Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 50.0 MPa-m1/2 . If, during service use, the plate is exposed to a tensile stress of 200 MPa (29,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.

Answer 1

Answer:

The answer is "$0.0199044586 \ m$".

Explanation:

The crucial stress essential for activating the spreading of the crack is $\sigma$, the hardness of the strain break is K, as well as the area long of a break is a, for dimensionless Y. Its equation of the length of its surface of the fracture is 50.1 MPa $\sqrt{m}$ on K, 200MPa on $\sigma$, and 1 on Y.

$a = \frac{1}{\pi} (\frac{K}{Y \ \sigma} )^2$

  $= \frac{1}{\pi} ( \frac{50.0}{1 \times 200} )^2 \\\\= \frac{1}{3.14} ( \frac{1}{4})^2 \\\\= \frac{1}{3.14} ( \frac{1}{16}) \\\\=\frac{1}{50.24}\\\\=0.0199044586 \ m$