## Fracture Stress

A brittle specimen made out of Silicone Carbide has a stress intensity of 3.5 MPa∙m^(1/2). What is the applied tensional force (MPa) if the specimen developed the interior crack shown in the figure?

##
__
__**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:

- $$Y=1$$ (since an interior crack was produced)
- $$K_{IC}=3.5\:MPa\cdot m^{1/2}$$
- $$a=\frac{0.2mm}{2}=0.1\:mm$$

Solving for applied stress:

$$$\sigma=\frac{K_{IC}}{Y \sqrt{\pi (a)}}$$$

$$$\sigma=\frac{3.5\:MPa\cdot\sqrt{ m}}{(1) \sqrt{\pi (0.0001m)}}=\frac{3.5MPa\sqrt{m}}{0.0177\sqrt{m}}=197.5\:MPa$$$

197.5 MPa