Question

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 $ 10-4 mm (10-5 in.) and a crack length of 2.5 $ 10-2 mm (10-3 in.) when a tensile stress of 170 MPa (25,000 psi) is applied

Answer 1

Solution :

The maximum stress is given by the equation :

$\sigma_m = 2\sigma_o (\frac{a}{P_t})^{1/2}$

$\sigma_m$ is the applied stress = 170 MPa

a is equal to half of the internal crack  $=\frac{2.5 \times 10^{-2}}{2} \ mm$    

$P_t$ is the radius of the curvature of the tip of the internal crack = $2.5 \times 10^{-4} \ mm$

So substituting we get,

$\sigma_m = 2\times 170 (\frac{\frac{2.5 \times 10^{-2}}{2}}{2.5 \times 10^{-4}})^{1/2}$

$\sigma_m$ = 1700 MPa