Conveyor Gearbox Output Shaft: Keyseat Fatigue Check
A 15 kW, 360 rpm motor drives a single-reduction gearbox whose output shaft carries a 200 mm pitch-diameter spur gear with a 20° pressure angle. The shaft is 45 mm diameter at the bearing seats, simply supported over a 500 mm span, with the gear at midspan and located by a keyseat cut with a standard end mill. Material is 4140 normalized with Sut = 700 MPa. A fully corrected endurance limit of 240 MPa has already been established for the keyseat section, and the fatigue notch factor there is Kf = 1.7. The shaft transmits the full 15 kW at 360 rpm. The conveyor it drives is started and stopped about 40 times per shift, and the measured breakaway torque is 2.5 times the running torque.
1. Determine the shaft torque, the tangential and radial gear forces, and the resultant bending moment at midspan.
Read the worked answer
Torque ω = 2π(360)/60 = 37.70 rad/s; T = P/ω = 15 000/37.70 = 397.9 N·m
Gear forces (pitch radius r = 0.100 m) Ft = T/r = 397.9/0.100 = 3 979 N Fr = Ft tan20° = 3 979 × 0.3640 = 1 448 N
Bending at midspan (load at midspan of a simple span, M = FL/4) From Ft: 3 979 × 0.500/4 = 497.4 N·m From Fr: 1 448 × 0.500/4 = 181.0 N·m M = √(497.4² + 181.0²) = √(247 400 + 32 760) = 529 N·m
Judgement point: the radial-plane moment is 34% of the tangential one. Omitting it understates M by about 6% and is a routine source of an undersized shaft or a wrong bearing reaction later in the calculation.
2. Using the given endurance limit and Kf, compute the fatigue factor of safety at the keyseat by the Goodman criterion, treating the torque as steady.
Read the worked answer
Section properties at d = 45 mm: Z = πd³/32 = 8.946 × 10⁻⁶ m³; polar modulus = 1.789 × 10⁻⁵ m³.
Nominal bending stress: σ = M/Z = 529.3/8.946 × 10⁻⁶ = 59.2 MPa (fully reversed, since the gear force rotates with the shaft). Nominal torsion: τ = T/Zp = 397.9/1.789 × 10⁻⁵ = 22.2 MPa (steady).
Alternating stress: σa = Kf × 59.2 = 1.7 × 59.2 = 100.6 MPa Mean stress (von Mises equivalent of steady shear): σm = √3 × 22.2 = 38.5 MPa
Modified Goodman: 1/n = σa/Se + σm/Sut = 100.6/240 + 38.5/700 = 0.419 + 0.055 = 0.474 n = 2.11
That is acceptable for steady operation, but it rests entirely on the assumption of a steady torque. It is not a licence to ignore the starting duty.
3. The shaft fractures at the keyseat after eight months; beach marks confirm fatigue. The keyseat was in fact cut with a sled-runner end mill (Kf ≈ 2.4) and the drive is started at full breakaway torque. Re-evaluate the alternating stress and state what you would change.
Read the worked answer
With breakaway torque at 2.5× and a sharp-cornered keyseat: σa = 2.4 × 59.2 = 142.1 MPa τa = 2.5 × 22.2 = 55.6 MPa → von Mises alternating = √(142.1² + 3 × 55.6²) = √(20 190 + 9 275) = 171.7 MPa With σm = 0, n = Se/σa = 240/171.7 = 1.40
At 40 starts per shift (roughly 30 000 load cycles a year) a factor of 1.4 on alternating stress is not enough margin at a stress concentration. Typical targets for a keyseat are 1.8–2.0. Changes, in order of cost-effectiveness:
- Specify a keyseat with a radiused run-out, or use a sled-runner cutter with the end radius left in place, dropping Kt from about 2.4 to 1.6.
- Increase the section to 50 mm - stress scales as 1/d³, giving a factor of (45/50)³ = 0.73, so n rises to about 1.9.
- Add a soft start or VFD ramp so breakaway torque never reaches 2.5× running torque.
Root cause is the combination in the prompt: sharp keyseat plus repeated high breakaway torque. Fixing only the diameter would still leave a fatigue-sensitive notch.