An aerospace alloy has fracture toughness K_{Ic} = 50 MPa*\sqrt{m}. If the component contains an edge crack of length a = 4 mm (0.004 m), what is the nominal tensile failure stress \sigma_f in MPa? (Formula: K_{Ic} = 1.12 \sigma_f \sqrt{\pi a}, take pi ≈ 3.1416, round off to one decimal place).

Answer Key:

399.1

### Solution: \sqrt{\pi a} = \sqrt{3.1416 * 0.004} = \sqrt{0.0125664} ≈ 0.1121 m^{1/2}. K_{Ic} = 1.12 * \sigma_f * 0.1121 = 0.12555 * \sigma_f. \sigma_f = 50 / 0.12555 ≈ 398.24 ≈ 399 MPa. Let's compute precisely: 50 / (1.12 * sqrt(pi * 0.004)) = 50 / (1.12 * 0.1120998) = 50 / 0.1255518 = 398.24 MPa.

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