Fatigue and Stress Concentration
The S-N curve, the endurance limit and its modifying factors, Goodman and Gerber lines, notches and how to design against fatigue.
On this page
- What fatigue is
- The S–N curve and the endurance limit
- From the lab specimen to the real part: the Marin factors
- Stress concentration and notch sensitivity
- Mean stress: the Goodman diagram
- Worked example: a shaft in fatigue
- Variable loads: Miner’s rule
- Cracks, fracture toughness and crack growth
- Designing against fatigue
- FAQ
What fatigue is
Most mechanical failures in service are not from a single overload but from fatigue: a part that carries a repeated (cyclic) load, well below its yield strength, develops a tiny crack at a point of high stress, the crack grows a little with every cycle, and finally the remaining section can no longer carry the load and the part breaks suddenly. A fatigue fracture surface shows beach marks (growth rings) spreading from the origin, with a rough final-fracture zone. Shafts, springs, gears, bolts, aircraft structures, bridges and the legs of a bicycle all fail this way; the loads that cause it are a rotating shaft (stress reverses every turn), vibration, pressure cycles and thermal cycling.
The S–N curve and the endurance limit
A fatigue test applies a constant stress amplitude S to a specimen and counts the cycles N to failure; plotting S against N gives the S–N (Wöhler) curve. For a given amplitude the life is finite; for steels the curve flattens at about 106 to 107 cycles at the endurance limit S′e: below it the part lasts indefinitely. For steels with a tensile strength up to about 1 400 MPa, S′e ≈ 0.5 Su; above it the value saturates at about 700 MPa. Aluminum, copper and most non-ferrous alloys have no endurance limit: the curve keeps falling, and the strength is quoted at 5 × 108 cycles (for 6061-T6, about 97 MPa against 310 MPa tensile strength).
| Regime | Cycles | Typical approach |
|---|---|---|
| Low-cycle fatigue | Up to 10³ to 10⁴ | Plastic strain; strain-life methods; approximate by the static strength |
| High-cycle, finite life | 10⁴ to 10⁶ (or 10⁷) | S–N curve: the stress for the life needed |
| High-cycle, infinite life | Above 10⁶ to 10⁷ (steels) | Keep the stress below the endurance limit with a safety factor |
| Very high cycle | 10⁸ and more | Fatigue can still occur in high-strength steels and non-ferrous metals |
From the lab specimen to the real part: the Marin factors
The endurance limit S′e is measured on a small, polished specimen in reversed bending. A real part is rougher, bigger, loaded differently and needed with higher reliability, so it is reduced by the Marin factors: Se = ka kb kc kd ke S′e.
| Factor | Meaning | Typical value |
|---|---|---|
| ka: surface finish | Rougher surfaces start cracks sooner: ka = a Sub (Su in MPa) | Ground: a = 1.58, b = −0.085; machined or cold drawn: 4.51, −0.265; hot rolled: 57.7, −0.718; as forged: 272, −0.995 |
| kb: size | A larger part has more highly stressed material and more flaws: kb = 1.24 d−0.107 for bending and torsion, 2.79 to 51 mm | 1 for axial loading; about 0.85 for 30 mm, 0.80 for 50 mm |
| kc: load type | Bending is the reference | Bending 1; axial 0.85; torsion 0.59 |
| kd: temperature | Strength changes with temperature | 1 at room temperature; reduce above about 450 °C |
| ke: reliability | S′e is the mean; higher reliability needs a lower value | 90 %: 0.897; 95 %: 0.868; 99 %: 0.814; 99.9 %: 0.753 |
| kf: other | Corrosion, plating, residual stress, fretting | Corrosion in a wet environment can remove the endurance limit entirely; chrome plating reduces it; shot peening raises it |
Stress concentration and notch sensitivity
A hole, a shoulder, a keyway, a thread, a weld toe or a scratch interrupts the flow of stress and raises it locally. The factor of increase is the theoretical stress concentration factor Kt (the peak stress divided by the nominal stress), taken from charts (Peterson’s). For fatigue the real effect is smaller because small notches in many materials are less damaging: the fatigue notch factor Kf = 1 + q (Kt − 1), with q the notch sensitivity (0 to 1; about 0.8 to 0.95 for steel with a notch radius of 2 to 3 mm, smaller for small radii and soft materials). The stress used in the fatigue calculation is the nominal stress times Kf.
| Feature | Typical Kt | Remarks |
|---|---|---|
| Circular hole in a wide plate in tension | 3.0 | The stress at the edge of the hole is three times the nominal |
| Shoulder fillet, D ÷ d = 1.5, r ÷ d = 0.1 (bending) | 1.7 | Rises sharply for small radii: r ÷ d = 0.02 gives 2.5 or more |
| Shoulder fillet in torsion | 1.4 | About the same geometry |
| Keyway (end-milled), bending | 2.1 | Torsion about 3.0; a sled-runner keyway is about 1.7 |
| Retaining ring groove | 3 to 5 | A common hidden fatigue origin |
| Thread root | 3 to 5 | Rolled threads, with compressive residual stress, are better than cut threads |
| Weld toe | 2 to 4 (and residual stress) | Welds typically lower the endurance limit to a quarter or a third |
| Machining scratch, sharp corner | Can exceed 3 | Round every corner; polish the highly stressed surface |
For example, a shoulder with Kt = 1.68 and notch sensitivity q = 0.85 in a steel of Su = 630 MPa has a fatigue notch factor Kf = 1 + 0.85 × (1.68 − 1) = 1.58: the nominal stress is multiplied by about 1.6 in the fatigue check. A hole in a plate loaded to 80 MPa nominal sees 240 MPa at its edge. A larger fillet radius is the cheapest and most effective cure; the geometry of the notch matters more than the material.
Mean stress: the Goodman diagram
Most real loads have a mean (steady) stress σm as well as an alternating amplitude σa: a belt-tensioned shaft, a bolt with preload, a spring with a pre-compression. A tensile mean stress reduces the allowable amplitude. The failure criteria plot the allowable amplitude against the mean stress.
| Criterion | Equation (safety factor n) | Character |
|---|---|---|
| Goodman | σa ÷ Se + σm ÷ Su = 1 ÷ n | A straight line: conservative and the standard for machine design |
| Gerber | n σa ÷ Se + (n σm ÷ Su)² = 1 | A parabola: fits ductile test data best, less conservative |
| Soderberg | σa ÷ Se + σm ÷ Sy = 1 ÷ n | Very conservative: guards against yielding too |
| Langer (first-cycle yield) | σa + σm = Sy ÷ n | Check that the part does not yield in the first cycle |
A compressive mean stress is not harmful (it is even helpful): use σm = 0 in the Goodman line for compression. Combined bending and torsion are reduced to equivalent amplitude and mean stresses with the von Mises formulas (see Torsion, Shear and Combined Loads): σ′a = √(σa² + 3 τa²), σ′m = √(σm² + 3 τm²), with each stress multiplied by its own Kf.
Worked example: a shaft in fatigue
- A 30 mm shaft of AISI 1045 (Su = 630 MPa, Sy = 530 MPa), machined, carries a rotating bending moment M = 120 N·m and a steady torque T = 90 N·m (the loads of Shafts, Keys and Couplings), at a shoulder with Kt = 1.68 (bending) and 1.40 (torsion); reliability 99 percent.
- Endurance limit. S′e = 0.5 × 630 = 315 MPa. ka = 4.51 × 630−0.265 = 0.817; kb = 1.24 × 30−0.107 = 0.862; kc = kd = 1; ke = 0.814. So Se = 0.817 × 0.862 × 0.814 × 315 = 181 MPa: less than 30 percent of the tensile strength.
- Stresses. Nominal bending stress σ = 32 M ÷ (π d³) = 45.3 MPa. With Kf = 1 + 0.85 × (1.68 − 1) = 1.578, the alternating stress is σa = 1.578 × 45.3 = 71.4 MPa (fully reversed). The torque is steady: τm = 16 T ÷ (π d³) = 17.0 MPa and with Kfs = 1.35 the equivalent mean stress is σ′m = √3 × 1.35 × 17.0 = 39.8 MPa.
- Goodman. 1 ÷ n = 71.4 ÷ 181 + 39.8 ÷ 630 = 0.396 + 0.063 = 0.459, so n = 2.18 for infinite life. (Ignoring the mean stress: n = 2.53.) The static (Langer) yield check gives n = 530 ÷ (71.4 + 39.8) = 4.77.
- Compare the finishes. A ground surface (ka = 1.58 Su−0.085 = 0.914) would raise the endurance limit by 12 percent; a hot-rolled, unmachined surface (ka = 0.564) would lower it by 31 percent. A bigger fillet radius (Kt falls to 1.4) would raise n much more cheaply than a bigger shaft.
Variable loads: Miner’s rule
Real parts see a mix of stress levels. Miner’s rule (linear damage accumulation) adds up the fraction of life used at each level: D = Σ ni ÷ Ni, with ni the applied cycles at stress level i and Ni the cycles to failure at that level (from the S–N curve). Failure is predicted when D = 1.
Worked example: three load levels
- From the S–N curve of a part: 250 MPa lasts 20 000 cycles, 200 MPa lasts 100 000 cycles, 150 MPa lasts 800 000 cycles.
- In service the part sees 5 000 cycles at 250 MPa, 20 000 at 200 MPa and 100 000 at 150 MPa per block. The damage is D = 5 000 ÷ 20 000 + 20 000 ÷ 100 000 + 100 000 ÷ 800 000 = 0.250 + 0.200 + 0.125 = 0.575 per block.
- The life is 1 ÷ 0.575 = 1.7 blocks. The 250 MPa cycles do 43 percent of the damage though they are only 4 percent of the cycles: the few high stresses dominate. Miner’s rule ignores the order of the loads and is only a guide, so use a factor of 0.3 to 1.0 on D and test critical parts.
Cracks, fracture toughness and crack growth
Real parts contain tiny flaws, and a crack in a stress field has a stress intensity factor K = Y σ √(π a) (MPa√m, with Y about 1.12 for an edge crack in a wide plate, a the crack length). The crack runs unstably when K reaches the material’s fracture toughness KIC (about 30 to 100 MPa√m for steels, 25 to 40 for aluminum, 1 for glass and ceramics). In fatigue the crack grows at a rate given by the Paris law da/dN = C (ΔK)m, where ΔK is the stress intensity range and C and m (m about 3 for steels) are material constants.
Worked example: how big a crack is critical?
- A steel plate (KIC = 50 MPa√m) carries a nominal tensile stress of 150 MPa and has a 2 mm edge crack: K = 1.12 × 150 × √(π × 0.002) = 13.3 MPa√m, well below KIC.
- The critical crack length is ac = (1 ÷ π) (KIC ÷ (1.12 σ))² = (1 ÷ π) × (50 ÷ 168)² = 0.028 m = 28 mm.
- Growing from 2 mm to 28 mm under a stress range of 150 MPa with C = 10-11 and m = 3 (da/dN in m, ΔK in MPa√m) takes about 124 268 cycles. Inspection and a leak-before-break design rest on this kind of calculation (damage-tolerant design in aircraft).
Designing against fatigue
- Reduce the stress concentrations: generous fillet radii (r ≥ 0.1 d), gradual changes of section, stress-relief grooves, no sharp corners, fine finishing of the surface at high-stress points; keep holes, keyways and grooves away from the peak-stress locations.
- Improve the surface: grinding, polishing, shot peening and cold rolling of fillets and threads induce compressive residual stress and can raise the endurance limit by 20 to 50 percent; nitriding and case hardening do the same.
- Lower the stresses: a larger section, a lighter load, a smoother load path, damping of vibrations.
- Preload bolts and joints so that cyclic loads change the stress little (see Bolted and Welded Joints).
- Control the mean stress: avoid large tensile mean stresses; compressive residual stress is welcome.
- Choose the material: clean steels, appropriate heat treatment; avoid corrosion; aluminum has no endurance limit: design for a finite life.
- Treat welds as the weakest link: use full-penetration, ground-flush welds in high-stress zones, detail category design (IIW, EN 1993-1-9) and inspect.
- Test and inspect: fatigue test prototypes, use strain gauges to measure the real stresses (see Force, Load Cells and Vibration), and inspect critical parts for cracks at intervals.
- Allow for the environment: corrosion fatigue, fretting at press fits and splines, temperature, hydrogen in high-strength steels.
Related
- Structures and Strength of Materials: the overview of this guide.
- Stress, Strain and Safety Factors and Torsion, Shear and Combined Loads: stresses and combined loads.
- Bolted and Welded Joints: preload against bolt fatigue, weld fatigue.
- Shafts, Keys and Couplings and Gear Strength and Design Calculations: shafts and gear teeth in fatigue.
- Worked Design Checks: complete checks.
FAQ
Is there a fatigue limit for aluminum?
No. Aluminum alloys keep losing strength with cycles; the fatigue strength is quoted at a given life such as 5 × 108 cycles. Design for a finite life or keep the stress very low.
Why is the endurance limit of a real part so much lower than that of a test specimen?
The specimen is small, polished and loaded in a simple way. A real part has a rougher surface, a larger volume of stressed material, notches and a different load, which the Marin factors and the notch factor take into account, often reducing the limit to a third of the specimen value.
What does a fatigue crack look like?
A smooth, flat fracture with curved beach marks centered on an origin (usually a surface notch or flaw), and a rough, last-to-fail region where the remaining section tore. There is little deformation.
Does shot peening really work?
Yes. It leaves a skin of compressive residual stress that closes surface cracks and delays their start, improving the fatigue strength of notched, machined or welded parts significantly. It must be controlled (intensity, coverage).
References
- Shigley, J. E. and Mischke, C. R., Mechanical Engineering Design, McGraw-Hill: fatigue failure resulting from variable loading (Marin factors, Goodman, Miner).
- Peterson, R. E., Stress Concentration Factors, Wiley; Pilkey, W. D., Peterson’s Stress Concentration Factors.
- Suresh, S., Fatigue of Materials, Cambridge University Press; ASM Handbook, vol. 19 (fatigue and fracture); Anderson, T. L., Fracture Mechanics.
Last reviewed: October 2026. Fatigue data scatter widely: treat the calculations as estimates and test critical parts.