Cracked Keyway on a Rotating Pump Shaft: Static Check Passes, Fatigue Does Not
A 60 kW, 600 rpm motor in a river-pumping station drives a centrifugal pump through a 70 mm diameter solid steel shaft (yield 340 MPa, ultimate 600 MPa). The pulley overhangs the bearing by 180 mm and, after re-tensioning, the belt pull is 9 kN. The contractor's static check shows the combined stress far below yield, and the shaft was accepted on that basis. After 14 months of continuous running, a fine transverse crack is found at the edge of a 12 mm wide keyway whose end is square and unradiused, at the section of maximum bending moment. Maintenance records show the belt was re-tensioned twice and the motor realigned only once. The pump runs 24 hours a day, so every surface point of the rotating shaft sees the bending stress reverse each revolution.
1. Compute the nominal torsional shear stress, bending stress, principal stresses and von Mises equivalent stress at the shaft surface, and state the static factor of safety on yield.
Read the worked answer
Step 1 - Torque and torsional shear. T = 60P/(2πN) = 60 x 60 000/(2π x 600) = 955 N·m. τ = 16T/(πd³) = 16 x 955 000/(π x 70³) = 14.2 MPa.
Step 2 - Bending at the keyway section. The overhung pulley makes the bending moment at the bearing M = 9 x 0.18 = 1.62 kN·m. σ = 32M/(πd³) = 32 x 1.62 x 10⁶/(π x 343 000) = 48.1 MPa (tension on one side, compression on the other).
Step 3 - Principal stresses. σ1,2 = 24.05 ± √(24.05² + 14.2²) = 24.05 ± 27.9, giving σ1 = 52.0 MPa and σ2 = −3.9 MPa; maximum shear = 27.9 MPa.
Step 4 - Distortion energy (von Mises). σeq = √(σ² + 3τ²) = √(48.1² + 3 x 14.2²) = √2918 = 54.0 MPa.
Judgement. The static factor of safety is 340/54 ≈ 6.3, and even the more conservative maximum-shear criterion (27.9 MPa) is far below the permissible 0.5 x 340/1.5 ≈ 113 MPa. The static check is therefore genuinely satisfactory - the contractor is not wrong on statics. What the check omits is that the bending stress is not steady: because the shaft rotates under a fixed transverse load, every fibre cycles from +48.1 to −48.1 MPa on each revolution, which is a pure reversed-stress problem, not a static one. Note also that no stress concentration has been applied; the keyway, not the nominal stress level, is where the answer lies.
2. Reconcile the satisfactory static check with the crack found at 14 months. Show the fatigue numbers and identify the dominant cause.
Read the worked answer
Nominal fatigue loading. Alternating bending σa = 48.1 MPa; mean bending = 0. Torsion is essentially steady, τm = 14.2 MPa. The cycle count at 600 rpm is 600 x 60 x 24 x 365 x 1.17 years ≈ 4.4 x 10⁸ reversals - far into the high-cycle regime, where fatigue, not yield, is the design criterion.
Stress concentration. A square-ended, unradiused keyway in a 70 mm shaft has Kt ≈ 2.0–2.5. Peak alternating stress at the keyway root = 2.0 x 48.1 to 2.5 x 48.1, i.e. 96–120 MPa.
Corrected endurance limit. Se′ = 0.5 Su = 300 MPa; surface factor (machined) ≈ 0.78; size factor at 70 mm ≈ 0.85; reliability (99%) ≈ 0.81; Se = 300 x 0.78 x 0.85 x 0.81 = 161 MPa. Dividing by the fatigue notch factor Kf ≈ 1.8–2.0 gives Se,corrected ≈ 80–90 MPa.
Conclusion. The peak alternating stress (96–120 MPa) exceeds the corrected endurance limit (80–90 MPa) at the keyway, so the shaft has finite life. This explains the location (keyway root), the orientation (transverse, i.e. normal to the reversing bending stress) and the timing (14 months ≈ 10⁸ cycles). The dominant cause is the unradiused keyway end combined with residual misalignment; the re-tensioning episodes added a steady bending component on top of an already marginal alternating stress. The 'adequate static factor of safety' argument is irrelevant at this cycle count.
3. Recommend remedial measures and state what inspection and testing you would order before restarting the pump.
Read the worked answer
Order of action matters. The shaft must not be returned to service on the strength of a static check, and the second unit should be inspected on the same reasoning.
Design/retrofit measures.
- Replace the shaft, and in the new shaft machine the keyway with a radiused or sled-runner end and a fillet; or eliminate the keyway by using an interference/shrink-fit coupling. This alone removes most of the Kf penalty.
- Increase the diameter. Going from 70 mm to 80 mm reduces bending stress by (70/80)³ = 0.67, i.e. to about 32 MPa, and the peak keyway stress to about 64–80 MPa - below the corrected endurance limit.
- Reduce the source of bending: fit a proper belt-tensioning device so the belt pull is set to the manufacturer's value rather than 'as tight as seems right', and record the tension.
- Correct alignment. Use laser alignment of pump and motor rather than a straight-edge; the realignment was clearly not repeated after the second re-tensioning.
Inspection and testing before restart.
- Dye-penetrant or magnetic-particle inspection of the whole shaft, not only the visible crack, to detect secondary cracks at the other keyway edge and at the bearing shoulder fillet.
- Dial-gauge check of shaft run-out and coupling eccentricity (target < 0.05 mm).
- Vibration signature in the 1x and 2x running-speed bands before and after alignment, to confirm that misalignment and not imbalance is the residual excitation.
- Confirm the actual belt pull by dynamometer, and update the maintenance schedule so alignment is repeated after every belt re-tensioning.