Fatigue Safety Factor Calculator (Goodman, Soderberg)

Understanding the fatigue safety factor calculator — concept and principle

Parts that are loaded and unloaded millions of times can crack at stresses well below their static strength. When a cyclic load also has a non-zero mean, the endurance limit alone is not enough to judge safety. This calculator combines alternating and mean stress with the Goodman, Soderberg and Gerber criteria to give a fatigue safety factor for infinite life, as in shafts, springs, bolts and connecting rods.

The safety factor is how far the load could grow along the dashed line before hitting the limit.

How it is calculated

Goodman: 1/n = σa/Se + σm/Su.
Soderberg: 1/n = σa/Se + σm/Sy.
Gerber: nσa/Se + (nσm/Su)² = 1, solved for n: n = ½(Su/σm)²(σa/Se)[−1 + √(1 + (2σmSe/(Suσa))²)].

σa is the alternating stress amplitude (half the range), σm the mean stress, Se the (corrected) endurance limit, Su the ultimate tensile strength and Sy the yield strength, all in MPa.

The principle

Plot alternating stress against mean stress. With zero mean, a part survives indefinitely if σa is below Se. With zero alternation, it fails at Su (or yields at Sy). Each criterion draws a boundary between these two points: Goodman a straight line to Su, Soderberg a straight line to Sy, Gerber a parabola to Su. The safety factor is how far the load line from the origin can be scaled before it touches the boundary. Test data for ductile steels mostly fall near the Gerber parabola, so Goodman is moderately conservative and Soderberg most conservative.

Worked example

Inputs: σa = 100 MPa, σm = 150 MPa, Se = 250 MPa, Su = 600 MPa, Sy = 450 MPa.
Goodman: 100/250 + 150/600 = 0.4 + 0.25 = 0.65, so n = 1.538462.
Soderberg: 0.4 + 150/450 = 0.7333, so n = 1.363636.
Gerber: 2σmSe/(Suσa) = 75,000/60,000 = 1.25; √(1 + 1.5625) = 1.60078; ½ × 16 × 0.4 × 0.60078 = 1.922499.

Each point is the calculator re-run with only “Alternating stress (MPa)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Alternating stress (MPa)	Goodman safety factor
25	2.857142857
50	2.222222222
75	1.818181818
100	1.538461538
125	1.333333333
150	1.176470588
200	0.9523809524

How to read the result

n greater than 1 predicts infinite life under these assumptions; many designers look for about 1.5 or more with Goodman. Also check first-cycle yielding: Sy/(σa + σm) = 450/250 = 1.8 here, which is acceptable. Gerber is the least conservative and should be used with reliable data.

Common mistakes

Using the polished-specimen endurance limit without surface, size, load, temperature and reliability factors (Marin factors).
Entering the stress range instead of the amplitude for σa.
Forgetting the stress concentration factor at notches, holes and shoulders.

Keep in mind: These criteria assume a positive mean stress and constant-amplitude loading. Real fatigue-critical parts must be designed to applicable standards with tested material data and checked by a qualified engineer.

Frequently asked questions

What is a typical endurance limit for steel?

For polished rotating-beam specimens of steel it is often estimated at about half the ultimate strength (for Su up to about 1,400 MPa), before correction factors.

Do aluminium alloys have an endurance limit?

Generally not a true one; designers use fatigue strength at a stated number of cycles, such as 5 × 10⁸.

What about compressive mean stress?

Compressive mean stress is generally less damaging, and these tensile-side formulas do not apply directly.

Sources: R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, McGraw-Hill · N. E. Dowling, Mechanical Behavior of Materials, Pearson

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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