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Anton [14]
3 years ago
5

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.
Chemistry
1 answer:
likoan [24]3 years ago
4 0

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.

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