Answer:
2195 MPa
Explanation:
Given that:
Radius of curvature ([tex]\rho_t[/tex]) = 3 × 10⁻⁴ mm,
Crack length ([tex]l_c[/tex]) = 2.5 × 10⁻² mm,
Therefore the length of surface crack (a) = crack length / 2 = 2.5 × 10⁻² mm / 2 = 1.25 × 10⁻² mm,
Tensile stress ([tex]\sigma_0[/tex]) = 170 MPa = 170 × 10⁶ Pa
The magnitude of the maximum stress that exists at the tip of an internal
crack ([tex]\sigma_m[/tex]) is given by the equation:
[tex]\sigma_m=2\sigma_0(\sqrt{\frac{a}{\rho t} })=2* 170*10^6(\sqrt{\frac{1.25*10^{-2}}{3*10^{-4}} } )=2195*10^6Pa=2195MPa[/tex]
Therefore the maximum stress is 2195 MPa
