MEC-26

Bearing life L10

L10 = (C/P)^p million revolutions. p = 3 ball, 10/3 roller.

Reading speed

BearingsISO 281

Governing equation

L_{10}=\left(\dfrac{C}{P}\right)^p\ \mathrm{10^6\ rev}

$P \in \left[0.2, 80\right]$

where

C: Dynamic capacity (kN)
P: Equivalent load (kN)
p: Life exponent (—)
n: Speed (rpm)
L_{10}: Life in revolutions (10⁶ rev)
L_{10h}: Life in hours (h)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

L_{10}=\left(\dfrac{C}{P}\right)^p\ \mathrm{10^6\ rev}

$P \in \left[0.2, 80\right]$

$P = 0.2$ $L_{10} = 2.17\times 10^{7}$

Along the curve

②

Domain of the variable — blue

P_{\min} = 0.2,\quad P_{\max} = 80

$P \in \left[0.2, 80\right]$

$P = 0.2$ $P = 0.2$

Along the curve

③

Response curve — green

L_{10}=\left(\dfrac{C}{P}\right)^p\ \mathrm{10^6\ rev}

$P \in \left[0.2, 80\right]$

$P = 0.2$ $L_{10} = 2.17\times 10^{7}$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}L_{10}}{\mathrm{d}P}

$P \in \left[0.2, 80\right]$

$P = 0.702$ $L_{10}^{\prime} = -2.15\times 10^{7}$

Along the curve

Lecture brief

Historical brief

Machine design grew from Coulomb torsion and Hertz contact (1881) through Soderberg fatigue and the heat-engine cycle. The lab writes shaft, bearing, contact and thermodynamic limits in SI. This sheet (MEC-26 — Bearing life L10) is the form associated with ISO 281. Working symbols: $C$ , $P$ , $p$ , $n$ $\rightarrow$ $L_{10}$ , $L_{10h}$ . Weibull reliability: L10 is the life 90 % of a population survives. Dynamic capacity C is catalogued.

Purpose

Purpose: compute

L_{10}

L_{10h}

from

C

P

p

n

in Mechanical via

L_{10}=\left(\dfrac{C}{P}\right)^p\ \mathrm{10^6\ rev}

L10 = (C/P)^p million revolutions. p = 3 ball, 10/3 roller. Use it when a real mechanical question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

C = 25.000\,\mathrm{kN}

P = 8.000\,\mathrm{kN}

p = 3.000\,\mathrm{—}

n = 1500.000\,\mathrm{rpm}

, the governing relation

L_{10}=\left(\dfrac{C}{P}\right)^p\ \mathrm{10^6\ rev}

yields

L_{10} = 30.52\,\mathrm{10⁶ rev}

L_{10h} = 339.1\,\mathrm{h}

. A rolling bearing, a load P, a life bar. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Dynamic capacity C25 kNEquivalent load P8 kNLife exponent p3 —Speed n1500 rpm

Outputs

Life in revolutions L_{10}30.52 10⁶ rev
Life in hours L_{10h}339.1 h

Reading speed

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Narration of this film

A rolling bearing, a load P, a life bar.

Weibull reliability: L10 is the life 90 % of a population survives. Dynamic capacity C is catalogued.

Reading speed

Watch on YouTube