Alarge plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 mpa (70.98 ksi). if the plate is exposed to a tensile stress of 345 mpa (50040 psi) during use, determine the minimum length of a surface crack that will lead to fracture. assume a value of 1.04 for y.

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =       ....................1

put here value we get

α  =  

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

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the answer is d

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