Question

Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of It has been determined that fracture results at a stress of 250 MPa (36,250 psi) when the maximum (or critical) internal crack length is 2.0 mm (0.08 in.). For this same component and alloy, will fracture occur at a stress level of 325 MPa (47,125 psi) when the maximum internal crack length is 1.0 mm (0.04 in.)? Why or why not?

Answer 1

Complete Question:

Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35MPa√m. It has been determined that fracture results at a stress of 250 MPa (36,250 psi) when the maximum (or critical) internal crack length is 2.0 mm (0.08 in.). For this same component and alloy, will fracture occur at a stress level of 325 MPa (47,125 psi) when the maximum internal crack length is 1.0 mm (0.04 in.)? Why or why not?

Answer:

Fracture will not occur

Reason: Because the fracture toughness = 32.20MPa√m which is less than the material toughness = 35MPa√m. The fracture toughness is not enough to make the material fracture.

Step-by-step explanation:

Critical Stress,
`\sigma_c = 250 MPa`

Internal crack length, a = 2 mm

a = l/2 = 2/2 = 1mm

a = 10⁻³ m

Toughness of the material,
`K_(Ic) = 35 MPa√(m)`

For the aluminium alloy to fracture, the material toughness,
`K_(Ic) < Y \sigma √(\pi a)`

Where
`Y \sigma √(\pi a)` is the fracture toughness

Where Y is a dimensionless parameter.

Using the critical stress crack propagation equation, let us find the dimensionless parameter:

`Y = (K_(Ic) )/(\sigma_(c) √(\pi a ) )`

`Y = \frac{35 }{250 \sqrt{\pi * 10^(-3) } }`

Y = 2.5

For a stress level of 325 MPa,
`\sigma = 325 MPa` at maximum internal crack length, l = 1.0 mm =
`10^(-3)` m

`a = (l)/(2) = (10^(-3))/(2)`

a = 0.0005 m

`Y \sigma √(\pi a) = 2.5 * 325 * √(\pi * 0.0005) \\Y \sigma √(\pi a) = 32.20 MPa√(m)`

Since
`Y \sigma √(\pi a)` is less than the material toughness at a stress level of 325 MPa, fracture will not occur