Shaft Fatigue at a Keyway Under Normal and Jam Loads
A 35 mm diameter shaft of 1045 cold-drawn steel (Sut = 630 MPa, Sy = 530 MPa) drives a gear through a parallel keyway. The keyway ends are milled with a standard end-mill radius and the shaft surface is turned. Bearing span is 300 mm, the gear sits 100 mm from the left bearing, and an overhung V-belt pulley applies 1,200 N at 80 mm outboard of the left bearing. A 15 kW, 1,750 rpm motor drives the shaft through a flexible coupling. At the keyway section, normal running gives a bending moment of 220 N.m and a torque of 82 N.m. The drawing being routed for your signature shows no shoulder radius at the keyway and no post-machining treatment. The end user states that belt tension can reach twice the normal value during a jam, and that the motor draws locked-rotor current until a shear pin set at 2.5 times rated torque releases.
1. Estimate the fatigue factor of safety at the keyway under normal running using the DE-Goodman criterion. Use Kf = 1.6 in bending and Kfs = 1.3 in torsion, 99% reliability, and size factor kb = 0.848. State each factor you apply.
Read the worked answer
Endurance limit (Marin). Se' = 0.5(630) = 315 MPa. Surface factor for machined/turned 1045: ka = 4.51(630)^-0.265 = 0.817. Size factor kb = 0.848 (given). Load factor kc = 1.0 (bending). Temperature kd = 1.0. Reliability ke = 0.814 (99%).
Se = 0.817 x 0.848 x 0.814 x 315 = 177.6 MPa
Stresses at d = 0.035 m (pi*d^3 = 1.347e-4 m^3).
- Alternating bending: sigma_a = 32 Kf M / (pi d^3) = 32(1.6)(220)/1.347e-4 = 83.6 MPa
- Mean torsion: tau_m = 16 Kfs T / (pi d^3) = 16(1.3)(82)/1.347e-4 = 12.7 MPa
- Von Mises mean: sigma_m' = sqrt(3)(12.7) = 21.9 MPa
DE-Goodman. 1/n = 83.6/177.6 + 21.9/630 = 0.4707 + 0.0348 = 0.5055
n = 1.98
Sanity check on the torque term: T = P/omega = 15,000/183.3 = 81.9 N.m, consistent with the stated 82 N.m. Note that the torsion contributes only 7% of the damage; this shaft is bending-dominated.
2. Check the one-cycle yielding condition for the jam event (belt tension doubled, torque at the 2.5x shear-pin setting). Does the jam condition govern the design?
Read the worked answer
Jam loads: M = 440 N.m, T = 2.5(82) = 205 N.m.
- Bending stress (peak, fully reversed at the notch): sigma = 32(1.6)(440)/1.347e-4 = 167 MPa
- Torsion: tau = 16(1.3)(205)/1.347e-4 = 31.7 MPa
- Von Mises peak = sqrt(167^2 + 3(31.7)^2) = sqrt(27,889 + 3,014) = 176 MPa
- ny = Sy/sigma' = 530/176 = 3.0
The jam condition does not govern statically; there is ample margin against yielding, and the shear pin releases before anything else is at risk.
However, if the doubled belt tension recurs, the fatigue result collapses: 1/n = 167/177.6 + (sqrt(3)(31.7))/630 = 0.940 + 0.087 = 1.027, giving n = 0.97, i.e., predicted failure. The governing question is therefore not "how strong is the shaft" but "how often does the 2x tension occur." A one-time jam is a static event; a mis-adjusted or auto-tensioned belt running continuously at 2x is a fatigue event and this shaft will not survive it. Confirm the tension schedule before signing.
3. On the basis of these numbers, would you sign the drawing as submitted? Identify the single most effective drawing change and the cheapest field action.
Read the worked answer
Do not sign as submitted. Two omissions are unquantified on the drawing and both attack Kf, which is the term that dominates n:
- Sharp keyway end, no radius. With a standard end-mill radius Kf = 1.6; a square-ended or worn-tool keyway can push Kt above 3, giving Kf near 2.3-2.5. Re-running with Kf = 2.4: sigma_a = 125 MPa, 1/n = 0.705 + 0.035, n = 1.35 - below the 1.5-2.0 typically required for a non-redundant drive. The drawing must specify a keyway end radius (or a sled-runner/end-milled relief) and prohibit a sharp corner.
- No post-machining treatment. The turned surface already costs 18% of Se via ka. Shot peening of the keyway region or a rolled fillet recovers several percent; it is cheaper than a diameter increase.
Cheapest effective field action: verify belt tension by the deflection method and install a proper tensioner. If the 2x figure is a jam transient, no hardware change is needed; if it is the as-found running tension, nothing on the drawing matters.
If a hardware change is required, the most effective single change is a larger shaft at the keyway or a larger keyway end radius, not a material upgrade - the shaft is bending-dominated and Se tracks Sut at only 0.5x, so geometry pays better than alloy.