Cracked Centrifuge Shaft: Design Factor Versus Fitness for Service
A centrifuge drive shaft made of 42CrMo4 (Sut = 1000 MPa, Sy = 850 MPa) has a machined shoulder of 30 mm diameter with a fillet radius of 1.5 mm. At the shoulder the shaft carries a steady torque of 180 N·m and a fully reversed bending moment of 220 N·m. The rotating-beam endurance limit is estimated as 0.5Sut; surface factor 0.80, size factor 0.85, reliability factor 0.814 for 90% reliability, and temperature factor 1.0. The fatigue notch factor at the fillet is Kf = 2.1. The shaft has run for eight years without failure. During a scheduled magnetic-particle inspection, a 0.8 mm long indication is found at the fillet, oriented transverse to the bending stress. The maintenance team proposes returning the shaft to service because 'the crack is smaller than the fillet radius and the shaft has a design factor above 1.' A replacement shaft has a lead time of six weeks; the plant has a standby unit.
1. Calculate the modified endurance limit and the factor of safety against fatigue at the shoulder using the Goodman criterion. Treat the torque as steady and the bending moment as fully reversed.
Read the worked answer
Section modulus: Z = πd³/32 = π(0.030)³/32 = 2.651×10^-6 m³.
Bending stress amplitude: σa = M/Z = 220 / 2.651×10^-6 = 83.0 MPa.
Torsional mean shear stress: τm = T/(2Z) = 180 / (5.302×10^-6) = 34.0 MPa. Von Mises mean stress: σm = √3 τm = 58.8 MPa.
Apply the notch factor to the alternating stress: Kf σa = 2.1 × 83.0 = 174.3 MPa.
Modified endurance limit: Se = 0.5(1000)(0.80)(0.85)(0.814)(1.0) = 276.8 MPa.
Goodman: 1/n = Kf σa/Se + σm/Sut = 174.3/276.8 + 58.8/1000 = 0.630 + 0.059 = 0.689.
n = 1.45. The yield check is Kf σa + σm = 233 MPa, well below Sy = 850 MPa, so fatigue at the notch governs.
2. Evaluate the maintenance team's argument and decide whether the shaft may return to service. Specify the engineering actions required.
Read the worked answer
The argument confuses a design factor for an uncracked notched member with fitness for service of a cracked member. Once a transverse indication exists at the highest-stress location, the failure mode changes from crack initiation to fatigue crack growth. A design factor of 1.45 does not cover a 0.8 mm crack, and the fillet radius is not a crack-tolerance limit.
A rough fracture-mechanics check makes the concern concrete. For fully reversed bending, Δσ ≈ 2 × 83 = 166 MPa. With a = 0.0008 m and Y ≈ 1.1, ΔK = Y Δσ √(πa) ≈ 9 MPa√m, which is above the typical threshold for steel. Growth is therefore credible.
The shaft should not return to service on the basis of the visual argument. Actions: size the indication by ultrasonic or alternating-current field measurement, determine whether it is a crack or a forging lap, perform a fitness-for-service assessment using the actual stress intensity factor and remaining life, and either replace the shaft or grind out the defect only if the remaining section and surface are re-qualified. Shot peening over an existing crack is not acceptable.
3. Compare two proposed remedies for the shoulder: shot peening to raise the endurance limit by 20%, or increasing the shoulder diameter from 30 mm to 34 mm. Which is technically sound for the cracked shaft, and what factor of safety would the larger diameter give?
Read the worked answer
Shot peening raises the initiation resistance by inducing a shallow compressive layer, but it does not arrest a crack that is already 0.8 mm deep. The compressive layer is typically only a few tenths of a millimetre, so peening the cracked shaft would be cosmetic and could even hide the indication during inspection. It is a valid preventive measure for a new, crack-free shaft, not a repair.
Increasing the diameter reduces nominal stress. Since Z scales with d³, stress scales as (30/34)³ = 0.687. The alternating stress becomes 83.0 × 0.687 = 57.0 MPa, and Kf σa = 119.7 MPa. The mean von Mises stress becomes 58.8 × 0.687 = 40.4 MPa.
Revised Goodman factor: 1/n = 119.7/276.8 + 40.4/1000 = 0.432 + 0.040 = 0.472, so n = 2.12.
A larger shoulder is technically sound only as a replacement or redesign; it cannot repair the existing cracked shaft. The correct immediate action is replacement or a qualified grind-out and re-inspection, with the larger fillet and diameter adopted for the replacement.