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

The maximum stress is given by the equation :

`$\sigma_m = 2\sigma_o ((a)/(P_t))^(1/2)$`

`$\sigma_m$` is the applied stress = 170 MPa

a is equal to half of the internal crack
`$=(2.5 * 10^(-2))/(2) \ mm$`

`$P_t$` is the radius of the curvature of the tip of the internal crack =
`$2.5 * 10^(-4) \ mm$`

So substituting we get,

`$\sigma_m = 2* 170 (((2.5 * 10^(-2))/(2))/(2.5 * 10^(-4)))^(1/2)$`

`$\sigma_m$` = 1700 MPa