Question

Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been determined that fracture results at a stress of 300 MPa whenthe maximum (or critical) internal crack length is 4.0 mm. For this same component and alloy, will fracture occur at a stress level of 260MPa when the maximum internal crack length is 6.0 mm

Answer 1

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

Answer 2

Answer:

The fracture toughness at 6 mm is greater than the fracture toughness of the material (42.46 MPa(m)^1/2 > 40 MPa(m)^1/2), thus, the material will fracture.

Explanation:

The half length of the internal crack is:

$a=\frac{L_{surface} }{2} =\frac{4}{2}=2 mm= 2x10^{-3}m$

The dimensionless parameter Y is:

$Y=\frac{K}{o\sqrt{\pi a} }$

Where K is the plane strain fracture toughness=40 MPa(m)^1/2.

o is the initiating crack propagation=300 MPa. Replacing values:

$Y=\frac{40}{300\sqrt{\pi 2x10^{-3} } }=1.68$

The half length of the internal crack when the length is 6 mm is:

a=6/2=3 mm=3x10^-3 m

The strain fracture toughness at the critical stress is:

$K=oY\sqrt{\pi a}=260*1.68*\sqrt{\pi *3x10^{-3} }=42.46 MPa\sqrt{m}$