A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-11-04T07:14:08+01:00

Answer:

a > 3 mm

so that critical flow is subject to detection

Explanation:

plane strain fracture toughness = 98.9 MPa root m

yield strength = 860 MPa

flaw detection apparatus = 3.0 mm

Y = 1

solution

we know critical stress formula that is

critical stress [tex]\sigma c = \frac{K}{Y\sqrt{\pi * a} }[/tex] .............................1

here K is design stress plane strain fracture toughness and a is length of surface creak

so here length of surface creak will be

length of surface creak a = [tex] \frac{1}{\pi} * (\frac{K}{Y*{\sigma}})^2[/tex]

put here value

a = [tex] \frac{1}{\pi} * (\frac{98.9\sqrt{m}}{1*{\frac{860}{2}}})^2[/tex]

a = 0.0168 m = 16.8 mm = 0.66 in

so

a > 3 mm

so that critical flow is subject to detection