## Fracture Toughness

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?

##
__
__**Hint**

**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**

**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