Critical Crack Length

#200 / Materials Science Easy

Consider a part is formed by bending a sheet of 4340 Steel, which has a Fracture Toughness of 46 MPa∙m^(1/2), and a yield strength of 470 MPa. What is the max crack length that can be tolerated for an exterior crack in mm?

Expand Hint
$$$K_{IC}=Y\cdot \sigma \cdot \sqrt{\pi (a)}$$$
where $$K_{IC}$$ is fracture toughness, $$\sigma$$ is applied stress, $$a$$ is crack length, and $$Y$$ is geometrical factor.
Hint 2
Fracture toughness is the stress intensity of when a brittle material will fail due to the combination of an applied stress and crack length.
$$$K_{IC}=Y\cdot \sigma \cdot \sqrt{\pi (a)}$$$
where $$K_{IC}$$ is fracture toughness, $$\sigma$$ is applied stress, $$a$$ is crack length, and $$Y$$ is geometrical factor.

Based on the problem statement:
  • $$\sigma=470\:MPa$$
  • $$Y=1.1$$ (since an exterior crack will be produced)
  • $$K_{IC}=46\:MPa\cdot m^{1/2}$$
Solving for crack length:
$$$a=\left (\frac{K_{IC}}{Y\cdot \sigma}  \right )^2\cdot \frac{1}{\pi}$$$
$$$a=\left (\frac{46MPa\cdot \sqrt{m}}{(1.1)(470MPa)} \right )^2\cdot \frac{1}{\pi}=\frac{0.0079m}{\pi }=0.0025=2.5\:mm$$$
2.5 mm


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