Fracture Toughness

#445 / Materials Science Easy

Consider a part is formed by bending a sheet of 52100 Steel, which has a Fracture Toughness of 14.3 MPa∙m^(1/2), and a yield strength of 586 MPa. What is the total max crack length that can be tolerated for an interior 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:
  • $$\sigma=586\:MPa$$
  • $$Y=1$$ (since an interior crack will be produced)
  • $$K_{IC}=14.3\:MPa\cdot m^{1/2}$$
Solving for crack length:
$$$a=\left (\frac{K_{IC}}{Y\cdot \sigma} \right )^2\cdot \frac{1}{\pi}$$$
$$$a=\left (\frac{14.3MPa\cdot \sqrt{m}}{(1)(586MPa)}  \right )^2\cdot \frac{1}{\pi}=\frac{0.000595m}{\pi}=0.0001896m=0.1896\:mm$$$
Because an interior crack has a total crack length of $$2 \times a$$ (based on the figure):
$$$0.1896mm\times 2=0.38\:mm$$$
0.38 mm


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