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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 mpa1m (75.0 ksi1in.). if the plate is exposed to a tensile stress of 345 mpa (50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture. assume a value of 1.0 for y.
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 mpa1m (75.0 ksi1in.). if the plate is exposed to a tensile stress of 345 mpa (50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture. assume a value of 1.0 for y.
using the equation σ =K/(Y √π a) ; a=1/π (K/Yσ)^2. The critical stress required for initiating crack propagation is σ, plain strain fracture toughness is K, surface length of the crack is a, and dimensionless parameter us Y.Substituting the given parameters to the equation. 82.4 MPa √m for K, 345 MPa for σ, and 1 for Y in the equation of surface length of the crack.a = 1/π (K / Yσ )^2 = 1/π (82.4 / 1*345 )^2 = 0.01815 m=18.15mm...
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