The rotating stress cycle

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This is the single most important fact about a rotating shaft. The bending moment can be perfectly steady and the material still sees a full tension–compression cycle every revolution — which is why shafts are a fatigue problem, not a strength problem.

Amplitude and midrange

Any fatigue calculation needs the load split into a part that alternates and a part that stays put:

σ_a = σ_max − σ_min2 σ_m = σ_max + σ_min2

For the standard rotating shaft with steady loads:

Action	alternating part	steady part	because
bending M	M_a = M	M_m = 0	the fibre rotates through the moment field
torque T	T_a = 0	T_m = T	the torque field rotates with the shaft
axial F	0	F	same reason

That asymmetry is the whole reason bending is so much more dangerous than torque on a shaft of the same nominal stress: bending gets the full fatigue penalty, torque only the mean-stress penalty.

Combining bending and torsion

The two act on the same element as a normal stress and a shear stress, so they are combined with the distortion-energy (von Mises) rule — separately for the alternating and the steady part:

σ′_a = √(σ_a² + 3τ_a²) σ′_m = √(σ_m² + 3τ_m²)

Written out for a round shaft, with the notch factors already inside:

σ′_a = 12W·√( 4(K_fM_a)² + 3(K_fsT_a)² )

where W = I/c is the section modulus — πd³/32 for a solid shaft, and the hollow version if there is a bore. That "2W" is the neat trick that makes the standard textbook formula (which is written with 16/πd³) work for hollow shafts too.

When the shaft does not rotate

An axle that carries load but does not turn, or a shaft that rotates with a load rotating with it, has a steady bending stress instead. That is a completely different — and much less severe — fatigue case. There is a switch for it in panel 01; use it honestly, because getting it wrong the optimistic way is the single biggest error you can make here.

⚠ If the load itself pulses (a reciprocating compressor, a rock crusher, anything with shock), neither case applies directly. Multiply the nominal torque by a service factor first — 1.25 for a smooth drive, 1.5–2 for moderate shock, 2–3 for heavy shock — and then design with the steady equations.