Answer
4.13 mm
Work Step by Step
Required: maximum internal crack length allowable for a Ti-6Al-4V titanium alloy (Table 8.1) component that is loaded to a stress one-half its yield strength. Assume that the value of Y is 1.50
Solution: Using Equation 8.7, and using the values from Table 8.1, $K_{Ic} = 55 MPa \sqrt m$, $σ = σ_{y}/2 = 910 MPa/2 = 455 MPa$,
$2a_{c} = \frac{2}{π} (\frac{K_{Ic}}{Yσ})^{2} = \frac{2}{π} (\frac{55 MPa \sqrt m}{(1.50)(455 MPa)})^{2} = 4.134 \times 10^{-3} m = 4.13 mm$
