In a steel with a yield stress of 500 MPa, the stress intensity factor at a crack tip is K = 30 MPa·√m. Estimating the size of the plane-stress plastic zone (its distance from the crack tip) with r_p = (1/2π)(K/σy)², which value is closest?

Answer: B. About 0.57 mm

K/σy = 30/500 = 0.060 √m, which squared gives 3.6×10⁻³ m; dividing by 2π ≈ 6.28 gives r_p ≈ 5.7×10⁻⁴ m, about 0.57 mm. In the elastic solution, the stress at the crack tip grows without limit in proportion to 1/√r, but in reality the region where it exceeds the yield stress becomes a plastic zone, whose size is proportional to (K/σy)². About 3.6 mm forgets to divide by 2π, and about 9.5 mm skips the squaring and treats 0.060/2π as a length. Linear elastic fracture mechanics, which judges fracture by K, can be used when the plastic zone is much smaller than the crack length and the plate thickness. Under plane strain, the plastic zone is about one third of this value.