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 1205 MPa(174800psi) . (a) If the largest surface crack is 0.8 mm (0.03150 in.) long, determine the critical stress ?c. (b) Will this specimen experience fracture? Assume that the parameter Y has a value of 0.99.

Answer 1

Explanation:

(a)  Formula for critical stress is as follows.

         $\sigma_{c} = \frac{k_{IC}}{\tau \sqrt{\pi \times a}}$

Here,  $K_{IC}$ = 54.8

          $\tau$ = 0.99

            a = 0.8 mm = $0.8 \times 10^{-3}$ m

Putting the given values into the above formula as follows.

          $\sigma_{c} = \frac{k_{IC}}{\tau \sqrt{\pi \times a}}$

                       = $\frac{54.8}{0.99 \times \sqrt{3.14 \times 0.8 \times 10^{-3}}}$

                       = 1107 MPa

Hence, value of critical stress is 1107 MPa.

(b)    Applied stress value is given as 1205 MPa and since it is more than the critical stress (1107 MPa) as a result, a fracture will occur.