A steel plate with a fracture toughness K_IC = 50 MPa·√m carries a tensile stress of 200 MPa. Using K = Yσ√(πa) with a geometry factor Y = 1.0, which value is closest to the critical crack length a at which fracture occurs?

Answer: A. About 20 mm

Setting K = K_IC gives a = (1/π)(K_IC/Yσ)². K_IC/σ = 50/200 = 0.25 √m, so a = 0.25²/π = 0.0625/3.14 ≈ 0.0199 m, about 20 mm. About 62.5 mm forgets to divide by π, and about 80 mm is the mistake of taking 0.25/π without squaring. The critical crack length is inversely proportional to the square of the stress, so halving the stress allows a crack four times as long. Comparing the crack sizes that nondestructive inspection can detect against this critical value is the basis of damage-tolerant design.