Crack Found at a Wing Fastener Hole: Life and Inspection Interval
At a C-check a 16 mm through-crack is found at a fastener hole in the lower wing skin of a transport aircraft (2024-T3, K_IC = 33 MPa√m, skin 3.2 mm thick). The hole sits well inside a 150 mm stringer bay, so the crack is treated as a centre crack in an infinite plate with geometry factor Y = 1.12. Limit-load stress in the skin is 120 MPa. The approved gust spectrum for the current role gives one cycle per flight at Δσ = 100 MPa and nine cycles at Δσ = 45 MPa. Fleet material data give da/dN = 1.0×10⁻¹⁰ (ΔK)³, with da/dN in m/cycle and ΔK in MPa√m. The maintenance programme calls for a repeat inspection of this hole every 6,000 flights, using bolt-hole eddy current with 90% probability of detecting an 8 mm crack. The operator wants to keep the aircraft in service and asks whether the crack can be released, and under what limits.
1. Compute the critical crack size at limit load and the fatigue life (in flights) from the found size to that critical size, using Miner's rule.
Read the worked answer
Critical size. K = Y σ √(πa) with σ = 120 MPa at limit load:
a_cr = (1/π)[K_IC/(Yσ)]² = (1/π)[33/(1.12 × 120)]² = (1/π)(0.2455)² = 0.0192 m = 19.2 mm (half-length)
The found crack has a = 8 mm, so K at the found size is 1.12 × 120 × √(π × 0.008) = 21.3 MPa√m - about 65% of K_IC, so residual strength at the found size is adequate, but the margin is thinning.
Growth integration. With m = 3, integrating da/dN = C(YΔσ√π)³ from a₁ to a₂ gives
N = 2/[C(YΔσ√π)³] × (a₁^(−1/2) − a₂^(−1/2))
and a₁^(−1/2) − a₂^(−1/2) = 1/√0.008 − 1/√0.0192 = 11.18 − 7.22 = 3.96 (m^(−1/2)).
Heavy cycles, Δσ = 100 MPa: (YΔσ√π)³ = (198.5)³ = 7.82×10⁶, C × that = 7.82×10⁻⁴, so N = (2/7.82×10⁻⁴) × 3.96 = 10,130 cycles.
Light cycles, Δσ = 45 MPa: (89.3)³ = 7.13×10⁵, C × that = 7.13×10⁻⁵, N = (2/7.13×10⁻⁵) × 3.96 = 111,100 cycles.
Miner per flight. D = 1/10,130 + 9/111,100 = 9.87×10⁻⁵ + 8.10×10⁻⁵ = 1.80×10⁻⁴.
Life = 1/D ≈ 5,560 flights. The current 6,000-flight inspection interval exceeds the entire remaining life, so the existing programme would allow the crack to reach critical size between inspections. That is not acceptable.
2. State the release decision you would make and the inspection interval you would set, showing the factors you applied.
Read the worked answer
Decision: release for continued service only with a shortened, mandated inspection interval and a recorded repair plan. Do not release under the existing 6,000-flight interval.
Interval calculation. Two factors apply to the calculated life of 5,560 flights from the 8 mm detection threshold to 19.2 mm:
- A factor of 2 to cover scatter in the growth-rate data and in the load spectrum (standard damage-tolerance practice for a single load path in the skin).
- A factor to cover the 90% probability of detection - with a 10% miss chance, the crack that survives one inspection may be larger than 8 mm at the next, so the interval must not be justified solely on the 8 mm starting size.
Interval ≤ 5,560/2 = 2,780 flights. Rounded to a maintenance-friendly figure: 2,500 flights or 18 months, whichever comes first, with the check repeated at that interval until the hole is repaired.
Supporting actions:
- Confirm the crack is a through-thickness crack at the hole, not a corner crack, by removing the fastener and inspecting the bore; a corner crack changes Y and the growth rate.
- Blend out only with engineering approval - blending removes material and increases the local stress, which can shorten life rather than extend it.
- Plan a permanent repair (cold-worked or interference-fit hole, or a bonded/riveted doubler) at the next heavy maintenance, and re-run the damage-tolerance calculation for the repaired configuration.
- Record the finding in the fleet database: if a second aircraft shows a similar crack at the same hole, the inspection threshold for the whole fleet must be reduced.
3. The operator proposes a route change with 2 heavy cycles (100 MPa) and 8 light cycles (45 MPa) per flight. Recompute the life and determine whether your interval still stands. What else would you change in the monitoring approach?
Read the worked answer
Recalculate the Miner damage per flight with the new mix, keeping the same integration results (N = 10,130 cycles at 100 MPa and 111,100 cycles at 45 MPa):
D = 2/10,130 + 8/111,100 = 1.974×10⁻⁴ + 7.20×10⁻⁵ = 2.69×10⁻⁴ per flight
Life = 1/D = 3,715 flights, against 5,560 on the present role - a 33% reduction for a change that looks modest on paper, because the damage depends on the cube of stress amplitude and the heavy cycles dominate.
Applying the same factor of 2: interval ≤ 3,715/2 = 1,860 flights. My earlier 2,500-flight interval no longer stands; it must come down to 1,800 flights for the new role, or the aircraft must stay on the present route until the hole is repaired.
This shows why intervals cannot be set from a nominal spectrum and then left alone. Changes required:
- Count cycles from the recorded data, not from the design spectrum. Use the flight data recorder to produce a rainflow count of the actual stress history per flight, and update the damage accumulation flight by flight.
- Confirm the stress conversion. The 100 MPa and 45 MPa figures come from a stress model of the wing; if the aircraft is flying heavier or faster than the model assumed, the conversion itself is wrong.
- Use the best available detection method. Bolt-hole eddy current with a rotating probe, plus a periodic high-frequency ultrasonic check of the bore, raises the probability of detection and therefore allows a larger assumed starting size, which partly offsets the shorter life.
- Add a crack gauge or a structural health monitoring coupon at the hole if the interval becomes operationally unworkable; a sensor that detects 1 mm of growth directly is worth more than a smaller interval nobody can schedule.