Answer
$σ_{m} = 2800 MPa $
Work Step by Step
Required:
magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of $1.9 \times 10^{-4} mm (7.5 \times 10^{-6} in.)$ and a crack length of $3.8 \times 10^{-2}mm (1.5 \times 10^{-3} in.)$ when a tensile stress of 140 MPa (20,000 psi) is applied.
Solution:
Using Equation 8.1,
$σ_{m} = 2σ_{0} (\frac{a}{ρ_{t}})^{1/2} = (2)(140 MPa) (\frac{\frac{3.8 \times 10^{-2}mm }{2}}{1.9 \times 10^{-4} mm})^{1/2} = 2800 MPa $
