Question

A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa (293400 psi). Assume that the parameter Y has a value of 1.14. (a) If the largest surface crack is 0.2 mm (0.007874 in.) long, determine the critical stress .

Answer 1

Answer:

Explanation:

The formula for critical stress is

$\sigma_c=\frac{K}{Y\sqrt{\pi a} }$

$\sigma_c =\texttt{critical stress}$

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

$\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa$

The fracture will not occur because this material can handle a stress of 2186.20Mpa  before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm