{"id":2480,"date":"2026-11-12T14:13:23","date_gmt":"2026-11-12T06:13:23","guid":{"rendered":"https:\/\/www.lycbearingcn.com\/news\/bearing-load-calculation-shaft-forces-to-l10\/"},"modified":"2026-11-12T14:13:36","modified_gmt":"2026-11-12T06:13:36","slug":"bearing-load-calculation-shaft-forces-to-l10","status":"publish","type":"post","link":"https:\/\/www.lycbearingcn.com\/ru\/news\/bearing-load-calculation-shaft-forces-to-l10\/","title":{"rendered":"Bearing Load Calculation: From Shaft Forces to Fr, Fa and L10"},"content":{"rendered":"<h1>Bearing Load Calculation: From Shaft Forces to Fr, Fa and L10<\/h1>\n<p>Every bearing selection guide starts with a table that says &#8220;determine your radial load Fr and<br \/>\naxial load Fa first&#8221;. Almost none of them tells you <em>where those two numbers come from<\/em>. In a<br \/>\nreal machine they are not measured, they are not in the catalogue, and they are not the forces on<br \/>\nthe gear or the belt. They are the <strong>reaction forces at the bearing seats<\/strong>, and getting<br \/>\nfrom a transmitted torque to those reactions takes a free-body diagram, two planes of static<br \/>\nequilibrium, and a decision about which bearing locates the shaft axially.<\/p>\n<p>This article runs that whole chain on one shaft, end to end, with numbers: torque and speed<br \/>\n&rarr; the three force components of a helical gear &rarr; tight-side and slack-side belt tensions<br \/>\n&rarr; support reactions in the vertical and horizontal planes &rarr; vector-summed radial loads<br \/>\n&rarr; axial load allocation &rarr; service factor &rarr; equivalent dynamic load <em>P<\/em> &rarr;<br \/>\nISO 281 <em>L<sub>10h<\/sub><\/em> &rarr; required dynamic load rating <em>C<\/em>. The load case we<br \/>\narrive at is exactly the one used in our <a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/bearing-type-selection-decision-matrix\/\">bearing type selection decision<br \/>\nmatrix<\/a>, so you can follow the arithmetic straight through to a bearing choice.<\/p>\n<p class=\"note\"><strong>About the numbers:<\/strong> the geometry, forces, ratings and factors in this<br \/>\nworked example are <em>illustrative magnitudes for teaching the method<\/em>, not catalogue values for<br \/>\nany specific product. Substitute the actual dimensions, duty and catalogue ratings before making a<br \/>\npurchasing decision.<\/p>\n<h2>1. Why the load case is the part people get wrong<\/h2>\n<p>A bearing fails because of the load <em>at the bearing<\/em>, not the load at the gear. Between the<br \/>\ntwo there is a lever system that can amplify, redistribute or cancel forces. A pulley hung outboard<br \/>\nof the last support is the classic example: a 5 kN belt pull can produce a 7.7 kN reaction at one<br \/>\nsupport and almost nothing at the other, because the moment arm outside the span has to be balanced<br \/>\nby an opposing force inside it. If you simply &#8220;divide the total load by two&#8221;, every downstream<br \/>\ncalculation is wrong.<\/p>\n<p>Three consequences make this worth doing properly:<\/p>\n<ul>\n<li><strong>Life is nonlinear.<\/strong> <em>L<sub>10<\/sub><\/em> scales with <em>(C\/P)<sup>p<\/sup><\/em>,<br \/>\n      with <em>p<\/em> = 3 for ball bearings and 10\/3 for roller bearings. A 20% error in the reaction<br \/>\n      force becomes roughly a 45&ndash;50% error in predicted life.<\/li>\n<li><strong>The two bearings rarely carry the same load.<\/strong> Sizing both from an average is<br \/>\n      simultaneously wasteful on the lightly loaded seat and unsafe on the heavily loaded one.<\/li>\n<li><strong>Axial load is not shared.<\/strong> In a locating\/floating arrangement the axial force<br \/>\n      goes entirely to one bearing. Splitting it 50\/50 understates <em>P<\/em> at the locating bearing<br \/>\n      and overstates it at the floating one.<\/li>\n<\/ul>\n<p>The remedy is a fixed sequence. Do not open a catalogue until step 7.<\/p>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/Deep-Groove-Ball-Bearing-1.png\" alt=\"Single row deep groove ball bearing\" \/><figcaption>A deep groove ball bearing sees the reaction at its seat, not the force applied at the gear or pulley. Establishing that reaction is the whole point of the calculation chain.<\/figcaption><\/figure>\n<h2>2. The calculation chain at a glance<\/h2>\n<p>Eight steps, each with one output that becomes the next step&#8217;s input. Keep them separate: most<br \/>\nmistakes come from collapsing two steps into one, or applying a factor twice.<\/p>\n<table>\n<thead>\n<tr>\n<th>Step<\/th>\n<th>What you compute<\/th>\n<th>Output<\/th>\n<th>Common trap<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1<\/td>\n<td>Torque from power and speed; gear force components<\/td>\n<td>Ft, Fr<sub>gear<\/sub>, Fa<\/td>\n<td>Using normal pressure angle without the helix correction<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>Belt effective pull and total radial pull on the pulley<\/td>\n<td>T1, T2, F<sub>belt<\/sub><\/td>\n<td>Confusing T1&minus;T2 with T1+T2<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>Static equilibrium in two perpendicular planes<\/td>\n<td>R<sub>Av<\/sub>, R<sub>Bv<\/sub>, R<sub>Ah<\/sub>, R<sub>Bh<\/sub><\/td>\n<td>Forgetting the overhung moment arm<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>Vector sum of the two planes at each seat<\/td>\n<td>Fr<sub>A<\/sub>, Fr<sub>B<\/sub><\/td>\n<td>Adding reactions arithmetically instead of geometrically<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>Axial load allocation (locating \/ floating)<\/td>\n<td>Fa per bearing<\/td>\n<td>Splitting Fa equally between supports<\/td>\n<\/tr>\n<tr>\n<td>6<\/td>\n<td>Application \/ shock factor and duty cycle<\/td>\n<td>f<sub>d<\/sub>, duty-weighted load<\/td>\n<td>Multiplying f<sub>d<\/sub> into reactions and again into P<\/td>\n<\/tr>\n<tr>\n<td>7<\/td>\n<td>Equivalent dynamic load P = f<sub>d<\/sub>(X&middot;Fr + Y&middot;Fa)<\/td>\n<td>P in kN<\/td>\n<td>Using X\/Y from the wrong Fa\/Fr band<\/td>\n<\/tr>\n<tr>\n<td>8<\/td>\n<td>L<sub>10h<\/sub> and required C from ISO 281<\/td>\n<td>Life in hours, C<sub>req<\/sub><\/td>\n<td>Using the ball exponent for a roller bearing<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>3. Build the mechanical model first<\/h2>\n<p>Our example is a horizontal geared drive shaft: a helical gear inside the bearing span, a V-belt<br \/>\npulley overhung beyond the right-hand support, and two bearing seats. Support <strong>A<\/strong> is<br \/>\nthe floating (non-locating) bearing; support <strong>B<\/strong> is the locating bearing and therefore<br \/>\ntakes the shaft&#8217;s axial load. The geometry is deliberately simple so every number can be checked by<br \/>\nhand.<\/p>\n<table>\n<thead>\n<tr>\n<th>Item<\/th>\n<th>Symbol \/ position<\/th>\n<th>Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Support A (floating)<\/td>\n<td>x<sub>A<\/sub><\/td>\n<td>0 mm<\/td>\n<\/tr>\n<tr>\n<td>Support B (locating)<\/td>\n<td>x<sub>B<\/sub><\/td>\n<td>300 mm<\/td>\n<\/tr>\n<tr>\n<td>Helical gear centre<\/td>\n<td>x<sub>G<\/sub><\/td>\n<td>100 mm (inside the span)<\/td>\n<\/tr>\n<tr>\n<td>Pulley centre<\/td>\n<td>x<sub>P<\/sub><\/td>\n<td>390 mm (90 mm overhung past B)<\/td>\n<\/tr>\n<tr>\n<td>Transmitted torque<\/td>\n<td>T<\/td>\n<td>687 N&middot;m<\/td>\n<\/tr>\n<tr>\n<td>Shaft speed<\/td>\n<td>n<\/td>\n<td>720 rpm (51.8 kW)<\/td>\n<\/tr>\n<tr>\n<td>Gear pitch diameter<\/td>\n<td>d<sub>g<\/sub><\/td>\n<td>200 mm<\/td>\n<\/tr>\n<tr>\n<td>Gear normal pressure angle \/ helix angle<\/td>\n<td>&alpha;<sub>n<\/sub> \/ &beta;<\/td>\n<td>20&deg; \/ 20&deg;<\/td>\n<\/tr>\n<tr>\n<td>Pulley pitch diameter<\/td>\n<td>d<sub>p<\/sub><\/td>\n<td>500 mm<\/td>\n<\/tr>\n<tr>\n<td>Belt tension ratio<\/td>\n<td>r = T1\/T2<\/td>\n<td>3.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We resolve forces into two perpendicular planes. The <strong>vertical plane V<\/strong> carries the<br \/>\ngear&#8217;s separating (radial) force and the belt pull, both acting downward. The <strong>horizontal<br \/>\nplane H<\/strong> carries the gear&#8217;s tangential force. Axial force acts along the shaft and is handled<br \/>\nseparately in step 5. Forces in the two planes never get added to each other &mdash; they get combined<br \/>\nas vectors at the very end.<\/p>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/nu312-bearing.png\" alt=\"NU design cylindrical roller bearing\" \/><figcaption>An NU cylindrical roller bearing is the natural choice for the floating seat: it carries radial load and lets the shaft move axially as it grows with temperature.<\/figcaption><\/figure>\n<h2>4. Step 1 &mdash; torque, power and the three gear forces<\/h2>\n<p>A helical gear does not apply one force to the shaft. It applies three mutually perpendicular<br \/>\ncomponents, and mixing them up is the single most common source of a wrong load case. With torque<br \/>\n<em>T<\/em> and gear pitch diameter <em>d<sub>g<\/sub><\/em>:<\/p>\n<table>\n<thead>\n<tr>\n<th>Component<\/th>\n<th>Formula<\/th>\n<th>Direction<\/th>\n<th>In our example<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Tangential (transmits the torque)<\/td>\n<td>F<sub>t<\/sub> = 2T \/ d<sub>g<\/sub><\/td>\n<td>Horizontal plane H<\/td>\n<td>2 &times; 687 \/ 0.200 = <strong>6.870 kN<\/strong><\/td>\n<\/tr>\n<tr>\n<td>Radial \/ separating<\/td>\n<td>F<sub>r,gear<\/sub> = F<sub>t<\/sub> &middot; tan&alpha;<sub>n<\/sub> \/ cos&beta;<\/td>\n<td>Vertical plane V<\/td>\n<td>6.870 &times; 0.3640 \/ 0.9397 = <strong>2.661 kN<\/strong><\/td>\n<\/tr>\n<tr>\n<td>Axial (thrust from the helix)<\/td>\n<td>F<sub>a<\/sub> = F<sub>t<\/sub> &middot; tan&beta;<\/td>\n<td>Along the shaft<\/td>\n<td>6.870 &times; 0.3640 = <strong>2.500 kN<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Two details matter. First, the radial component uses the <em>normal<\/em> pressure angle divided by<br \/>\ncos&beta;; using tan&alpha;<sub>n<\/sub> alone understates it by about 6% at a 20&deg; helix. Second,<br \/>\nthe axial component exists only because the teeth are helical &mdash; a spur gear has<br \/>\n<em>F<sub>a<\/sub><\/em> = 0 and the whole axial-load question disappears. Reversing the direction of<br \/>\nrotation reverses which side of the bearing sees the thrust, but not its magnitude.<\/p>\n<p>Power is a useful cross-check: <em>T<\/em> &middot; &omega; = 687 &times; (2&pi; &times; 720\/60)<br \/>\n= 51.8 kW. If a customer quotes power and speed instead of torque, invert this first:<br \/>\n<em>T<\/em> = 9550 &middot; P<sub>kW<\/sub> \/ n<sub>rpm<\/sub> N&middot;m.<\/p>\n<p>Combined-load types such as <a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/tapered-roller-bearings\/\">tapered roller bearings<\/a> and<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/spherical-roller-bearings\/\">spherical roller bearings<\/a> exist precisely because a shaft often<br \/>\nhas to carry all three components at once; see our<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/tapered-roller-bearing-selection-guide-and-luoyang-inno-bearings-customized-service\/\">tapered roller bearing selection guide<\/a> for how paired arrangements<br \/>\nhandle the axial part.<\/p>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/Single-Row-Tapered-Roller-Bearing.png\" alt=\"Single row tapered roller bearing\" \/><figcaption>A tapered roller bearing turns the combined radial-plus-axial load case into line contact along the raceways \u2014 which is why its X\/Y factors differ so much from a ball bearing&#8217;s.<\/figcaption><\/figure>\n<h2>5. Step 2 &mdash; belt tension: T1&minus;T2 is not T1+T2<\/h2>\n<p>Here is the step that changes the answer more than any other, and the one almost every supplier<br \/>\nblog omits. A belt transmits torque through the <em>difference<\/em> between tight-side tension T1 and<br \/>\nslack-side tension T2, but the shaft feels their <em>sum<\/em>. Both numbers are needed, and they are<br \/>\nnot interchangeable.<\/p>\n<p style=\"text-align:center;font-size:16px\"><strong>F<sub>e<\/sub> = T1 &minus; T2 = T \/ (d<sub>p<\/sub>\/2)<br \/>\n&nbsp;&nbsp;&nbsp;&nbsp;and&nbsp;&nbsp;&nbsp;&nbsp; F<sub>belt<\/sub> = T1 + T2<\/strong><\/p>\n<table>\n<thead>\n<tr>\n<th>Quantity<\/th>\n<th>Calculation<\/th>\n<th>Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Effective circumferential pull F<sub>e<\/sub><\/td>\n<td>687 \/ 0.250<\/td>\n<td>2.748 kN<\/td>\n<\/tr>\n<tr>\n<td>Slack-side tension T2<\/td>\n<td>F<sub>e<\/sub> \/ (r &minus; 1) = 2.748 \/ 2.2<\/td>\n<td>1.249 kN<\/td>\n<\/tr>\n<tr>\n<td>Tight-side tension T1<\/td>\n<td>r &times; T2 = 3.2 &times; 1.249<\/td>\n<td>3.997 kN<\/td>\n<\/tr>\n<tr>\n<td><strong>Radial pull on the shaft<\/strong> F<sub>belt<\/sub><\/td>\n<td>T1 + T2<\/td>\n<td><strong>5.246 kN<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Note the ratio: the shaft sees 5.246 kN of radial load while transmitting only 2.748 kN of useful<br \/>\ncircumferential force &mdash; nearly twice as much. The excess is installed pre-tension, and it is<br \/>\npresent even at zero torque, which is why a belt drive can brinell a bearing while idling. Over-tension<br \/>\nis a maintenance fault, not a design input, and step 12 quantifies exactly what it costs in life.<\/p>\n<p>The same logic applies to chain drives (without the elastic tension ratio) and to any overhung<br \/>\nload. If the drive is a coupling instead, there is no deliberate radial pull at all, but there is<br \/>\nusually a residual misalignment force &mdash; measure it rather than assume zero.<\/p>\n<h2>6. Step 3 &mdash; static equilibrium in two planes<\/h2>\n<p>Now treat the shaft as a simply supported beam with two supports 300 mm apart. Work each plane<br \/>\nseparately, using &Sigma;F = 0 and &Sigma;M = 0 about support A. All distances are in mm from A.<\/p>\n<p><strong>Vertical plane V<\/strong> &mdash; gear separating force 2.661 kN at x = 100 mm, belt pull<br \/>\n5.246 kN at x = 390 mm:<\/p>\n<p style=\"text-align:center;font-size:15px\">\nR<sub>Av<\/sub> + R<sub>Bv<\/sub> = 2.661 + 5.246 = 7.907 kN<br \/>\nR<sub>Bv<\/sub> &times; 300 = 2.661 &times; 100 + 5.246 &times; <strong>390<\/strong> = 2312.0 kN&middot;mm<br \/>\n&rarr; R<sub>Bv<\/sub> = <strong>7.707 kN<\/strong>, &nbsp; R<sub>Av<\/sub> = <strong>0.200 kN<\/strong>\n<\/p>\n<p><strong>Horizontal plane H<\/strong> &mdash; only the gear tangential force 6.870 kN at x = 100 mm:<\/p>\n<p style=\"text-align:center;font-size:15px\">\nR<sub>Bh<\/sub> &times; 300 = 6.870 &times; 100 = 687.0 &rarr; R<sub>Bh<\/sub> = <strong>2.290 kN<\/strong>,<br \/>\n&nbsp; R<sub>Ah<\/sub> = <strong>4.580 kN<\/strong>\n<\/p>\n<table>\n<thead>\n<tr>\n<th>Plane<\/th>\n<th>Applied loads<\/th>\n<th>Reaction at A<\/th>\n<th>Reaction at B<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Vertical V<\/td>\n<td>2.661 kN @ 100 mm; 5.246 kN @ 390 mm<\/td>\n<td>0.200 kN<\/td>\n<td>7.707 kN<\/td>\n<\/tr>\n<tr>\n<td>Horizontal H<\/td>\n<td>6.870 kN @ 100 mm<\/td>\n<td>4.580 kN<\/td>\n<td>2.290 kN<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The 390 mm term is the whole story. Because the pulley sits <em>outside<\/em> the span, its moment<br \/>\nabout A is larger than a load inside the span would produce, and the only way to balance it is a large<br \/>\nupward reaction at B plus a very small reaction at A. If you wrongly used 300 mm for the pulley<br \/>\nposition, R<sub>Bv<\/sub> would come out at 6.133 kN instead of 7.707 &mdash; a 20% understatement of<br \/>\nthe vertical reaction at the critical bearing. Support A&#8217;s vertical reaction is nearly zero not<br \/>\nbecause there is no load, but because the overhung load is trying to lift it.<\/p>\n<h2>7. Step 4 &mdash; combine the planes into the real Fr<\/h2>\n<p>The four reactions are components, not answers. A bearing sees the resultant of the vertical and<br \/>\nhorizontal components at its seat:<\/p>\n<p style=\"text-align:center;font-size:16px\"><strong>Fr = &radic;(R<sub>v<\/sub><sup>2<\/sup> + R<sub>h<\/sub><sup>2<\/sup>)<\/strong><\/p>\n<table>\n<thead>\n<tr>\n<th>Seat<\/th>\n<th>R<sub>v<\/sub><\/th>\n<th>R<sub>h<\/sub><\/th>\n<th>Resultant Fr<\/th>\n<th>Comment<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>A (floating)<\/td>\n<td>0.200 kN<\/td>\n<td>4.580 kN<\/td>\n<td><strong>4.58 kN<\/strong><\/td>\n<td>Dominated by the gear tangential force<\/td>\n<\/tr>\n<tr>\n<td>B (locating)<\/td>\n<td>7.707 kN<\/td>\n<td>2.290 kN<\/td>\n<td><strong>8.04 kN &asymp; 8.0 kN<\/strong><\/td>\n<td>Dominated by the overhung belt pull; governs the selection<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Adding the components arithmetically at B would give 9.997 kN, a 24% overstatement; averaging all<br \/>\napplied loads across two supports would give 7.389 kN, an 8% understatement of the governing bearing.<br \/>\nBoth are wrong in different directions, which is why an &#8220;about right&#8221; load case is not good enough<br \/>\nonce the life exponent is applied.<\/p>\n<p>We will size from bearing B, the heavier seat, and round to <strong>Fr = 8.0 kN<\/strong>. That is<br \/>\nthe number carried into the <a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/bearing-type-selection-decision-matrix\/\">worked selection example<\/a> in our companion<br \/>\nguide. In practice you would also run the same arithmetic for A and check that the lighter seat still<br \/>\nmeets minimum load and fit requirements.<\/p>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/22220-bearing.png\" alt=\"Spherical roller bearing\" \/><figcaption>At the governing seat, a self-aligning spherical roller bearing is the type that survives both the 8 kN resultant and the frame deflection in the companion selection example.<\/figcaption><\/figure>\n<h2>8. Step 5 &mdash; who carries the axial load?<\/h2>\n<p>The 2.500 kN gear thrust acts along the shaft. In a locating\/floating arrangement it is carried<br \/>\nentirely by <strong>one<\/strong> bearing, the locating one, and then passed into the housing. The<br \/>\nfloating bearing is allowed to slide axially so that thermal growth does not preload the pair.<\/p>\n<table>\n<thead>\n<tr>\n<th>Seat<\/th>\n<th>Role<\/th>\n<th>Fr<\/th>\n<th>Fa<\/th>\n<th>Typical type<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>A<\/td>\n<td>Floating \/ non-locating<\/td>\n<td>4.58 kN<\/td>\n<td><strong>0 kN<\/strong><\/td>\n<td><a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/cylindrical-roller-bearings\/\">NU or N cylindrical roller<\/a>, or a deep groove ball bearing with a sliding outer-ring fit<\/td>\n<\/tr>\n<tr>\n<td>B<\/td>\n<td>Locating<\/td>\n<td>8.04 kN<\/td>\n<td><strong>2.500 kN<\/strong><\/td>\n<td><a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/spherical-roller-bearings\/\">Spherical roller<\/a>, paired <a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/tapered-roller-bearings\/\">tapered roller<\/a>, or <a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/angular-contact-ball-bearings\/\">paired angular contact<\/a> depending on alignment and stiffness<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Two exceptions are worth flagging. If both bearings are arranged as an opposed tapered-roller or<br \/>\nangular-contact pair, the axial load is shared through the pair&#8217;s internal load distribution and the<br \/>\n<em>P<\/em> calculation uses pair-specific factors &mdash; not a simple 50\/50 split. If there is no<br \/>\naxial load at all (spur gear, no thrust), the locating bearing still needs to be defined for shaft<br \/>\npositioning, but <em>Fa<\/em> = 0 and the <em>X\/Y<\/em> choice collapses to the pure-radial case.<\/p>\n<p>Also remember that axial load changes <em>Fa\/Fr<\/em>, which decides the <em>X<\/em> and <em>Y<\/em><br \/>\nfactors in step 7. Here <em>Fa\/Fr<\/em> = 2.5 \/ 8.0 = 0.313. Pure axial cases &mdash; vertical shafts,<br \/>\nhooks, screw presses &mdash; are a different calculation entirely and use<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/thrust-ball-bearings\/\">thrust ball or thrust roller bearings<\/a>; see our<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/what-is-thrust-bearing-and-where-it-is-used\/\">thrust bearing overview<\/a>.<\/p>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/Large-Thrust-Ball-Bearing.png\" alt=\"Large thrust ball bearing\" \/><figcaption>Thrust bearings are designed for a predominantly axial load case; they cannot be dropped into a shaft that also carries 8 kN of radial reaction without a separate radial bearing.<\/figcaption><\/figure>\n<h2>9. Step 6 &mdash; nominal loads versus design loads<\/h2>\n<p>Everything so far is a static, steady-state nominal load. Real machines add shock, start-stop<br \/>\ncycles, and duty segments where load and speed both vary. ISO 281 handles this with an application<br \/>\nfactor f<sub>d<\/sub> applied when forming <em>P<\/em>, not when solving the reactions. Applying it in<br \/>\nboth places would double-count it: our P would be 21.31 kN instead of 16.39 kN, and the predicted<br \/>\nlife would collapse by more than half for no physical reason.<\/p>\n<table>\n<thead>\n<tr>\n<th>Operating character<\/th>\n<th>Illustrative f<sub>d<\/sub><\/th>\n<th>Typical examples<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Smooth, steady, no shock<\/td>\n<td>1.0 &ndash; 1.2<\/td>\n<td>Fans, light conveyors, generators<\/td>\n<\/tr>\n<tr>\n<td>Moderate shock, frequent starts<\/td>\n<td>1.2 &ndash; 1.5<\/td>\n<td>Geared drives, mixers, machine tools (<strong>our example: 1.3<\/strong>)<\/td>\n<\/tr>\n<tr>\n<td>Heavy shock, reversing<\/td>\n<td>1.5 &ndash; 2.0<\/td>\n<td>Crushers, rolling mills, excavators<\/td>\n<\/tr>\n<tr>\n<td>Severe impact \/ very heavy<\/td>\n<td>2.0 &ndash; 3.0<\/td>\n<td>Presses, pile drivers, severe mining duty<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>These bands are illustrative guidance, not a standard table; use the factors published by the<br \/>\nbearing manufacturer for your application class. If the machine runs a genuine duty cycle &mdash; say<br \/>\n40% of the time at full load and 60% at half load &mdash; the rigorous route is a cumulative damage<br \/>\ncalculation: compute <em>P<\/em> for each segment, obtain each segment&#8217;s <em>L<sub>10h<\/sub><\/em>, and<br \/>\ncombine them with the Palmgren&ndash;Miner rule 1\/L = &Sigma;(u<sub>i<\/sub>\/L<sub>i<\/sub>), where<br \/>\nu<sub>i<\/sub> is the fraction of time spent in segment <em>i<\/em>. A single &#8220;average load&#8221; is a<br \/>\nconservative shortcut only when the exponent is applied to the weighted cube (or 10\/3 power) of the<br \/>\nloads, never to the loads themselves.<\/p>\n<h2>10. Step 7 &mdash; equivalent dynamic load P<\/h2>\n<p>ISO 281 reduces a combined radial and axial load to one equivalent radial load that would produce<br \/>\nthe same fatigue life:<\/p>\n<p style=\"text-align:center;font-size:17px\"><strong>P = f<sub>d<\/sub> &middot; (X&middot;F<sub>r<\/sub> + Y&middot;F<sub>a<\/sub>)<\/strong><\/p>\n<p><em>X<\/em> and <em>Y<\/em> depend on the bearing type and on whether <em>Fa\/Fr<\/em> exceeds a<br \/>\nlimiting value <em>e<\/em>. For the governing bearing B, using the same illustrative factors as the<br \/>\ncompanion selection example (<em>e<\/em> = 0.23, <em>X<\/em> = 0.67, <em>Y<\/em> = 2.9 for a spherical<br \/>\nroller bearing):<\/p>\n<ul>\n<li><em>Fa\/Fr<\/em> = 0.313 &gt; <em>e<\/em> = 0.23, so the axial term stays in with both factors.<\/li>\n<li>P = 1.3 &times; (0.67 &times; 8.0 + 2.9 &times; 2.5) = 1.3 &times; (5.36 + 7.25) = 1.3 &times; 12.61 = <strong>16.39 kN<\/strong>.<\/li>\n<\/ul>\n<p>The axial component contributes 7.25 kN of the 12.61 kN bracket &mdash; more than the radial<br \/>\ncomponent &mdash; because <em>Y<\/em> = 2.9 penalises axial load heavily in this bearing family. This<br \/>\nis the step where a wrong <em>Fa<\/em> from step 5 does the most damage. Had we split the axial load<br \/>\nequally (1.25 kN each), P would have been only 11.96 kN and the life prediction would have been<br \/>\nroughly 2.6&times; too optimistic.<\/p>\n<p>For a deep groove ball bearing, <em>e<\/em> and <em>Y<\/em> are themselves functions of<br \/>\n<em>Fa\/C<sub>0<\/sub><\/em>, so you need a trial bearing before you can finish <em>P<\/em>. The full<br \/>\nthree-candidate comparison &mdash; deep groove ball, paired tapered roller and spherical roller,<br \/>\nincluding their <em>X\/Y<\/em> values &mdash; is worked out in our<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/bearing-type-selection-decision-matrix\/\">bearing type selection decision matrix<\/a>; this article stops at the load<br \/>\ncase that feeds it. Factors <em>e<\/em>, <em>X<\/em> and <em>Y<\/em> must always come from the actual<br \/>\nmanufacturer&#8217;s catalogue, never from memory.<\/p>\n<h2>11. Step 8 &mdash; L10h and the capacity you actually need<\/h2>\n<p>ISO 281 basic rating life in million revolutions is <em>L<sub>10<\/sub><\/em> = (C\/P)<sup>p<\/sup>,<br \/>\nwith <em>p<\/em> = 3 for ball bearings and 10\/3 for roller bearings. Converting to hours at speed<br \/>\n<em>n<\/em>:<\/p>\n<p style=\"text-align:center;font-size:17px\"><strong>L<sub>10h<\/sub> = (10<sup>6<\/sup> \/ 60n) &middot; (C\/P)<sup>p<\/sup><\/strong><\/p>\n<p>At 720 rpm, 10<sup>6<\/sup>\/(60 &times; 720) = <strong>23.15 h per million revolutions<\/strong>.<br \/>\nWith the illustrative <em>C<\/em> = 178 kN for a 22214 spherical roller bearing, <em>P<\/em> = 16.39 kN<br \/>\nand <em>p<\/em> = 10\/3:<\/p>\n<p style=\"text-align:center;font-size:16px\">L<sub>10h<\/sub> = 23.15 &times; (178 \/ 16.39)<sup>10\/3<\/sup> = <strong>65,623 h<\/strong><\/p>\n<p>Rearranged for a 20,000 h target: <em>C<sub>req<\/sub><\/em> = P &middot; (20,000 \/ 23.15)<sup>1\/p<\/sup><br \/>\n= 16.39 &times; 7.60 = <strong>124.6 kN<\/strong>. The candidate with <em>C<\/em> = 178 kN clears the<br \/>\ntarget by roughly three times, which is where the companion guide takes over and checks the<br \/>\nconstraints that are <em>not<\/em> about life &mdash; 0.4&deg; of frame misalignment, a 150 mm outside<br \/>\ndiameter envelope, speed and lubrication.<\/p>\n<table>\n<thead>\n<tr>\n<th>Stage<\/th>\n<th>Value<\/th>\n<th>Where it came from<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Torque \/ speed<\/td>\n<td>687 N&middot;m \/ 720 rpm<\/td>\n<td>Step 1 input<\/td>\n<\/tr>\n<tr>\n<td>Fr \/ Fa at bearing B<\/td>\n<td>8.0 \/ 2.5 kN<\/td>\n<td>Steps 1&ndash;5<\/td>\n<\/tr>\n<tr>\n<td>f<sub>d<\/sub><\/td>\n<td>1.3<\/td>\n<td>Step 6<\/td>\n<\/tr>\n<tr>\n<td>P<\/td>\n<td>16.39 kN<\/td>\n<td>Step 7<\/td>\n<\/tr>\n<tr>\n<td>L<sub>10h<\/sub> at C = 178 kN<\/td>\n<td>65,623 h<\/td>\n<td>Step 8<\/td>\n<\/tr>\n<tr>\n<td>C<sub>req<\/sub> for 20,000 h<\/td>\n<td>124.6 kN<\/td>\n<td>Step 8, rearranged<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>That chain &mdash; from a quoted torque to a required catalogue rating &mdash; is the deliverable.<br \/>\nIf a customer changes the pulley diameter, the belt pre-tension, the gear helix angle or the support<br \/>\nspan, every row below the change has to be recomputed; none of them can be reused on assumption.<\/p>\n<h2>12. Sensitivity: belt pre-tension moves life more than torque does<\/h2>\n<p>This is the calculation competitors never show, and the one that usually changes a maintenance<br \/>\ndecision. Keep torque, geometry and <em>Fa<\/em> fixed, and vary only the belt tension ratio<br \/>\n<em>r<\/em> = T1\/T2. A lower ratio means more installed pre-tension for the same transmitted torque<br \/>\n&mdash; the shaft never sees a newton of extra useful work, but it carries a lot more radial load.<\/p>\n<table>\n<thead>\n<tr>\n<th>r = T1\/T2<\/th>\n<th>Belt pull T1+T2 (kN)<\/th>\n<th>R<sub>Bv<\/sub> (kN)<\/th>\n<th>Fr<sub>B<\/sub> (kN)<\/th>\n<th>P (kN)<\/th>\n<th>L<sub>10h<\/sub> (h)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>2.0 (over-tensioned)<\/td>\n<td>8.244<\/td>\n<td>11.604<\/td>\n<td>11.828<\/td>\n<td>19.73<\/td>\n<td><strong>35,402<\/strong><\/td>\n<\/tr>\n<tr>\n<td>2.5<\/td>\n<td>6.412<\/td>\n<td>9.223<\/td>\n<td>9.503<\/td>\n<td>17.70<\/td>\n<td>50,799<\/td>\n<\/tr>\n<tr>\n<td><strong>3.2 (design case)<\/strong><\/td>\n<td>5.246<\/td>\n<td>7.707<\/td>\n<td>8.040<\/td>\n<td>16.43<\/td>\n<td><strong>65,159<\/strong><\/td>\n<\/tr>\n<tr>\n<td>4.0<\/td>\n<td>4.580<\/td>\n<td>6.841<\/td>\n<td>7.214<\/td>\n<td>15.71<\/td>\n<td>75,649<\/td>\n<\/tr>\n<tr>\n<td>5.0 (loose, slip risk)<\/td>\n<td>4.122<\/td>\n<td>6.246<\/td>\n<td>6.652<\/td>\n<td>15.22<\/td>\n<td><strong>84,067<\/strong><\/td>\n<\/tr>\n<tr>\n<td>No belt load (reference only)<\/td>\n<td>0<\/td>\n<td>0.887<\/td>\n<td>2.456<\/td>\n<td>11.56<\/td>\n<td>210,002<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Going from r = 5.0 to r = 2.0 &mdash; a plausible range of how the same belt might be tensioned in<br \/>\nthe field &mdash; cuts predicted life from 84,067 h to 35,402 h, a <strong>58% reduction<\/strong>,<br \/>\nwith no change in transmitted power. The rows use the unrounded <em>Fr<sub>B<\/sub><\/em> = 8.040 kN,<br \/>\nwhich is why the design-case row reads 65,159 h while the round-number chain in step 11 gives<br \/>\n65,623 h; the difference is 0.7% and is purely rounding.<\/p>\n<p>The practical conclusions: specify a tension ratio, measure belt tension at installation and<br \/>\nre-check it after run-in, and never &#8220;tighten it until it stops squeaking&#8221;. If a bearing at an<br \/>\noverhung pulley fails early with no change in duty, the belt tension is the first thing to measure.<br \/>\nLubrication and clearance matter too &mdash; see our notes on<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/the-importance-of-bearing-clearance\/\">bearing clearance<\/a> and<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/things-to-note-when-choosing-bearing-lubricant\/\">choosing a bearing lubricant<\/a> &mdash; but they cannot recover a life deficit<br \/>\ncreated by a 3 kN surplus of radial load.<\/p>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/Inch-series-four-row-tapered-roller-bearings-1.png\" alt=\"Four row tapered roller bearing\" \/><figcaption>In heavy-duty rolling mill and crusher shafts the same chain applies with much larger forces; see our <a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/selecting-the-right-rolling-mill-bearings-is-crucial-for-steel-mills-luoyang-inno-bearings-safeguards-production\/\">rolling mill bearing notes<\/a> for that load class.<\/figcaption><\/figure>\n<h2>13. Eight mistakes that quietly halve (or double) the answer<\/h2>\n<p>Each row below was computed from the same example, so the size of the error is visible. The last<br \/>\ncolumn shows what the mistake does to the governing bearing&#8217;s load or life.<\/p>\n<table>\n<thead>\n<tr>\n<th>Mistake<\/th>\n<th>Wrong value<\/th>\n<th>Correct value<\/th>\n<th>Effect<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Use T1+T2 as the transmitted force<\/td>\n<td>5.246 kN &#8220;circumferential&#8221;<\/td>\n<td>2.748 kN<\/td>\n<td>Nearly 2&times; overstates torque \/ understates pre-tension<\/td>\n<\/tr>\n<tr>\n<td>Average all applied loads over two supports<\/td>\n<td>7.389 kN each<\/td>\n<td>A 4.58 \/ B 8.04 kN<\/td>\n<td>8% low at B, 61% high at A<\/td>\n<\/tr>\n<tr>\n<td>Add the two planes arithmetically<\/td>\n<td>9.997 kN at B<\/td>\n<td>8.040 kN<\/td>\n<td>24% high; life understated by ~2&times;<\/td>\n<\/tr>\n<tr>\n<td>Use 300 mm instead of 390 mm for the overhung pulley<\/td>\n<td>R<sub>Bv<\/sub> = 6.133 kN<\/td>\n<td>7.707 kN<\/td>\n<td>20% low on the vertical reaction<\/td>\n<\/tr>\n<tr>\n<td>Split the axial load 50\/50<\/td>\n<td>1.250 kN each<\/td>\n<td>2.500 \/ 0 kN<\/td>\n<td>P understated ~27%; life overstated ~2.6&times;<\/td>\n<\/tr>\n<tr>\n<td>Apply f<sub>d<\/sub> to reactions and again to P<\/td>\n<td>P = 21.31 kN<\/td>\n<td>16.39 kN<\/td>\n<td>Life understated by ~55%<\/td>\n<\/tr>\n<tr>\n<td>Use the ball exponent p = 3 for a roller bearing<\/td>\n<td>wrong (C\/P)<sup>3<\/sup><\/td>\n<td>(C\/P)<sup>10\/3<\/sup><\/td>\n<td>Life error grows with C\/P<\/td>\n<\/tr>\n<tr>\n<td>Assume steady full load instead of a duty cycle<\/td>\n<td>one P for all hours<\/td>\n<td>segment-wise P and Miner&#8217;s rule<\/td>\n<td>Can overstate or understate by multiples<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that the errors do not cancel: several of them push in the same direction. A load case built<br \/>\nfrom two or three shortcuts is not &#8220;approximately right&#8221;, it is simply a different machine.<\/p>\n<h2>14. What to measure before you calculate<\/h2>\n<p>The arithmetic is only as good as the inputs. This is the shortlist we ask customers for when a<br \/>\nload case has to be rebuilt from an existing machine rather than designed from scratch.<\/p>\n<table>\n<thead>\n<tr>\n<th>Input<\/th>\n<th>Why it is needed<\/th>\n<th>How it is usually obtained<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Power \/ torque and speed, plus duty segments<\/td>\n<td>Sets F<sub>t<\/sub> and L<sub>10h<\/sub><\/td>\n<td>Motor nameplate, drive log, or measured current<\/td>\n<\/tr>\n<tr>\n<td>Gear: pitch diameter, &alpha;<sub>n<\/sub>, &beta;, hand of helix<\/td>\n<td>Splits torque into Ft, Fr, Fa<\/td>\n<td>Gear drawing or tooth count + module<\/td>\n<\/tr>\n<tr>\n<td>Pulley \/ sprocket diameter, overhang, belt type and tension<\/td>\n<td>Dominates the reaction at the adjacent bearing<\/td>\n<td>Layout drawing; tension gauge at installation<\/td>\n<\/tr>\n<tr>\n<td>Distance between bearing centres and all load positions<\/td>\n<td>Moment arms for &Sigma;M = 0<\/td>\n<td>Assembly drawing<\/td>\n<\/tr>\n<tr>\n<td>Which bearing locates axially; thermal growth allowance<\/td>\n<td>Decides Fa allocation<\/td>\n<td>Arrangement drawing; see <a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/spherical-cylindrical-roller-bearings-the-core-solution-for-industrial-transmission\/\">roller bearing arrangements<\/a><\/td>\n<\/tr>\n<tr>\n<td>Shock \/ start frequency, reversing, ambient temperature<\/td>\n<td>f<sub>d<\/sub>, clearance class, lubrication<\/td>\n<td>Operating history<\/td>\n<\/tr>\n<tr>\n<td>Available envelope: bore, OD, width<\/td>\n<td>Eliminates candidates after life is computed<\/td>\n<td>Housing measurement<\/td>\n<\/tr>\n<tr>\n<td>Existing bearing designation and failure appearance<\/td>\n<td>Sanity check on the computed load direction<\/td>\n<td>Photos of the failed part<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For structures that rotate slowly under a tilting moment rather than a spinning shaft, the whole<br \/>\nmodel changes: the load case is built from axial force, radial force and tilting moment at the<br \/>\nmounting plane, not from gear and belt reactions. Our<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/how-to-choose-the-right-slewing-bearing\/\">slewing bearing selection overview<\/a> and<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/crossed-roller-vs-ball-slewing-bearings\/\">crossed roller vs ball slewing comparison<\/a> cover that case, and the<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/what-is-the-difference-between-rolling-bearings-and-slewing-bearings\/\">rolling bearing vs slewing bearing boundary<\/a> explains where the<br \/>\ntwo approaches diverge.<\/p>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/cross-roller-slewing-bearing-2.png\" alt=\"Crossed roller slewing bearing\" \/><figcaption>Slewing rings are sized from axial force, radial force and tilting moment at the bolt circle \u2014 a different load chain from the shaft reactions computed here.<\/figcaption><\/figure>\n<h2>Send us the shaft layout, get the load case back<\/h2>\n<p>The chain above is what our engineers run before quoting a bearing: gear and belt data, support<br \/>\ngeometry, which seat locates axially, duty cycle, target life and envelope. Give us those inputs and<br \/>\nwe will return the computed <em>Fr<\/em> and <em>Fa<\/em> at each seat, the equivalent dynamic load<br \/>\n<em>P<\/em>, the resulting <em>L<sub>10h<\/sub><\/em> and the shortlist of types that fit &mdash; including<br \/>\nwhat would have to change in the shaft or frame if none of them does.<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/contact-us\/\">Send your operating data to our engineering team<\/a>.<\/p>\n<p>You can also browse <a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/deep-groove-ball-bearings\/\">deep groove ball<\/a>,<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/tapered-roller-bearings\/\">tapered roller<\/a>,<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/cylindrical-roller-bearings\/\">cylindrical roller<\/a>,<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/spherical-roller-bearings\/\">spherical roller<\/a>,<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/angular-contact-ball-bearings\/\">angular contact<\/a>,<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/thrust-ball-bearings\/\">thrust<\/a> and<br \/>\n<a href=\"https:\/\/www.lycbearingcn.com\/ru\/products\/thin-section-bearings\/\">thin-section<\/a> bearings for the families that appear in this load chain.<\/p>\n<h2>Related reading<\/h2>\n<ul>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/bearing-type-selection-decision-matrix\/\">Bearing Type Selection: A Decision Matrix with a Worked Example<\/a> &mdash; takes this exact load case (Fr 8.0 kN, Fa 2.5 kN, 720 rpm, 20,000 h) through P, L10h and three candidate bearings.<\/li>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/tapered-roller-bearing-selection-guide-and-luoyang-inno-bearings-customized-service\/\">Tapered Roller Bearing Selection Guide<\/a> &mdash; paired arrangements, preload and endplay for the locating seat.<\/li>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/spherical-cylindrical-roller-bearings-the-core-solution-for-industrial-transmission\/\">Spherical &amp; Cylindrical Roller Bearings<\/a> &mdash; the two high-capacity roller families and where they belong in an arrangement.<\/li>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/characteristics-and-differences-between-angular-contact-bearings-and-deep-groove-ball-bearings\/\">Angular Contact vs Deep Groove Ball Bearings<\/a> &mdash; how the X\/Y factors differ when Fa\/Fr rises.<\/li>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/the-importance-of-bearing-clearance\/\">\u0412\u0430\u0436\u043d\u043e\u0441\u0442\u044c \u0437\u0430\u0437\u043e\u0440\u0430 \u0432 \u043f\u043e\u0434\u0448\u0438\u043f\u043d\u0438\u043a\u0430\u0445<\/a> &mdash; what fits and temperature do to the clearance you selected.<\/li>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/the-types-of-ball-bearings\/\">The Types of Ball Bearings<\/a> &mdash; a closer look at the ball-bearing families.<\/li>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/crossed-roller-vs-ball-slewing-bearings\/\">Crossed Roller vs Ball Slewing Bearings<\/a> &mdash; the decision-matrix approach applied to slewing rings.<\/li>\n<li><a href=\"https:\/\/www.lycbearingcn.com\/ru\/news\/how-to-choose-the-right-slewing-bearing\/\">How to Choose the Right Slewing Bearing<\/a> &mdash; the alternative load chain for slow-rotating structures.<\/li>\n<\/ul>\n<figure>\n  <img decoding=\"async\" src=\"https:\/\/www.lycbearingcn.com\/wp-content\/uploads\/2026\/06\/Angular-Contact-Ball-Bearing.png\" alt=\"Angular contact ball bearing\" \/><figcaption>Paired angular contact bearings are one option for the locating seat, but their X\/Y factors and preload sensitivity must be checked against the computed Fa\/Fr.<\/figcaption><\/figure>\n<h2>Frequently Asked Questions<\/h2>\n<h3>How do you calculate the radial load on a bearing?<\/h3>\n<p>Draw a free-body diagram of the shaft, resolve all gear, belt, chain or coupling forces into two perpendicular planes, solve static equilibrium (sum of forces and moments about one support) in each plane, then take the vector sum of the two reaction components at that bearing seat: Fr = sqrt(Rv^2 + Rh^2). The result is the reaction at the seat, not the force applied at the gear.<\/p>\n<h3>Is belt tension T1+T2 or T1-T2?<\/h3>\n<p>Both, for different purposes. T1-T2 is the effective circumferential force that transmits torque (T1-T2 = T\/(d\/2)). T1+T2 is the total radial pull the shaft and bearings actually carry. With a tension ratio of 3.2, a belt transmitting 2.748 kN puts 5.246 kN of radial load on the shaft.<\/p>\n<h3>Which bearing takes the axial load?<\/h3>\n<p>In a locating\/floating arrangement the axial load goes entirely to the locating bearing; the floating bearing is free to move axially to accommodate thermal growth. Splitting axial load equally between the two supports understates P at the locating bearing and can overstate life by more than two times. Opposed tapered or angular contact pairs use pair-specific factors instead.<\/p>\n<h3>What is equivalent dynamic load P?<\/h3>\n<p>P is the single radial load that would give the same fatigue life as the actual combined radial and axial loading. It is P = fd*(X*Fr + Y*Fa), where X and Y depend on the bearing type and on whether Fa\/Fr exceeds the limit e. Because life scales with (C\/P)^p, an error in P is amplified.<\/p>\n<h3>How does an overhung pulley affect bearing loads?<\/h3>\n<p>An overhung load acts outside the bearing span, so its moment about the far support is larger than the same load inside the span. The near support reaction rises sharply while the far support reaction can approach zero or reverse. In our example a 5.246 kN belt pull at 90 mm overhang produces a 7.707 kN vertical reaction at the adjacent bearing.<\/p>\n<h3>Should I apply a service factor to the reaction forces?<\/h3>\n<p>Apply the application factor fd once, when forming the equivalent dynamic load P. Applying it to the static reactions and again to P double-counts the shock allowance and can understate life by more than half. For a real duty cycle, compute P per segment and combine lives with Miner&#x27;s rule.<\/p>\n<h3>How do I get torque from motor power and speed?<\/h3>\n<p>T (N.m) = 9550 * P(kW) \/ n(rpm). For 51.8 kW at 720 rpm that is 687 N.m. Torque then gives the gear tangential force Ft = 2T\/d, from which the radial and axial components follow using the normal pressure angle and helix angle.<\/p>\n<h3>Why does more belt tension shorten bearing life?<\/h3>\n<p>Extra pre-tension adds radial load without adding transmitted power. In the worked example, lowering the tension ratio from 5.0 to 2.0 raises belt pull from 4.122 to 8.244 kN, raises the governing bearing load from 6.652 to 11.828 kN, and cuts calculated L10h from 84,067 to 35,402 h.<\/p>\n<p><script type=\"application\/ld+json\">{\"@context\": \"https:\/\/schema.org\", \"@type\": \"FAQPage\", \"mainEntity\": [{\"@type\": \"Question\", \"name\": \"How do you calculate the radial load on a bearing?\", \"acceptedAnswer\": {\"@type\": \"Answer\", \"text\": \"Draw a free-body diagram of the shaft, resolve all gear, belt, chain or coupling forces into two perpendicular planes, solve static equilibrium (sum of forces and moments about one support) in each plane, then take the vector sum of the two reaction components at that bearing seat: Fr = sqrt(Rv^2 + Rh^2). The result is the reaction at the seat, not the force applied at the gear.\"}}, {\"@type\": \"Question\", \"name\": \"Is belt tension T1+T2 or T1-T2?\", \"acceptedAnswer\": {\"@type\": \"Answer\", \"text\": \"Both, for different purposes. T1-T2 is the effective circumferential force that transmits torque (T1-T2 = T\/(d\/2)). T1+T2 is the total radial pull the shaft and bearings actually carry. With a tension ratio of 3.2, a belt transmitting 2.748 kN puts 5.246 kN of radial load on the shaft.\"}}, {\"@type\": \"Question\", \"name\": \"Which bearing takes the axial load?\", \"acceptedAnswer\": {\"@type\": \"Answer\", \"text\": \"In a locating\/floating arrangement the axial load goes entirely to the locating bearing; the floating bearing is free to move axially to accommodate thermal growth. Splitting axial load equally between the two supports understates P at the locating bearing and can overstate life by more than two times. 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In our example a 5.246 kN belt pull at 90 mm overhang produces a 7.707 kN vertical reaction at the adjacent bearing.\"}}, {\"@type\": \"Question\", \"name\": \"Should I apply a service factor to the reaction forces?\", \"acceptedAnswer\": {\"@type\": \"Answer\", \"text\": \"Apply the application factor fd once, when forming the equivalent dynamic load P. Applying it to the static reactions and again to P double-counts the shock allowance and can understate life by more than half. For a real duty cycle, compute P per segment and combine lives with Miner's rule.\"}}, {\"@type\": \"Question\", \"name\": \"How do I get torque from motor power and speed?\", \"acceptedAnswer\": {\"@type\": \"Answer\", \"text\": \"T (N.m) = 9550 * P(kW) \/ n(rpm). For 51.8 kW at 720 rpm that is 687 N.m. Torque then gives the gear tangential force Ft = 2T\/d, from which the radial and axial components follow using the normal pressure angle and helix angle.\"}}, {\"@type\": \"Question\", \"name\": \"Why does more belt tension shorten bearing life?\", \"acceptedAnswer\": {\"@type\": \"Answer\", \"text\": \"Extra pre-tension adds radial load without adding transmitted power. In the worked example, lowering the tension ratio from 5.0 to 2.0 raises belt pull from 4.122 to 8.244 kN, raises the governing bearing load from 6.652 to 11.828 kN, and cuts calculated L10h from 84,067 to 35,402 h.\"}}]}<\/script><br \/>\n<script type=\"application\/ld+json\">{\"@context\": \"https:\/\/schema.org\", \"@type\": \"Article\", \"headline\": \"Bearing Load Calculation: From Shaft Forces to Fr, Fa and L10\", \"description\": \"Bearing load calculation, fully worked end to end: helical gear and belt forces, two-plane reaction equilibrium, Fr and Fa, equivalent load P and L10 life.\", \"author\": {\"@type\": \"Organization\", \"name\": \"LUOYANG INNO BEARING CO., LTD\"}, \"publisher\": {\"@type\": \"Organization\", \"name\": \"LUOYANG INNO BEARING CO., LTD\"}, \"mainEntityOfPage\": {\"@type\": \"WebPage\", \"@id\": \"https:\/\/www.lycbearingcn.com\/news\/bearing-load-calculation-shaft-forces-to-l10\/\"}}<\/script><\/p>","protected":false},"excerpt":{"rendered":"<p>Bearing load calculation, fully worked end to end: helical gear and belt forces, two-plane reaction equilibrium, Fr and Fa, equivalent load P and L10 life.<\/p>","protected":false},"author":1,"featured_media":2334,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[13,12,11,14],"_links":{"self":[{"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/posts\/2480"}],"collection":[{"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/comments?post=2480"}],"version-history":[{"count":1,"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/posts\/2480\/revisions"}],"predecessor-version":[{"id":2481,"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/posts\/2480\/revisions\/2481"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/media\/2334"}],"wp:attachment":[{"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/media?parent=2480"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/categories?post=2480"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.lycbearingcn.com\/ru\/wp-json\/wp\/v2\/tags?post=2480"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}