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,  $r_{t}$ =  $3.5\times 10^{-4}$ mm

crack length = $4.5\times 10^{-2}$ mm

half of internal crack length, a = $\frac{crack length}{2} = \frac{4.5\times 10^{-2}}{2}$

a =  $2.25\times 10^{-2}$

Formula Used:

$\sigma _{max} = 2\times\sigma \sqrt{\frac{a}{r_t}}$

Solution:

Using the given formula:

$\sigma _{max} = 2\times170 \sqrt{\frac{2.25\times 10 ^{-2}}{3.5\times 10^{-4}}}$

$\sigma _{max}$ = 2726 MPa (395372.9 psi)