level for internal crack length of 4m

so half internal crack is = length of surface crack /2

4/2 = 2mm ( 2 × 10⁻³)

from the previous formula critical stress crack propagation equation

∝ = K / Y√πα

∝ = 26 / 1.9973 √(π × 2 × 10⁻³)

∝ = 164.2253 Mpa