What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5.5 × 10-4 mm (2.165 × 10-5 in.) and a crack length of 5 × 10-2 mm (1.969 × 10-3 in.) when a tensile stress of 220 MPa (31910 psi) is applied?

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-12-04T07:28:51+01:00

Answer:

magnitude of the maximum stress is 3263 MPa

Explanation:

given data

radius of curvature = 5.5 × [tex]10^{-4}[/tex] mm

crack length = 5 × [tex]10^{-2}[/tex] mm

tensile stress = 220 MPa

to find out

magnitude of the maximum stress

solution

we know that magnitude of the maximum stress is express as

magnitude of the maximum stress = [tex]2\sigma_o ( \frac{\alpha }{2 \rho} )^{0.5}[/tex] ..........................1

here σo is tensile stress and α is crack length and ρ is radius of curvature

so put all value in equation 1 we get

magnitude of the maximum stress = [tex]2*220 ( \frac{5.5*10^{-2}}{2 *5*10^{-4}} )^{0.5}[/tex]

solve it we get

magnitude of the maximum stress = 3263 MPa

so magnitude of the maximum stress is 3263 MPa