●Bearing Load Calculation

4. Bearing Load Calculation

To compute bearing loads, the forces which act on the 6
shaft being supported by the bearing must be 19.1×10 ・H
Kt ＝ N
determined. These forces include the inherent dead
weight of the rotating body (the weight of the shafts and
components themselves), loads generated by the
working forces of the machine, and loads arising from
＝
Dp・n

1.95×10 ・H
Dp・n
6
｛kgf｝
｝……（4.2）

transmitted power. Ks ＝ Kt・tanα（Spur gear）………（4.2a）

tanα
It is possible to calculate theoretical values for these ＝ Kt・ cosβ（Helical gear）……（4.2b）
loads; however, there are many instances where the
load acting on the bearing is usually determined by the
Kr ＝ √Kt ＋Ks ………………………（4.3）
2 2
nature of the load acting on the main power
transmission shaft. Ka ＝ Kt・tanβ（Helical gear） ……（4.4）
4.1 Load acting on shafts where,
Kt ：Tangential gear load (tangential force), N
4.1.1 Load factor Ks ：Radial gear load (separating force), N
There are many instances where the actual operational Kr ：Right angle shaft load (resultant force of
shaft load is much greater than the theoretically tangential force and separating force), N
calculated load, due to machine vibration and/or shock. Ka：Parallel load on shaft, N
This actual shaft load can be found by using formula H ：Transmission force , kW
(4.1). n ：Rotational speed, r/min
Dp：Gear pitch circle diameter, mm
K ＝ fw・Kc ……………………………（4.1） α：Gear pressure angle
where, β：Gear helix angle
K ：Actual shaft load N｛kgf｝
fw ：Load factor (Table 4.1)
Kc：Theoretically calculated value N｛kgf｝

4.1.2 Gear load

The loads operating on gears can be divided into three
main types according to the direction in which the load is Ks
applied; i.e. tangential (Kt), radial (Ks), and axial (Ka).
The magnitude and direction of these loads differ
according to the types of gears involved. The load
Kt
calculation methods given herein are for two general-use
gear and shaft arrangements: parallel shaft gears, and Fig. 4.1 Spur gear loads
cross shaft gears. For load calculation methods
regarding other types of gear and shaft arrangements,
please consult NTN Engineering.

(1)Loads acting on parallel shaft gears

The forces acting on spur and helical parallel shaft Ks Ka
gears are depicted in Figs. 4.1, 4.2, and 4.3. The load
magnitude can be found by using or formulas (4.2),
through (4.4).
Kt

Table 4.1 Load factor fw Fig. 4.2 Helical gear loads

Amount fw
of shock
Application
Kt
Very little or Electric machines, machine tools,
no shock 1.0∼1.2 measuring instruments.
Kr Ks
Railway vehicles, automobiles,
rolling mills, metal working machines,
paper making machines, rubber mixing
Light shock 1.2∼1.5 machines, printing machines, aircraft,
Dp

textile machines, electrical units, office

machines.

Heavy shock Crushers, agricultural equipment,

1.5∼3.0 construction equipment, cranes.

Fig. 4.3 Radial resultant forces

A-19
●Bearing Load Calculation

Because the actual gear load also contains vibrations where,

and shock loads as well, the theoretical load obtained by
the above formula should also be adjusted by the gear Ksp，Ksg ：Pinion and gear separating force, N
factor fz as shown in Table 4.2. Kap，Kag：Pinion and gear axial load, N

Table 4.2 Gear factor fz For spiral bevel gears, the direction of the load varies
Gear type fz depending on the direction of the helix angle, the direction
of rotation, and which side is the driving side or the driven
Precision ground gears side. The directions for the separating force (Ks) and axial
(Pitch and tooth profile errors of less than 0.02 mm) 1.05∼1.1
load (Ka) shown in Fig. 4.5 are positive directions. The
Ordinary machined gears direction of rotation and the helix angle direction are
(Pitch and tooth profile errors of less than 0.1 mm) 1.1∼1.3 defined as viewed from the large end of the gear. The
gear rotation direction in Fig. 4.5 is assumed to be
clockwise (right).
(2)Loads acting on cross shafts
Gear loads acting on straight tooth bevel gears and
K tp
spiral bevel gears on cross shafts are shown in Figs. 4.4
and 4.5. The calculation methods for these gear loads are
shown in Table 4.3. Herein, to calculate gear loads for Ka p
straight bevel gears, the helix angle β= 0. Ks p
Ka g
Ks g
The symbols and units used in Table 4.3 are as follows:

Kt ：Tangential gear load (tangential force), N Kt g

Ks ：Radial gear load (separating force), N
Ka ：Parallel shaft load (axial load), N
H ：Transmission force, kW Fig. 4.4 Loads on bevel gears
n ：Rotational speed, r/min
Dpm ：Mean pitch circle diameter, mm Kt
α ：Gear pressure angle
β ：Helix angle
δ ：Pitch cone angle Ka
D pm
Ks 2
In general, the relationship between the gear load and δ β

the pinion gear load, due to the right angle intersection of

the two shafts, is as follows:

Ksp＝Kag…………………（4.5）
Kap＝Ksg…………………（4.6） Fig. 4.5 Bevel gear diagram

Table 4.3 Loads acting on bevel gears Unit N

Rotation
direction Clockwise Counter clockwise Clockwise Counter clockwise
Pinion
Helix
direction Right Left Left Right
6 6
19.1×10 ・H 1.95×10 ・H
Tangential load Kt Kt=
Dpm・n , Dpm・n

Driving side Ks=Kt tanα cosδ + tanβsinδ Ks=Kt tanα cosδ - tanβsinδ
cosβ cosβ
Separating force Ks

Driven side Ks=Kt tanα cosδ - tanβsinδ Ks=Kt tanα cosδ + tanβsinδ
cosβ cosβ

Driving side Ka=Kt tanα sinδ - tanβcosδ Ka=Kt tanα sinδ + tanβcosδ
cosβ cosβ
Axial load Ka

Driven side Ka=Kt tanα sinδ + tanβcosδ Ka=Kt tanα sinδ - tanβcosδ
cosβ cosβ

A-20
●Bearing Load Calculation

4.1.2 Chain / belt shaft load

The tangential loads on sprockets or pulleys when 4.2 Bearing load distribution
power (load) is transmitted by means of chains or belts For shafting, the static tension is considered to be
can be calculated by formula (4.7). supported by the bearings, and any loads acting on the
6 shafts are distributed to the bearings.
19.1 ×10 ・H
Kt＝ N
Dp・n

｝
For example, in the gear shaft assembly depicted in
……………（4.7） Fig. 4.7, the applied bearing loads can be found by using
1.95×10 ・H
6 formulas (4.10) and (4.11).
＝ ｛kgf｝
Dp・n a+b d
FrA＝ F1＋ F2 ……………（4.10）
where, b c+d
Kt ：Sprocket/pulley tangential load, N a c
FrB＝− F1＋ F2 ……………（4.11）
H ：Transmitted force, kW b c+d
Dp：Sprocket/pulley pitch diameter,mm where,
FrA：Radial load on bearing A, N
For belt drives, an initial tension is applied to give FrB：Radial load on bearing B, N
sufficient constant operating tension on the belt and F1, F2：Radial load on shaft, N
pulley. Taking this tension into account, the radial loads
acting on the pulley are expressed by formula (4.8). For
chain drives, the same formula can also be used if
vibrations and shock loads are taken into consideration.

Kr＝f b・Kt…（4.8）
where,
Kr：Sprocket or pulley radial load, N a b
f b：Chain or belt factor (Table 4.3) Bearing A Bearing B

Table. 4.4 chain or belt factor f b

FrA FrB
Chain or belt type fb
Chain (single) 1.2∼1.5 F! F@
V-belt 1.5∼2.0 c d
Timing belt 1.1∼1.3
Flat belt (w / tension pulley) 2.5∼3.0 Fig. 4.7 Gear shaft
Flat belt 3.0∼4.0

se side
F1 Loo

Dp
Kr

F2 Tens
ion side

Fig. 4.6 Chain / belt loads

A-21
●Bearing Load Calculation

4.3 Mean load (3) Linear fluctuating load

The mean load, Fm, can be approximated by formula
The load on bearings used in machines under normal
(4.14).
circumstances will, in many cases, fluctuate according to
a fixed time period or planned operation schedule. The Fmin＋2Fmax
Fm＝ …（4.14）
load on bearings operating under such conditions can be 3
converted to a mean load (Fm), this is a load which gives
bearings the same life they would have under constant
operating conditions.
F
(1) Fluctuating stepped load
The mean bearing load, Fm, for stepped loads is Fmax
calculated from formula (4.12). F1 , F2 ....... Fn are the Fm
loads acting on the bearing; n1, n2.......nn and t1, t2.......
tn are the bearing speeds and operating times Fmin
respectively.
p
（Fi ni ti） 1/p
Σ
〔
Fm＝
（ni ti）〕
Σ
…………………（4.12） t

where:
Fig. 4.10 Linear fluctuating load
p＝3 For ball bearings
p＝10/3 For roller bearings
(4) Sinusoidal fluctuating load
The mean load, Fm, can be approximated by formulas
F (4.15) and (4.16).
F1
case (a) Fm＝0.75 Fmax ………（4.15）
F2 Fm case (b) Fm＝0.65 Fmax ………（4.16）

Fn
F
n1 t1 n2t2 nn tn
Fmax
Fig. 4.8 Stepped load
Fm

(2) Consecutive series load

Where it is possible to express the function F(t) in t
terms of load cycle to and time t, the mean load is （a）
found by using formula (4.13).
1 to p 1/p F
〔
Fm＝ ∫ F
to o
（t）d t 〕………………（4.13）
Fmax
where:
p＝3 For ball bearings Fm
p＝10/3 For roller bearings

t
（b）
F
Fig. 4.11 Sinusoidal variable load
F(t)
Fm

0 to 2to t

Fig. 4.9 Time function series load

A-22
●Bearing Load Calculation

4.4 Equivalent load where,

Por：Static equivalent radial load, N
Fr ：Actual radial load, N
4.4.1 Dynamic equivalent load
Fa ：Actual axial load, N
When both dynamic radial loads and dynamic axial
Xo ：Static radial load factor
loads act on a bearing at the same time, the hypothetical
Yo ：Static axial load factor
load acting on the center of the bearing which gives the
The values for Xo and Yo are given in the respective
bearings the same life as if they had only a radial load or
bearing tables.
only an axial load is called the dynamic equivalent load.
(2) Static equivalent axial load
For radial bearings, this load is expressed as pure
For spherical thrust roller bearings the static equivalent
radial load and is called the dynamic equivalent radial
axial load is expressed by formula (4.21).
load. For thrust bearings, it is expressed as pure axial
Poa＝Fa＋2.7Fr…（4.21）
load, and is called the dynamic equivalent axial load.
where,
Poa：Static equivalent axial load, N
(1) Dynamic equivalent radial load
Fa ：Actual axial load, N
The dynamic equivalent radial load is expressed by
Fr ：Actual radial load, N
formula (4.17).
Provided that Fr / Fa ≦ 0.55 only.
where,
Pr：Dynamic equivalent radial load, N
Fr：Actual radial load, N
Fa：Actual axial load, N
X ：Radial load factor
Y ：Axial load factor
The values for X and Y are listed in the bearing tables.

(2) Dynamic equivalent axial load

As a rule, standard thrust bearings with a contact angle
of 90˚ cannot carry radial loads. However, self-aligning
thrust roller bearings can accept some radial load. The
dynamic equivalent axial load for these bearings is
given in formula (4.18).
Pa＝Fa＋1.2Fr………………（4.18）
where,
Pa：Dynamic equivalent axial load, N
Fa：Actual axial load, N
Fr ：Actual radial load, N
Provided that Fr / Fa ≦ 0.55 only.

4.4.2 Static equivalent load

The static equivalent load is a hypothetical load which
would cause the same total permanent deformation at the
most heavily stressed contact point between the rolling
elements and the raceway as under actual load
conditions; that is when both static radial loads and static
axial loads are simultaneously applied to the bearing.

For radial bearings this hypothetical load refers to pure

radial loads, and for thrust bearings it refers to pure
centric axial loads. These loads are designated static
equivalent radial loads and static equivalent axial loads
respectively.

(1) Static equivalent radial load

For radial bearings the static equivalent radial load can
be found by using formula (4.19) or (4.20). The greater
of the two resultant values is always taken for Por.
Por＝Xo Fr＋Yo Fa… （4.19）
Por＝Fr …………… （4.20）

A-23
●Bearing Load Calculation

4.4.3 Load calculation for angular ball bearings and

tapered roller bearings
For angular ball bearings and tapered roller bearings
the pressure cone apex (load center) is located as shown
in Fig. 4.12, and their values are listed in the bearing
tables. α α
Fa
Load center Fa Load center
When radial loads act on these types of bearings the Fr
component force is induced in the axial direction. For this Fr
reason, these bearings are used in pairs (either DB or DF
arrangements). For load calculation this component force
must be taken into consideration and is expressed by
formula (4.22). a a
0.5Fr
Fa ＝ Y …………………（4.22）
Fig. 4.12 Pressure cone apex
The equivalent radial loads for these bearing pairs are
given in Table 4.5.

Table 4.5 Bearing arrangement and dynamic equivalent load

Bearing arrangement Load condition Axial load Equivalent radial load

DB Brg1 Brg2 Fa1＝ 0.5Fr2 ＋ Fa Pr1＝XFr1＋Y1 0.5Fr2 ＋ Fa

0.5Fr1 0.5Fr2 Y2 Y2
arrangement ≦ ＋ Fa
Y1 Y2
Fa Fa2＝ 0.5Fr2 Pr2＝Fr2
Y2
Fr 1 Fr 2
DF Brg2 Brg1 Fa1＝ 0.5Fr1 Pr1＝Fr1
0.5Fr1 0.5Fr2 Y1
arrangement ＞ ＋ Fa
Fa Y1 Y2
Fr 2 Fr 1
Fa2＝ 0.5Fr1 − Fa Pr2＝XFr2＋Y2 0.5Fr1 − Fa
Y1 Y1
DB Brg1 Brg2 Fa1＝ 0.5Fr1 Pr1＝Fr1
0.5Fr2 0.5Fr1 Y1
arrangement ≦ ＋ Fa
Y2 Y1
Fa Fa2＝ 0.5Fr1 ＋ Fa Pr2＝XFr2＋Y2 0.5Fr1 ＋ Fa
Fr 1 Fr 2 Y1 Y1

DF Brg2 Brg1 Fa1＝ 0.5Fr2 − Fa Pr1＝XFr1＋Y1 0.5Fr2 − Fa

arrangement 0.5Fr2 0.5Fr1 Y2 Y2
＞ ＋ Fa
Fa Y2 Y1
Fa2＝ 0.5Fr2 Pr2＝Fr2
Fr 2 Fr 1 Y2
Note 1: The above are valid when the bearing internal clearance and preload are zero.
2: Radial forces in the opposite direction to the arrow in the above illustration are also regarded as positive.

A-24
●Bearing Load Calculation

200
Grease Mainly oil lubrication
4.5 Allowable axial ioad for cylindrical roller lubrication or
oil lubrication shows grease
lubrication
bearings

Ins
Allowable face pressure

tan
150
Cylindrical roller bearings having flanges on both the

ta
xia
inner and outer rings can be loaded with a certain axial

l lo
 Pz MPa 

ad
force at the same time. Unlike the basic dynamic load

In
te
m
rating with is determined by the development of rolling 100

itt
en
ta
fatigue, a permissible dynamic axial load of a rolling

xia
lo l
ad
cylindrical roller bearing is determined by heat generation, No
rm
50 al
seizure, etc., at the sliding contact surfaces of the guide ax
ial
loa
d
flanges and end faces of the rollers. The allowable axial
load is approximated by the formula below which is based
on past experience and experiments. 0
0 5 10 15 20 25 30
2 4
Pt = k・d ・Pz …………………(4.23) dp・n ×10 mm・rpm
where, dp：Pitoh circle diameter of rollers mm
dp≒(Bearing bore diameter
Pt ：Allowable axial load during rotation N｛kgf｝   + Bearing outer diameter)/2
k ：Coefficient determined by internal bearing n：Revolution per minute r/min
geometry (Please refer to Table 4.6) Fig. 4.13 Allowable face pressure of rib
d ：Bore diameter of the bearings mm
Pz：Allowable face pressure (bearing stress) of the
2
collar MPa (Please refer to Fig. 4.13)｛kgf/mm ｝
Table 4 Value of coefficient k and allowable axial (Fa max)

However, if the ratio axial load/radial load is large, Bearing type k Fa max

normal rolling motion of the roller cannot be achieved. NJ，NUP10

Therefor, a value exceeding Fa max shown in Table 4.6 NJ，NUP，NF，NH2， 0.040 0.4Fr
NJ，NUP，NH22
should not be used.
NJ，NUP，NF，NH3，
Moreover, when applying axial loads, the following 0.065 0.4Fr
NJ，NUP，NH23
guidelines are important; NJ，NUP，NH2E，
0.050 0.4Fr
(1) Be carful to specify proper radial internal clearance. NJ，NUP，NH22E
(2) Use a lubricant containing an extreme pressure NJ，NUP，NH3E，
0.080 0.4Fr
additive. NJ，NUP，NH23E
(3) The shaft and housing abutment height must be NJ，NUP，NH4， 0.100 0.4Fr
enough to cover those of the flanges. SL01-48 0.022 0.2Fr
(4) In case of severe axial loads, increase the mounting SL01-49 0.034 0.2Fr
accuracy and perform test running of the bearing. SL04-50 0.044 0.2Fr

In cases of axial loads being placed on large cylindrical

roller bearings (for example, bearing diameters of 300mm
or more), large axial loads being on the bearing under low
speed consult conditions, or forces bearing applied,
please consult with NTN Engineering. For cylindrical roller
bearings subjected to high axial use Type HT, Please
consult NTN Engineering.

A-25
●Bearing Load Calculation

4.6 Bearing rated life and load calculation ――――――――――――――――――――――――――――――――――――

examples (Example 3)
Determine the optimum model number for a
cylindrical roller bearing operating at 450 r/min,
In the examples given in this section, for the purpose of
with a radial load Fr of 200 kN, and which must
calculation, all hypothetical load factors as well as all
have a life of over 20,000 hours.
calculated load factors may be presumed to be included
――――――――――――――――――――――――――――――――――――
in the resultant load values.
From Fig. 3.1 the life factor fh = 3.02 (L10h at 20,000),
―――――――――――――――――――――――――――――――――――― and the speed factor fn = 0.46 (n = 450 r/min). To find the
(Example 1)
required basic dynamic load rating, Cr, formula (3.3) is
What is the rating life in hours of operation (L10h)
used.
for deep groove ball bearing 6208 operating at
650 r/min, with a radial load Fr of 3.2 kN ? fh 3.02
Cr＝ Pr ＝ ×200
―――――――――――――――――――――――――――――――――――― fn 0.46
From formula (4.17) the dynamic equivalent radial load: ＝1 313kN｛134,000kgf｝
Pr＝Fr＝3.2kN｛326kgf｝
From the bearing table, the smallest bearing that fulfills
The basic dynamic rated load for bearing 6208 (from
all the requirements is NU2336 (Cr = 1380 kN).
bearing table) is 29.1 kN, and the speed factor (fn) for ball
bearings at 650 r/min (n) from Fig. 4.1 is 0.37. The life ――――――――――――――――――――――――――――――――――――
factor, fh, from formula (3.3) is: (Example 4)
What are the rated lives of the two tapered roller
Cr 29.1 bearings supporting the shaft shown in Fig. 4.14
f h＝fn ＝0.37× ＝3.36
Pr 3.2 Bearing @ is an 4T-32206 with a Cr = 54.5 kN,
Therefore, with fh = 3.36 from Fig. 3.1 the rated life, L10h, and bearing ! is an 4T-32205 with a Cr = 42.0 kN.
is approximately 19,000 hours. The spur gear shaft has a pitch circle diameter Dp of
―――――――――――――――――――――――――――――――――――― 150 mm, and a pressure angle α of 20˚. The gear
(Example 2) transmitted force HP = 150 kW at 2,000 r/min
What is the life rating L10h for the same bearing and (speed factor n).
conditions as in Example 1, but with an additional ――――――――――――――――――――――――――――――――――――
axial load Fa of 1.8 kN ?
――――――――――――――――――――――――――――――――――――
To find the dynamic equivalent radial load value for Pr, Bearings1 Bearings2
the radial load factor X and axial load factor Y are used. (4T-32206) (4T-32205)
The basic static load rating, Cor, for bearing 6208 is 17.8
kN.
150

Fa 1.8
＝ ＝0.10
Cor 17.8

Therefore, from the bearing tables e= 0.29.

For the operating radial load and axial load: 70 100
170
Fa 1.8
＝ ＝0.56＞e=0.29 Fig. 4.14 Spur gear diagram
Fr 3.2

From the bearing tables X = 0.56 and Y = 1.48, and The gear load from formulas (4.1), (4.2a) and (4.3) is:
from formula (4.17) the equivalent radial load, Pr, is: 6
19.1×10 ・H 19,100×150
Pr＝XFr＋YFa＝0.56×3.2＋1.48×1.8 Kt ＝ ＝
Dp・n 150×2,000
＝4.46 kN｛455kgf｝ ＝9.55kN｛974kgf｝
Ks ＝Kt・tanα＝9.55×tan20˚
From Fig. 3.1 and formula (3.3) the life factor, fh, is:
＝3.48kN｛355kgf｝
Cr 29.1
f h＝fn ＝ 0.37× ＝ 2.41 Kr ＝√Kt2 ＋Ks2 ＝√9.552 ＋3.482
Pr 4.46
＝10.16kN｛1,040kgf｝
Therefore, with life factor fh = 2.41, from Fig. 5.1 the The radial loads for bearings ! and @ are:
rated life, L10h, is approximately 7,000 hours.
100 100
Fr1 ＝ Kr ＝ ×10.16＝5.98kN｛610kgf｝
170 170

A-26
●Bearing Load Calculation

70 70 The equivalent radial load, Pr, for each operating condition

Fr2 ＝ Kr＝ ×10.16＝4.18kN｛426kgf｝ is found by using formula (4.17) and shown in Table 4.8.
170 170
Because all the values for Fri and Fai from the bearing tables
0.5Fr1 0.5Fr2 are greater than Fa / Fr > e＝ 0.18, X ＝ 0.67 e Y2 ＝ 5.50.
＝1.87＞ ＝1.31
Y1 Y2
Pri ＝ XFri ＋Y2 Fai ＝ 0.67Fri ＋ 5.50Fai
The equivalent radial load is:
Table 4.8
Pr1 ＝Fr1 ＝5.98kN｛610kgf｝ Condition No. Equivalent radial load. Pri
0.5Fr1 i kN｛ kgf ｝
Pr2 ＝XFr2 ＋Y2 1 17.7｛ 1805 ｝
Y1
2 30.0｛ 3060 ｝
＝0.4×4.18＋1.67×1.87
3 46.4｛ 4733 ｝
＝4.66kN｛475kgf｝ 4 55.3｛ 5641 ｝
From formula (3.3) and Fig. 3.1 the life factor, fh, for each 5 75.1｛ 7660 ｝
bearing is:
From formula (4.12) the mean load, Fm, is:
Cr1 10/3
f h1＝ fn ＝0.293×54.5／5.98＝2.67 Σ
（Pri・ni・φi）3/10
Pr1 〔
Fm =
Σ
（n・i φi ）
〕 ＝48.1kN｛4,906kgf｝
Cr2
f h2＝ fn ＝0.293×42.0／4.66＝2.64 ――――――――――――――――――――――――――――――――――――
Pr2
(Example 6)
Therefore: a2 ＝ 1.4（4T-tapered roller bearings shown in Find the Basic rated life and limit of allowable axial
B-136) load when operated following.
Lh1 ＝13,200×a2 Provided that intermittent axial load and oil lubricant.
＝13,200×1.4
Fr＝10kN｛1,020kgf｝
＝18,480 ore
Lh2 ＝12,700×a2 n ＝2,000 r/min
＝12,700×1.4 ――――――――――――――――――――――――――――――――――――
＝17,780 ore The equivalent radial load is:
Pr＝Fr＝10kN｛1,020kgf｝
The combined bearing life, Lh, from formula (3.6) is:
The speed factor of cylindrical roller bearing, fn , at n＝
2,000 r/min
1
Lh＝ 33.3 3/10
1 1 1/e 〔
fn ＝
2,000 〕 ＝0.293
〔 Lh1
e＋
L h2
e 〕
The life factor, f h, from formula (3.4)
1 124
＝ f h＝0.293× ＝3.63
1 1 8/9 10
〔 18,4809/8 ＋
17,7809/8
〕 There fore the basic rated life, L10h ,from formula (3.3)
＝9,780 hour L10h ＝500×3.63 ≒24,000

And next, allowable axial load of cylindrical roller bearing is

―――――――――――――――――――――――――――――――――――― shown in a heading 4.5.
(Example 5)
Find the mean load for spherical roller bearing 23932
The value of coefficient, k, show in table 4.6. k ＝ 0.065
(La = 320 kN) when operated under the fluctuating
conditions shown in Table 4.7. d p＝
（60＋130） /2＝95mm，n＝2,000 r/min
―――――――――――――――――――――――――――――――――――― Take into consideration that intermittent axial load.
4 4
Table 4.7 dp・n×10 ＝19×10
The allowable face pressure of the collar, Pt , from
Condition Operating Radial load Axial load Revolution
time Fig.4.13.
No. φi Fri Fai ni
i ％ kN｛ kgf ｝ kN｛ kgf ｝ rpm Pt ＝40MPa
1 5 10｛ 1020 ｝ 2｛ 204 ｝ 1200 There fore the allowable axial load, Pz, following
2
2 10 12｛ 1220 ｝ 4｛ 408 ｝ 1000 Pz ＝0.065×60 ×40＝936N｛95.5kgf｝
3 60 20｛ 2040 ｝ 6｛ 612 ｝ 800 and meet a demand Fa max＜0.4×10,000＝4,000N from table
4.6.
4 15 25｛ 2550 ｝ 7｛ 714 ｝ 600
5 10 30｛ 3060 ｝ 10｛ 1020 ｝ 400

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