Solution :
The maximum stress is given by the equation :
[tex]$\sigma_m = 2\sigma_o (\frac{a}{P_t})^{1/2}$[/tex]
[tex]$\sigma_m$[/tex] is the applied stress = 170 MPa
a is equal to half of the internal crack [tex]$=\frac{2.5 \times 10^{-2}}{2} \ mm$[/tex]
[tex]$P_t$[/tex] is the radius of the curvature of the tip of the internal crack = [tex]$2.5 \times 10^{-4} \ mm$[/tex]
So substituting we get,
[tex]$\sigma_m = 2\times 170 (\frac{\frac{2.5 \times 10^{-2}}{2}}{2.5 \times 10^{-4}})^{1/2}$[/tex]
[tex]$\sigma_m$[/tex] = 1700 MPa
