## Critical Crack Length

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?

##
__
__**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:

- $$\sigma=470\:MPa$$
- $$Y=1.1$$ (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