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 \(98.9 \mathrm{MPa} \sqrt{\mathrm{m}}(90 \mathrm{ksi} \sqrt{\mathrm{in} .})\) and a yield strength of \(860 \mathrm{MPa}(125,000 \mathrm{psi})\). The flaw size resolution limit of the flaw detection apparatus is \(3.0 \mathrm{~mm}(0.12 \mathrm{in}\).). If the design stress is one-half the yield strength and the value of \(Y\) is \(1.0\), determine whether a critical flaw for this plate is subject to detection.

As the design stress is given as one-half of the yield strength, we can easily find this value by dividing the yield strength by two: Design Stress, \(\sigma = \frac{1}{2} \times 860 \mathrm{MPa} = 430 \mathrm{MPa}\)

We have the formula for plane-strain fracture toughness \(K_Ic = Y \sigma \sqrt{\pi a}\), with the following values: \(K_Ic = 98.9 \mathrm{MPa} \sqrt{\mathrm{m}}\), \(Y = 1.0\), and \(\sigma = 430 \mathrm{MPa}\). Now, we can rearrange the formula to solve for the critical flaw size, \(a\): \(a = \frac{K_Ic^2}{(Y \sigma \sqrt{\pi})^2}\) Substituting the given values into the formula: \(a = \frac{(98.9 \mathrm{MPa} \sqrt{\mathrm{m}})^2}{(1.0 \times 430 \mathrm{MPa} \sqrt{\pi})^2}\) \(a ≈ 7.97 \times 10^{-6} \mathrm{m^2}\) Now, we need to find the square root to express the critical flaw size in terms of length: \(\sqrt{a} ≈ \sqrt{7.97 \times 10^{-6} \mathrm{m^2}} = 2.82 \times 10^{-3} \mathrm{m} = 2.82 \mathrm{mm}\)

We found that the critical flaw size is \(2.82 \mathrm{mm}\), while the flaw detection resolution limit is given as \(3.0 \mathrm{mm}\). Since the critical flaw size is smaller than the resolution limit, unfortunately, the critical flaw is not subject to detection using the given flaw detection apparatus.