Respuesta :
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.
Explanation:
Critical Stress, [tex]\sigma_c = 250 MPa[/tex]
Internal crack length, a = 2 mm
a = l/2 = 2/2 = 1mm
a = 10⁻³ m
Toughness of the material, [tex]K_{Ic} = 35 MPa\sqrt{m}[/tex]
For the aluminium alloy to fracture, the material toughness, [tex]K_{Ic} < Y \sigma \sqrt{\pi a}[/tex]
Where [tex]Y \sigma \sqrt{\pi a}[/tex] is the fracture toughness
Where Y is a dimensionless parameter.
Using the critical stress crack propagation equation, let us find the dimensionless parameter:
[tex]Y = \frac{K_{Ic} }{\sigma_{c} \sqrt{\pi a } }[/tex]
[tex]Y = \frac{35 }{250 \sqrt{\pi * 10^{-3} } }[/tex]
Y = 2.5
For a stress level of 325 MPa, [tex]\sigma = 325 MPa[/tex] at maximum internal crack length, l = 1.0 mm = [tex]10^{-3}[/tex] m
[tex]a = \frac{l}{2} = \frac{10^{-3}}{2}[/tex]
a = 0.0005 m
[tex]Y \sigma \sqrt{\pi a} = 2.5 * 325 * \sqrt{\pi * 0.0005} \\Y \sigma \sqrt{\pi a} = 32.20 MPa\sqrt{m}[/tex]
Since [tex]Y \sigma \sqrt{\pi a}[/tex] is less than the material toughness at a stress level of 325 MPa, fracture will not occur
