Mechanical / fatigue

Fatigue life

A part that survives one load can still fail after a million smaller ones. The stress-life method builds an S-N line from the ultimate strength and the endurance limit, corrects the limit for surface, size and load, and uses Goodman to account for a mean stress. Miner then converts the cycles into damage.

VerdictPasseska0.869kb0.862

Endurance limit

187.2 MPa

0.869 surface × 0.862 size × 1.00 load

Equivalent amplitude

250.0 MPa

after Goodman at 0 MPa mean

Cycles to failure

71,905

at this amplitude

Damage

0.3477

2.9 blocks to failure

From ultimate strength to allowable amplitude

The laboratory endurance limit, then every factor that makes a real part weaker than a polished specimen.

187 MPa

From ultimate strength to allowable amplitude
Quantity	Value	Note
Ultimate strength	500 MPa
Endurance limit, uncorrected	250.0 MPa	0.5 Su, flat at 700 above 1400
Surface factor ka	0.869	Shigley Table 6-2
Size factor kb	0.862	at 30 mm
Load factor kc	1.00	bending
Endurance limit, corrected	187.2 MPa	the product
S-N coefficient a	854.8	Sf = a·N^b
S-N exponent b	-0.1099
Strength at 10³ cycles	400 MPa	f × Su
Mean stress	0 MPa
Goodman factor	1.000	1 − σm/Su
Equivalent amplitude	250.0 MPa	σa / Goodman factor
Cycles to failure	71905	from the S-N line
Damage at the applied cycles	0.3477	Miner, n/N
Strength at design life	260.2 MPa	at 5e+4 cycles

The surface factor is usually the largest single correction: a machined part loses about 13% of the polished endurance limit at 500 MPa, and an as-forged part loses far more. Polishing is a fatigue treatment, not a cosmetic one.

The S-N curve

A straight line on log-log axes from 10³ cycles to the endurance limit, then flat forever.

What each factor costs

The endurance limit as each correction is applied in turn.

Canonical equations

The exact relationships implemented by this engine.

Se = ka·kb·kc·(0.5 Su) · Shigley, §6-9.

Sf = a·N^b · a = (fSu)²/Se, b = −⅓log(fSu/Se).

σa_eq = σa/(1 − σm/Su) · Tensile mean stress reduces the allowable amplitude.

D = Σ nᵢ/Nᵢ · Failure at D = 1.

Show model notation and computationOpen the symbolic form, the variable definitions, and the step-by-step derivation when you need the detail.

A clear path from inputs to results.

Step-by-step methodWorked exampleAssumptions explained

The formula

S_e = k_a k_b k_c (0.5\,S_u) \;;\; S_f = a\,N^b \;;\; D = \sum \frac{n_i}{N_i}

Reference: Shigley's Mechanical Engineering Design, §6-9 to §6-12 for the Marin factors and the S-N construction; ASTM E739 for the stress-life method.

Worked example

Machined 500 MPa steel, 30 mm section, 250 MPa amplitude

Inputs

Ultimate strength 500 MPa, machined surface, bending.
30 mm section, 250 MPa alternating stress, zero mean.
25,000 cycles applied against a 50,000-cycle design life.

Calculation path

1
The uncorrected endurance limit is 0.5 × 500 = 250 MPa.
2
The surface factor is 4.51 × 500^−0.265 = 0.869 and the size factor is 1.24 × 30^−0.107 = 0.862.
3
The corrected endurance limit is 0.869 × 0.862 × 250 = 187.2 MPa.
4
With f = 0.8, the S-N coefficients are a = (0.8 × 500)² / 187.2 = 854.8 and b = −0.1099.
5
At 250 MPa the cycles to failure are (250 / 854.8)^(1/−0.1099) = 71,905.
6
Applying 25,000 cycles accumulates 25,000 / 71,905 = 0.348 of the damage.
7
The strength at the 50,000-cycle design life is 260.2 MPa, above the 250 MPa applied, so the design passes.

Result

Se 187.2 MPa · N 71,905 cycles · damage 0.348 · passes.

Assumptions & limitations

Constant-amplitude loading with a known S-N curve.

The endurance limit is 0.5 Su, flat above 1400 MPa, with Marin factors from Shigley.

Goodman mean-stress correction, which is conservative for ductile steels.

Miner's rule for damage accumulation, which ignores sequence effects.

No residual stress, corrosion or fretting, each of which can dominate the real behaviour.

Frequently asked questions

Does every material have an endurance limit?

No. Steels and titanium do, which is why their S-N curve flattens at about 10⁶ cycles. Aluminium, copper and most non-ferrous alloys do not, so their curve keeps falling and the design has to be for a finite life rather than infinite life.

What is the most important Marin factor?

The surface factor, for most machined parts. A polished specimen is the laboratory baseline; a machined or as-forged surface loses a significant fraction of the endurance limit before any other factor is applied.

What about variable-amplitude loading?

You need rainflow counting to reduce the load history to a set of cycles, then a Miner sum over the resulting histogram. That is the standard approach in ASTM E1049 and in every fatigue software package.

Why does a mean tensile stress matter?

It shifts the whole cycle upwards, so the same amplitude reaches the ultimate strength sooner. Goodman converts the combination into an equivalent fully-reversed amplitude, which lets one S-N curve cover every mean stress.

How accurate is Miner's rule?

It is a linear approximation and can be out by a factor of two or more. It ignores the fact that a few large cycles early in life can either retard or accelerate later crack growth, and it assumes damage accumulates at the same rate regardless of order.