Bearing Life Below Target: Bad Batch or Wrong Design Point?
A 50 mm line shaft in a split pillow block runs at 1,200 rpm under a steady 6.0 kN radial load. The selected 6210 deep-groove ball bearing has a basic dynamic load rating C = 35 kN, and the design target is 20,000 hours at 90% reliability. Two bearings have failed, at 2,700 h and 3,100 h. Maintenance blames a bad batch and wants to change suppliers. Vibration data from the failed units showed 1x running speed with BPFO sidebands, not the high-frequency broadband signature of poor lubrication. Grease type and re-lubrication intervals were verified correct. A dial indicator found 0.25 mm of offset between the two pillow block faces, and the housing bores were line-bored at the same time. No axial load is present.
1. Compute the L10 life of the 6210 under the stated load. What does the comparison with the observed failures tell you about the root cause?
Read the worked answer
L10 = (C/P)^3 x 10^6 revolutions = (35/6.0)^3 x 10^6 = 5.833^3 x 10^6 = 198.5 x 10^6 rev
In hours: L10h = 198.5e6 / (60 x 1,200) = 198.5e6 / 72,000 = 2,757 hours
The observed failures at 2,700 h and 3,100 h bracket the predicted L10 almost exactly. A bearing that fails at its calculated L10 is performing as designed, not failing prematurely. L10 by definition means 10% of a population is expected to fail by that time, so two failures at ~2,800 h in a small population is entirely consistent with the rating.
The evidence supports a design error, not a quality escape: (a) life matches prediction, (b) the failure signature is BPFO with 1x, i.e., raceway spalling from normal fatigue under load, not lubrication distress, (c) no axial load, so the radial-only calculation is valid. Changing suppliers will reproduce the same 2,700 h failures. Reject the bad-batch hypothesis.
2. What dynamic load rating is required to actually reach 20,000 hours at 90% reliability, and what rating is required if the end user quietly needs 99% reliability? Use a1 = 0.21 for 99%.
Read the worked answer
20,000 h at 90%. Required revolutions: L = 20,000 x 60 x 1,200 = 1.44e9 rev.
C_req = P x (L/10^6)^(1/3) = 6.0 x (1,440)^(1/3) = 6.0 x 11.29 = 67.8 kN
That is nearly double the 35 kN of the 6210. This is not a small correction; it is a different bearing size class, and a 68 kN bearing will not fit the existing 50 mm shaft seat or the existing pillow block. Check shaft and housing envelope before promising the life.
20,000 h at 99%. The catalogue rating is defined at 90%, so the required L10 must be inflated by dividing by a1: L10,req = 1.44e9 / 0.21 = 6.857e9 rev C_req = 6.0 x (6,857)^(1/3) = 6.0 x 19.0 = 114 kN
Note the non-linearity: going from 90% to 99% reliability costs 68% more capacity, not 10%. This is the number that should drive the conversation with the end user, not the 20,000 h figure alone.
3. The 0.25 mm offset between pillow block faces is offered as the explanation for the failures. Evaluate this, and list the practical options in order of cost.
Read the worked answer
Evaluate the misalignment claim. A 6210 is not self-aligning. Its permissible misalignment is on the order of 0.0017-0.004 rad, i.e., roughly 0.1-0.23 degrees for this size. A 0.25 mm face offset over a typical pillow block centre distance of 200-300 mm corresponds to about 0.05-0.07 degrees, which is within the permissible range - and, more importantly, misalignment produces edge loading and cage/housing distress signatures, not a clean BPFO spall at the exact predicted L10. The offset is worth correcting and may shave some life, but it is not the root cause. Do not let it become the explanation that closes the investigation.
Options, cheapest first:
- Accept and document. If 20,000 h is a wish rather than a requirement, restate the maintenance interval at ~2,500 h and stock the 6210. Zero capital.
- Reduce speed or load. Any 20% speed reduction gives 1.2x life; check whether the 1,200 N belt-derived load can be reduced.
- Upsize to C ~ 68 kN for 20,000 h at 90% - new pillow blocks, new shaft seat, likely a 60-65 mm shaft.
- Self-aligning spherical roller bearing if the 0.25 mm offset is expected to grow with thermal growth or foundation settling. Costs more capacity per unit of shaft diameter but tolerates the misalignment properly.
Fix the alignment regardless of which option is chosen, and record the 99% reliability case as an open item with the end user.