Question

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?

Answer 1

magnitude of the maximum stress is 3263 MPa

Step-by-step explanation:

given data

radius of curvature = 5.5 ×
`10^(-4)` mm

crack length = 5 ×
`10^(-2)` 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 =
`2\sigma_o ( (\alpha )/(2 \rho) )^(0.5)` ..........................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 =
`2*220 ( (5.5*10^(-2))/(2 *5*10^(-4)) )^(0.5)`

solve it we get

magnitude of the maximum stress = 3263 MPa

so magnitude of the maximum stress is 3263 MPa