Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 89 MPa (81.00 ksi). If the plate is exposed to a tensile stress of 336 MPa (48730 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 0.92 for Y.

Answer 1

Explanation:

The given data is as follows.

         $K_{k}$ = 89 MPa,     $\sigma$ = 336 MPa

           Y = 0.92

Now, we will calculate the length of critical interior flaw as follows.

     $a_{c} = \frac{1}{\pi}(\frac{K_{k}}{\sigma Y})^{2}$

                 = $\frac{1}{\pi}(\frac{89}{336 \times 0.92})^{2}$

                 = $\frac{656.38}{3.14}$

                = 209.04 mm

Thus, we can conclude that minimum length of a surface crack that will lead to fracture is 209.04 mm.