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 (23.7 ksi ). It has been determined that fracture results at a stress of 112 MPa (16,240 psi) when the maximum internal crack length is 8.6 mm (0.34 in.). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm

133.88 MPa approximately 134 MPa

Explanation:

Given

Plane strains fracture toughness, k = 26 MPa

Stress at which fracture occurs, σ = 112 MPa

Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m

Critical internal crack length, l' = 6 mm = 6*10^-3 m

We know that

σ = K/(Y.√πa), where

112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]

112 MPa = 26 MPa / Y.√(3.142 * 0.043)

112 = 26 / Y.√1.35*10^-2

112 = 26 / Y * 0.116

Y = 26 / 112 * 0.116

Y = 26 / 13

Y = 2

σ = K/(Y.√πa), using l'instead of l and, using Y as 2

σ = 26 / 2 * [√3.142 * (6*10^-3/2)]

σ = 26 / 2 * √(3.142 *3*10^-3)

σ = 26 / 2 * √0.009426

σ = 26 / 2 * 0.0971

σ = 26 / 0.1942

σ = 133.88 MPa

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