Inputs
For each step the equivalent load P and the time share q, plus the speed n when it varies; and the bearing type for the life exponent.
Rolling bearings · ISO 281
Duty cycle: mean equivalent bearing load
Condense a duty cycle of up to ten load steps into the single constant load that consumes the bearing at the same rate. Because rating life goes with P to the power −p, the averaging is not arithmetic but uses the life exponent; if the speed changes too, each step is weighted by its share of revolutions.
Only equivalent dynamic loads P = X·Fr + Y·Fa already formed for each step are condensed. Standstill periods, shock loads outside the spectrum, changing lubrication or contamination between steps and temperature effects are not covered.
PΣ Pᵢᵖ · qᵢ
02
Mean load
Set the duty cycle and calculate.
Inputs and method
Bearing duty cycle
Condense several load steps into the single constant load that consumes the bearing at the same rate.
Calculation
Because rating life goes with P to the power −p, the averaging is a p-th power mean: P = (Σ Pᵢᵖ · qᵢ/100)^(1/p). If the speed changes as well, each step is additionally weighted by nᵢ/nm, since damage accumulates per revolution; the mean speed is nm = Σ nᵢ · qᵢ/100. The exponent is p = 3 for ball and p = 10/3 for roller bearings.
Example
Three steps of 10,000 N at 20 %, 6,000 N at 50 % and 3,000 N at 30 % give P = 6,812 N for a ball bearing – well above the arithmetic mean of 5,900 N. If the steps also run at 3,000, 1,500 and 750 min⁻¹, the value rises to 7,871 N at nm = 1,575 min⁻¹.
Sources and limits: Life exponents per ISO 281:2007 and DIN ISO 281:2010-10; averaging formulae per Roloff/Matek, Maschinenelemente Formelsammlung, 15th edition, section 14, no. 14 to 17. Only equivalent loads already formed are condensed; standstill periods, shock loads outside the spectrum and changing lubrication conditions are not covered.
Bearing duty cycle in detail
Condense several load steps into the single constant load that consumes the bearing at the same rate.
What is a duty cycle and why condense it?
A duty cycle describes how load and speed of a bearing are distributed across operation. The ISO 281 rating-life formula, however, knows only one load and one speed. The mean equivalent load closes that gap: it is the constant load under which the bearing fatigues at the same rate as under the whole spectrum – formed with the same exponent by which load enters the rating life.
Formula and variables
P = ( Σ Pᵢᵖ · qᵢ/100 )^(1/p)
Varying speed: P = ( Σ Pᵢᵖ · nᵢ/nm · qᵢ/100 )^(1/p)Mean speed: nm = Σ nᵢ · qᵢ/100Linear load cycle: P = (Pmin + 2·Pmax)/3Life exponent: p = 3 (ball), p = 10/3 (roller)
| Symbol / input | Meaning |
|---|---|
| Pᵢ, qᵢ | Equivalent load and time share of the individual step. |
| nᵢ, nm | Speed of the step and the mean speed formed from them. |
| p | ISO 281 life exponent, which is also the averaging exponent. |
| P | Mean equivalent load – the input for the rating-life calculator. |
Choose the inputs correctly
For each step the equivalent load P and the time share q, plus the speed n when it varies; and the bearing type for the life exponent.
How to use the calculator
For each operating step enter the equivalent load P already formed for it and its time share; the shares have to add up to exactly 100 %. Choose the bearing type so the correct life exponent is used. If the bearing runs at different speeds in the steps, tick the box and add the speeds. The result is exactly the pair of P and nm that then belongs in the rating-life calculator.
Worked example
Three steps of 10,000 N at 20 %, 6,000 N at 50 % and 3,000 N at 30 % give P = 6,812 N for a ball bearing – well above the arithmetic mean of 5,900 N. If the steps also run at 3,000, 1,500 and 750 min⁻¹, the value rises to 7,871 N at nm = 1,575 min⁻¹.
How should the result be read?
The mean equivalent load and the mean speed are the pair for the rating-life calculation – individually they say little. What is informative is the comparison with the arithmetic mean: a ratio near 1 means an even spectrum, while a markedly larger ratio shows that a few heavy steps govern the life. Their magnitude and duration are then the values worth establishing most carefully.
Loads in newtons, time shares in per cent, speeds in min⁻¹. The life exponent and the revolution shares are dimensionless.
Typical applications
Gearboxes with a load spectrum from the operating programme, presses and stamping machines with idle and working phases, vehicle and wind-turbine drives with a documented duty cycle, and any bearing sizing where the load assumption exists for several operating points rather than one rated point.
Assumptions, limits and common mistakes
Life exponents per ISO 281:2007 and DIN ISO 281:2010-10; averaging formulae per Roloff/Matek, Maschinenelemente Formelsammlung, 15th edition, section 14, no. 14 to 17. Only equivalent loads already formed are condensed; standstill periods, shock loads outside the spectrum and changing lubrication conditions are not covered.
Common mistake: Do not enter radial and axial forces directly: each step needs its own equivalent load P = X·Fr + Y·Fa, which comes from the preceding calculator. Do not average arithmetically – that systematically underestimates the loading. And with varying speed do not omit the speed weighting, or fast steps are given too little weight.
Frequently asked questions
Why can loads not simply be averaged arithmetically?
Because rating life does not depend linearly on load but goes with P to the power −p. Doubling the load cuts the life of a ball bearing to an eighth, not to a half. An arithmetic mean therefore weights heavy steps far too weakly; the calculator carries it along only for comparison, to make exactly that difference visible.
Why is there an extra weighting when the speed varies?
Because a rolling bearing is damaged per overrolling, not per second. A step running twice as fast accumulates twice as many overrollings in the same time and therefore contributes twice as much. The factor nᵢ/nm represents exactly that; without it the fast step would be undervalued.
What belongs in as the P of a step?
The dynamically equivalent load P = X·Fr + Y·Fa already formed for that operating point. The X and Y factors come from the bearing catalogue and depend on the ratio Fa/Fr – which can change between steps, so each step needs its own evaluation. The linked calculator performs that step.
Why must the time shares add up to exactly 100 %?
Because otherwise a different operating regime would silently be assumed: a sum below 100 % means undescribed operating time, above 100 % means double counting. Both distort the averaging, so the calculator rejects a deviating sum instead of quietly normalising it.
How do I handle standstill periods?
They do not belong in as a step with a very small load, because no fatigue occurs at standstill. Compute the spectrum over the actual running time and convert the resulting life in operating hours to calendar time afterwards. Standstill can, however, cause other damage – false brinelling under vibration, for instance – which this analysis does not capture.