Given:
applied tensile stress, [tex]\sigma[/tex] = 170 MPa
radius of curvature of crack tip, [tex]r_{t}[/tex] = [tex]3.5\times 10^{-4}[/tex] mm
crack length = [tex]4.5\times 10^{-2}[/tex] mm
half of internal crack length, a = [tex]\frac{crack length}{2} = \frac{4.5\times 10^{-2}}{2}[/tex]
a = [tex]2.25\times 10^{-2}[/tex]
Formula Used:
[tex]\sigma _{max} = 2\times\sigma \sqrt{\frac{a}{r_t}}[/tex]
Solution:
Using the given formula:
[tex]\sigma _{max} = 2\times170 \sqrt{\frac{2.25\times 10 ^{-2}}{3.5\times 10^{-4}}}[/tex]
[tex]\sigma _{max}[/tex] = 2726 MPa (395372.9 psi)
