Tech Note 2, Iss. 1
Calculation of wheel bearings loads Factors necessary for bearing load calculation in wheel applications vary considerably. To obtain an exact expression for the resultant bearing loads is consequently impossible. Values based on experience with previous bearing arrangements are usually used for new designs. The calculation for front and rear hub bearing arrangements are normally similar; no consideration is taken of the fact that one pair of wheels is driven. Only in special circumstances is account taken of the power transmitted. The following survey is based on a hub bearing arrangement, incorporating two taper roller bearings as shown in Fig. 1, but the method of calculation also applies to other bearing arrangements having two pressure centres, for example, using two single row angular contact ball bearings or one double row. The static hub load K for a loaded vehicle is obtained from
K = K ' − K '' where K ' is the force between tyre and road (half axle load)
K '' is the weight of one wheel When rough operating conditions have to be considered, the static load K is increased by 20 per cent. The effect of the camber angle (in Fig. 1, the angle of the line III with the horizontal plane) may usually be ignored, as well as the influence of the driving and rolling resistance forces, which are small compared to the vertical forced caused by gravity. However, where the camber angle is considerable the axial component of the static load K must be taken into account. In addition to the static load it is necessary to consider other loads, both radial and axial, which are due, for instance, to road surface irregularities. The resultant bearing loads obviously depend on wheel diameter and the 1 / 11
SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
distance between the bearings and are taken into account by an additional force fK. On a straight route the following radial bearing loads should thus be considered, the axial loads ignored.
FrI1 = ε 1 K + ε 2 fK FrII1
= (1 − ε 1 ) K ± ε 2 fK
2 .1
Indexes I and II denote the inner and outer bearing respectively and with reference to Fig. 1, ε1 = a / l and ε 2 = RH / l . In the second equation the plus
sign applies when ε1 < 1 and the minus sign when ε1 > 1 . The value recommended for the coefficient f is 0.05 for private cars and commercial vehicles.
When cornering a centrifugal force K d kg acts at the centre of gravity of the vehicle; see Fig. 2a. Its magnitude is obtained from
K
2 d v = 1 × 127 G r
where: G is the maximum weight of the vehicle less the weight of the wheel, kg v is the speed of the vehicle in km/h r is the radius of the curve in the road, m
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
Fig. 2a also shows the effect the forces G and K d have on the front and rear wheels.
The forces on the front wheel and distances referred to below are given in Fig. 2b. Equilibrium is obtained when
)K K i = (1 − 2 bh KGd ) K K ae + K ai = xK d K e = (1 + 2 bh
Kd G)
2 .2
When the vehicle is about to overturn, the forces Ki and Kai on the inside b wheel are zero, which gives Ke = 2 K, and, since K = xG 2 then K ae = h K Equation 2.2 may also be applied to the rear wheels if factor x is replaced by (1 – x). If conditions of friction are assumed to be the same for the outside and the inside wheel the axial forces will be
K ae =
Kd
K ai =
Kd
G
G
h Kd
(1 + 2 b
G
h Kd
(1 − 2 b
G
) K ) K
2.3
Using the radial and axial forces obtained in Equation 2.2 and 2.3 the bearing loads when negotiating corners can be calculated as follows. For the outside wheel hub bearings
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
FrI 2 = ε 1 K e + ε 2 K ae
FrII 2 = (1 − ε 1 ) K e − ε 2 K ae K a = K ae
2 .4
Where Kae (direction I-II, case 2b or 2c, from 'Axial loading of taper roller bearings' table in taper bearing section of SKF General Catalogue. For the inside wheel hub bearing
FrI 3 = ε 1 K i − ε 2 K ai
FrII3 = (1 − ε 1 ) K i + ε 2 K ai K a = K ai
2 .5
Where Kai (direction II-I, case 1b or 1c, from 'Axial loading of taper roller bearings' table in taper bearing section of SKF General Catalogue. Normally it is assumed the Kd/G = 0,25, which applies when for instance driving at a speed of 40 km/h round a curve having a radius of 50 m or at 20 km/h round a curve having a radius of 12 m. If, in addition, h/b = 0,5 Equations 2.2 and 2.3 give
K e = 1.25 K K i = 0.75 K K ae = 0.31K K ai = 0.19 K
2.6
The equivalent bearing loads for bearing I when driving straight ahead (PI1), round a left-hand curve (PI2) and round a right-hand curve (PI3) are used to determine the mean equivalent load PIm. If it is assumed that 90 per cent of the route is straight, 5 per cent curved to the left and 5 percent curved to the right, therefore
PIm = 3 0.90 PI13 + 0.05 PI32 + 0.05 PI33 [N ]
2 .7
For bearing II the loads are calculated similarly. Finally, if the rolling radius of the wheels is RH mm, the bearing life will be
Ls = 2πRH ( CP ) p [km]
2 .8
This simplified wheel calculation is offered as a guide for bearing selection. SKF employ advanced calculation programs to more accurately determine bearing life and performance.
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
Example 1 A truck front wheel hub using a pair of taper roller bearings:Inboard, 32310: CI=161000N, eI=0.35, XI=0.4, YI=1.7 Outboard, 32307: CII=89700N, eII=0.31, XII=0.4, YII=1.9 Static hub load, K = 20000N Wheel radius, RH = 400mm Distance between pressure centres, l = 100mm Load line, a = 90mm (from outboard pressure centre towards car centre) Thus
ε1 =
a = 0.9 l
Assuming
&
Kd = 0.25 G
ε2 = &
RH =4 l
h = 0.5 b
Calculate the lives of the bearings. When driving straight ahead - Equation 2.1 gives
FrI 1 = (0.9 × 20000) + (4 × 0.05 × 20000) = 22000 N FrII 1 = ((1 − 0.9)20000) + (4 × 0.05 × 20000) = 6000 N Referring to 'Axial loading of taper roller bearings' table from the General Catalogue,
FrI 1 FrII 1 > YI YII
and
Ka = 0
therefore case 1a applies, so the internal
axial load becomes
FaI 1 =
0.5 FrI 1 0.05 × 22000 = = 6470 N = FaII 1 YI 1.7
Hence
FaI 1 6470 = = 0.29 (< eI ) FrI 1 22000
and
FaII 1 6470 = = 1.08 (> eII ) FrII 1 6000
The equivalent bearing loads become
PI 1 = FrI 1 = 22000 N PII 1 = X II FrII 1 + YII FaII 1 = (0.4 × 6000) + (1.9 × 6470) = 14690 N When cornering - outside wheel - Equation 2.6 gives 5 / 11
SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
K e = 1.25 × 20000 = 25000 N K ae = 0.31 × 20000 = 6250 N From Equations 2.4
FrI 2 = 0.9 × 25000 + 4 × 6250 = 47500 N FrII 2 = (1 − 0.9) × 25000 − 4 × 6250 = (−)22500 N Referring to 'Axial loading of taper roller bearings' table from the General Catalogue, case 2c applies since
FrI 2 FrII 2 > YI YII
and
F F 0.5 rI 2 − rII 2 = 8050 > K ae YII YI
Consequently
FaI 2 =
0.5 × FrI 2 0.5 × 47500 = = 13970 N YI 1.7
FaII 2 = FaI 2 − K ae = 13970 − 6250 = 7720 N Hence
FaI 2 13970 = = 0.29 (< eI ) FrI 2 47500
and
FaII 2 7720 = = 0.34 (> eII ) FrII 2 22500
The following equivalent loads are obtained
PI 2 = FrI 2 = 47500 N PII 2 = X II FrII 2 + YII FaII 2 = (0.4 × 22500) + (1.9 × 7720) = 23670 N When cornering - inside wheel - Equation 2.6 gives
K i = 0.75 × 20000 = 15000 N K ai = 0.19 × 20000 = 3750 N From Equations 2.5
FrI 3 = 0.9 × 15000 + 4 × 3750 = (−)1500 N FrII 3 = (1 − 0.9) × 15000 − 4 × 3750 = 16500 N Referring to 'Axial loading of taper roller bearings' table from the General Catalogue, case 1c applies since
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
FrII 3 FrI 3 > YII YI
and
F F 0.5 rII 3 − rI 3 = 3900 > K ai YI YII
Consequently
FaII 3 =
0.5 × FrII 3 0.5 × 16500 = = 4340 N YII 1.9
FaI 3 = FaII 3 − K ai = 4340 − 3750 = 590 N Hence
FaI 3 590 = = 0.39 (> eI ) FrI 3 1500
and
FaII 3 4340 = = 0.26 (< eII ) FrII 3 16500
The following equivalent loads are obtained
PI 3 = X II FrI 3 + YII FaI 3 = (0.4 × 1500) + (1.7 × 590) = 1600 N PII 3 = FrII 3 = 16500 N Summary of equivalent dynamic loads in N Inboard Brg I
Outboard Brg II
Straight ahead driving
22000
14690
Cornering, Outer Wheel
47500
23670
Cornering, Inner Wheel
1600
16500
The mean equivalent bearing load is determined using equation 2.7
PIm = 3 0.9 × 22000 3 + 0.05 × 47500 3 + 0.05 × 1600 3 = 25090 N PI Im = 3 0.9 × 14690 3 + 0.05 × 23670 3 + 0.05 × 16500 3 = 15590 N
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
The corresponding bearing lives are 10 3
10
C 161000 3 LI = π 2 RH I = π 2 × 400 = 1,234,000 Km P 25090 Im 10 3
10
C 89700 3 LII = π 2 RH II = π 2 × 400 = 858,000 Km P 15590 I Im Example 2 A single seater racer car uses a double row angular contact bearing (HBU1). This will be treated as 2 separate rows with the following data:Designation BA2B 633816 CI and CII rating = 27600 per row (44900N per unit) eI, eII = 0.86,
XI, XII = 0.38,
YI, YII = 0.72
Static hub, K = 800N Wheel radius, RH = 250mm Distance between pressure centres, l = 51mm Load line, a = 27mm (from outboard pressure centre towards car centre) Distance from road to vehicle centre of gravity, h = 300mm Track distance, b = 1800mm Thus
ε1 =
Assuming
R a = 0.53 , ε 2 = H = 4.9 l l
&
h = 0.167 b
Kd = 0.25 G
Calculate the lives of the bearings. When driving straight ahead - Equation 2.1 gives
FrI 1 = (0.53 × 800) + (4.9 × 0.05 × 800) = 620 N FrII 1 = ((1 − 0.53)800) + (4.9 × 0.05 × 800) = 572 N Referring to 'Axial loading of single row angular contact ball bearings' table from the General Catalogue, FrI 1 > FrII 1 and K a = 0 therefore case 1a applies and the internal axial load becomes
FaI 1 = eI × FrI 1 = 0.86 × 620 = 533 N = FaII 1
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
Hence
FaI 1 533 = = 0.86 (= eI ) FrI 1 620
and
FaII 533 = = 0.93 (> eII ) FrII 1 572
The equivalent bearing loads become
PI 1 = FrI 1 = 620 N PII 1 = X II FrII 1 + YII FaII 1 = (0.38 × 572) + (0.72 × 533) = 601N When cornering - outside wheel - Equation 2.2 and 2.3 gives
K e = (1 + (2 × 0.167 × 0.25) )800 = 867 N
K ae = 0.25(1 + (2 × 0.167 × 0.25) )800 = 217 N From Equations 2.4
FrI 2 = 0.53 × 867 + 4.9 × 217 = 1523 N FrII 2 = (1 − 0.53) × 867 − 4.9 × 217 = (−)656 N Referring to 'Axial loading of single row angular contact ball bearings' table from the General Catalogue, case 2c applies since
FrI 2 > FrII 2
and
0.86(FrI 2 − FrII 2 ) = 746 > K ae
Consequently
FaI 2 = eI × FrI 2 = 0.86 × 1523 = 1310 N FaII 2 = FaI 2 − K ae = 1310 − 217 = 1093 N Hence
FaI 2 1310 = = 0.86 (= eI ) FrI 2 1523
and
FaII 2 1093 = = 1.66 (> eII ) FrII 2 656
The following equivalent loads are obtained
PI 2 = FrI 2 = 1523 N PII 2 = X II FrII 2 + YII FaII 2 = (0.38 × 656) + (0.72 × 1093) = 1036 N When cornering - inside wheel - Equation 2.2 and 2.3 gives
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1
K i = (1 − (2 × 0.167 × 0.25))800 = 733 N K ai = 0.25(1 − (2 × 0.167 × 0.25))800 = 183 N From Equations 2.5
FrI 3 = 0.53 × 733 − 4.9 × 183 = (−)508 N FrII 3 = (1 − 0.53) × 733 + 4.9 × 183 = 1241N Referring to 'Axial loading of single row angular contact ball bearings' table from the General Catalogue, case 1c applies since
FrII 3 > FrI 3
and
0.86(FrII 3 − FrI 3 ) = 630 > K ai
Consequently
FaII 3 = eII × FrII 3 = 0.86 × 1241 = 1067 N FaI 3 = FaII 3 − K ai = 1067 − 183 = 884 N Hence
FaI 3 884 = = 1.74 (> eI ) FrI 3 508
and
FaII 3 1067 = = 0.86 (= eII ) FrII 3 1241
The following equivalent loads are obtained
PI 3 = X II FrI 3 + YII FaI 3 = (0.38 × 508) + (0.72 × 884) = 830 N PII 3 = FrII 3 = 1214 N Summary of equivalent dynamic loads in N Inboard Brg I
Outboard Brg II
Straight ahead driving
620
601
Cornering, Outer Wheel
1523
1036
Cornering, Inner Wheel
830
1214
The mean equivalent bearing load is determined using equation 2.7
PIm = 3 0.9 × 620 3 + 0.05 × 15233 + 0.05 × 830 3 = 749 N PI Im = 3 0.9 × 6013 + 0.05 × 1036 3 + 0.05 × 1214 3 = 703 N The corresponding bearing lives are
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SKF(UK)Ltd - Racing Unit
Tech Note 2, Iss. 1 3
3
C 27600 LI = π 2 RH I = π 2 × 250 = 78,685,000 Km P 749 Im 3
3
C 27600 LII = π 2 RH II = π 2 × 250 = 95,302,000 Km 703 PI Im
Please contact the Racing Unit for more information:
[email protected].
© Copyright SKF 2003 Every care has been taken to ensure the accuracy of the information contained in this document but no liability can be accepted for any loss or damage whether direct, indirect or consequential arising out of the use of the information contained herein.
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SKF(UK)Ltd - Racing Unit
