Fatigue

Fatigue is failure under repeated loading at stresses below yield. Learn S–N curves, endurance limit, mean-stress effects (Goodman), Marin factors, crack growth and how to design for millions of cycles.

Bend a paper clip back and forth and it snaps after a few dozen cycles, even though a single bend doesn't break it. Machine parts fail the same way, just over millions of cycles. Around 80–90 % of mechanical failures in service are fatigue.

Fatigue cracks start at stress raisers — holes, fillets, weld toes, scratches — grow slowly with each cycle, and then the remaining section fractures suddenly.

An analogy

Water dripping on stone: each drop is harmless, but millions of drops wear a hole. Each load cycle does tiny, cumulative damage.

Why parts fail below yield

Cyclic stress causes microscopic slip that forms a crack at the surface. The crack grows a tiny amount each cycle, leaving 'beach marks' on the fracture surface, until the remaining section can't carry the load.

S–N curves and endurance limit

Testing specimens at different stress amplitudes gives the S–N (Wöhler) curve. Steels show a knee near 10⁶ cycles — below the endurance limit they last indefinitely. Aluminium has no true limit, so designs use a strength at, say, 5 × 10⁸ cycles.

Real parts are weaker than specimens

Marin factors reduce the endurance limit for surface finish (machined ≈ 0.8, forged ≈ 0.4–0.5), size, loading type, temperature and reliability. Notches reduce it further by K_f. A real shaft may have only 30–40 % of the specimen endurance limit.

Mean stress, variable amplitude and fracture mechanics

Tensile mean stress reduces fatigue life — Goodman, Gerber and Soderberg lines account for it. Variable-amplitude service loads are counted with rainflow counting and summed with Miner's rule. Crack growth is modelled with Paris' law da/dN = C(ΔK)^m, which sets inspection intervals for aircraft (damage tolerance).

Formulas

Worked example: Life from the S–N curve

Given: Steel S_u = 600 MPa, S_e = 300 MPa (10⁶); 0.9 S_u at 10³ cycles; Alternating stress 400 MPa

Find: Cycles to failure.

Answer: ≈ 3.4 × 10⁴ cycles

Worked example: Goodman safety factor

Given: σ_a = 80 MPa, σ_m = 150 MPa; S_e = 200 MPa, S_u = 600 MPa

Find: Fatigue factor of safety.

Answer: n ≈ 1.54

Common mistakes

Applications

Key takeaways

Going further