Question

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3.5 × 10-4 mm (1.378 × 10-5 in.) and a crack length of 4.5 × 10-2 mm (1.772 × 10-3 in.) when a tensile stress of 170 MPa (24660 psi) is applied?

Answer 1

Given:

applied tensile stress,
`\sigma` = 170 MPa

radius of curvature of crack tip,
`r_(t)` =
`3.5* 10^(-4)` mm

crack length =
`4.5* 10^(-2)` mm

half of internal crack length, a =
`(crack length)/(2) = (4.5* 10^(-2))/(2)`

a =
`2.25* 10^(-2)`

Formula Used:

`\sigma _(max) =  2*\sigma \sqrt{(a)/(r_t)}`

Solution:

Using the given formula:

`\sigma _(max) = 2*170 \sqrt{(2.25* 10 ^(-2))/(3.5* 10^(-4))}`

`\sigma _(max)` = 2726 MPa (395372.9 psi)