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.

Explanation:

The given data is as follows.

          = 89 MPa,     = 336 MPa

           Y = 0.92

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

                 =

                 =

                = 209.04 mm

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

answer

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