Fatigue: S-N curves, Miner's rule and damage tolerance

Why it matters

An airframe sees millions of load cycles in its life: gusts, manoeuvres, one ground–air–ground cycle per flight and one pressurisation cycle per flight. Each is far below the static strength, yet together they grow cracks from rivet holes and corners. Fatigue caused some of the most important accidents in aviation history and is now a design case for every primary structure, handled through S–N data, cumulative damage rules and damage tolerance.

Key ideas

Cyclic stress. A constant-amplitude cycle between σ_max and σ_min has amplitude σ_a = (σ_max − σ_min)/2, mean σ_m = (σ_max + σ_min)/2 and stress ratio R = σ_min/σ_max. R = −1 is fully reversed, R = 0 is zero-to-tension (typical of fuselage pressurisation).

S–N curve. A plot of stress amplitude (or maximum stress) against the number of cycles to failure N, usually log–log, from tests on specimens at a given R. In the finite-life region it is close to a straight line on log–log axes (Basquin): N·S^m = C. Steels and titanium show a fatigue (endurance) limit near 10⁶–10⁷ cycles; aluminium alloys do not, so a fatigue strength at a stated life (for example 5 × 10⁸ cycles) is quoted. S–N data show large scatter; design curves are lowered statistically and life predictions are divided by a scatter factor (often 3–5, from the applicable rules).

Mean stress. A tensile mean stress shortens life. To use fully reversed S–N data for a cycle with σ_m > 0, convert to an equivalent fully reversed amplitude with the Goodman line: σ_a/σ_ar + σ_m/σ_u = 1 (Gerber and Soderberg are alternatives).

Stress concentration. Holes, fillets and joints raise the local stress by K_t (about 3 at a circular hole in a wide plate). Fatigue cracks almost always start there, so detail design (cold-worked holes, generous radii, interference fasteners, avoiding eccentric joints) matters more than nominal stress.

Cumulative damage (Palmgren–Miner). Under a spectrum of stress levels, each level i applied n_i times uses a fraction n_i/N_i of life. Failure is predicted when D = Σ n_i/N_i = 1. The rule ignores load sequence: high loads early can slow later crack growth (overload retardation), so real damage at failure ranges roughly from 0.3 to 3. It is nonetheless the standard method for safe-life estimates.

Design philosophies.

• Safe life: the part is retired after a demonstrated life divided by a scatter factor, whether or not cracks have appeared. Used where inspection is impractical, e.g. landing gear.
• Fail safe: multiple load paths (multi-spar wings, crack stoppers, tear straps) so that failure of one member leaves enough strength to fly safely until detected.
• Damage tolerance: assume a crack of detectable size exists from the start, predict its growth (fracture mechanics topic), and set inspection intervals so that it is found before it reaches critical size. Required for primary structure of transport aircraft.

Practical sources of cycles. Ground–air–ground (one large cycle per flight), pressurisation (one per flight, R ≈ 0 in the hoop direction), gusts and manoeuvres (many smaller cycles), taxiing, and acoustic and vibration loads.

Formulas

σ_a = (σ_max − σ_min)/2, σ_m = (σ_max + σ_min)/2, R = σ_min/σ_max

• Stress amplitude, mean stress (Pa) and stress ratio.

N·S^m = C

• Basquin-type S–N line; S: stress amplitude (Pa), N: cycles to failure, m and C: material constants for given R, surface and K_t (take from test data).

σ_a/σ_ar + σ_m/σ_u = 1

• Goodman relation; σ_ar: equivalent fully reversed amplitude (Pa), σ_u: ultimate tensile strength (Pa).

D = Σ n_i / N_i

• Miner's rule; n_i: applied cycles at level i, N_i: cycles to failure at that level from the S–N curve. Failure predicted at D = 1.

Life = 1 / D_per_block

• Number of repeats of a load block (for example flights) to failure, before applying the scatter factor.

σ_peak = K_t·σ_nominal

• Local stress at a notch (Pa); K_t: elastic stress concentration factor.

Worked examples

Example 1 (standard): Miner's rule over a flight spectrum. Given: each flight applies 1 cycle at 150 MPa (N = 2 × 10⁵), 10 cycles at 80 MPa (N = 5 × 10⁶) and 100 cycles at 40 MPa (N = 1 × 10⁸). Scatter factor 3.

1. Damage per flight: D = 1/(2 × 10⁵) + 10/(5 × 10⁶) + 100/(1 × 10⁸).
2. Terms: 5.0 × 10⁻⁶ + 2.0 × 10⁻⁶ + 1.0 × 10⁻⁶ = 8.0 × 10⁻⁶.
3. Predicted mean life: 1/D = 125 000 flights.
4. Safe life: 125 000/3 = 41 667 flights. Answer: mean life 125 000 flights; safe life ≈ 41 700 flights. The single large ground–air–ground cycle does most of the damage.

Example 2 (GATE level): mean stress and Basquin. Given: fully reversed S–N line N·S⁴ = C passing through S = 300 MPa at N = 10⁵. A lug sees cycles between σ_min = 20 MPa and σ_max = 200 MPa; σ_u = 480 MPa; Goodman correction.

1. Constant: C = 10⁵ × 300⁴ = 8.1 × 10¹⁴ (MPa⁴).
2. Cycle: σ_a = (200 − 20)/2 = 90 MPa, σ_m = (200 + 20)/2 = 110 MPa, R = 0.1.
3. Goodman: σ_ar = σ_a/(1 − σ_m/σ_u) = 90/(1 − 110/480) = 90/0.7708 = 116.8 MPa.
4. Life: N = C/σ_ar⁴ = 8.1 × 10¹⁴/116.8⁴ = 8.1 × 10¹⁴/1.858 × 10⁸ = 4.36 × 10⁶ cycles.
5. Ignoring the mean stress (using 90 MPa) would give 8.1 × 10¹⁴/90⁴ = 1.23 × 10⁷ cycles, nearly three times too long. Answer: σ_ar = 116.8 MPa, N ≈ 4.4 × 10⁶ cycles.

Common mistakes

• Using σ_max instead of σ_a with an S–N curve plotted in amplitude, or the reverse.
• Ignoring tensile mean stress, which makes fatigue predictions unconservative.
• Assuming aluminium alloys have an endurance limit.
• Treating Miner's D = 1 as exact; it is an estimate with large scatter and sequence effects.
• Forgetting the scatter factor when turning a mean life into a safe life.
• Using nominal stress at holes and fillets instead of applying K_t or notched S–N data.

For GATE AE

Expect Miner's rule numericals with two or three stress levels, amplitude/mean/R calculations, Goodman corrections, reading an S–N relation, and conceptual questions on safe-life, fail-safe and damage-tolerant design. Practise converting a spectrum to damage per flight and then to a life.

Quick check

1. A cycle runs from 0 to 200 MPa. What are σ_a, σ_m and R?
2. A part takes 4000 cycles at a level with N = 20 000 and 6000 cycles at a level with N = 60 000. What is D?
3. Name the design philosophy that assumes a crack exists from the start.
4. Does a compressive mean stress usually lengthen or shorten fatigue life? Answers: 1. 100 MPa, 100 MPa, R = 0; 2. D = 0.2 + 0.1 = 0.3; 3. damage tolerance; 4. lengthen.