ROLLING CONTACT BEARINGS   Home  


Notes:  bearings_roller.pdf



What SORT of bearing should you choose?  

Use this chart data to initiate the bearing selection process.  See the link "Bearing Engineering and Knowledge Application Menu" for additional bearings knowledge information. 

Sizing of bearing is usually done through supplier's catalogues or application experts. Typical selection procedures are given in the notes.

Legend:

A    = Excellent
B    = Good
C    = Satisfactory
F    = Poor
NR = Inconvenient or not recommended
-     = Not applicable


Bearing Type Pure Radial Load Pure Axial load Combined Load Moment Load High Speed High Running Accuracy High Stiffness Quiet Running Low Friction Comp. for Errors of Alignment During Operation Comp. for Errors of Alignment  (Initial) Location Bearing Installation Non-location Bearing Installation Axial Disp. Possible in Bearing 
Ball Single - Row  C C F A A C A A F F B C NR
C Ball Single Double - Row C C C C C C C C B NR NR C C NR
Ball Self - Aligning C F F NR B B F B B A B C C NR
Ball Angular Contact C C B F B A C B B F F B NR NR
Ball Angular Contact Back - to - Back B C B C C B B C C NR NR B C NR
Ball Four - Point Contact F C C C B C C C C NR NR B F NR
Cylindrical Roller N, NU B NR NR NR A B B B B F F NR A A
Cylindrical Roller NJ, NUP B C C NR A B B C B F F C C C
Cylindrical Roller Double Row A NR NR C A A A B B NR NR NR A A
Full Complement Cylinder Roller A C F NR F C A F F F F C C C
Full Cylinder Roller Double  - Row A C F C F C A F F NR NR C C C
Needle Roller B NR NR NR C C B C F NR NR NR A A
Spherical Roller A C A NR C C B C C A B B C NR
Taper Roller B B A NR C B B C C F F B NR NR
Taper Roller 
( Face - to - Face)
A B A F C C A C C F F A C NR
Thrust Ball NR C NR NR C B C F C NR NR C NR NR
Thrust Bal with Spherical Housing Washer NR C NR NR C C C F C NR - C NR NR
Thrust Cylinder Roller NR B NR NR F B B F F NR NR C NR NR
Thrust Needle Roller NR B NR NR F C B F F NR NR C NR NR
Thrust Spherical Roller NR A C NR C C B F C A B B NR NR
(From Engineers Edge)

There are many types of rolling-element bearings, each tuned for a specific kind of load and with specific advantages and disadvantages.

 
Ball bearings
Ball bearings use spheres instead of cylinders. Clever use of surface tension allows balls of high accuracy to be made much more cheaply than comparable cylinders. Ball bearings can support both radial (perpendicular to the shaft) and axial loads (parallel to the shaft). For lightly-loaded bearings, balls offer lower friction than rollers. Ball bearings can operate when the bearing races are misaligned.

 
Roller bearings
Common roller bearings use cylinders of slightly greater length than diameter. Roller bearings typically have higher radial load capacity than ball bearings, but a low axial capacity and higher friction under axial loads. If the inner and outer races are misaligned, the bearing capacity often drops quickly compared to either a ball bearing or a spherical roller bearing.

 
Needle bearing
Needle roller bearings use very long and thin cylinders. Since the rollers are thin, the outside diameter of the bearing is only slightly larger than the hole in the middle. However, the small-diameter rollers must bend sharply where they contact the races, and thus the bearing fatigues relatively quickly.


Tapered roller bearing

Tapered roller bearings use conical rollers that run on conical races. Most roller bearings only take radial loads, but tapered roller bearings support both radial and axial loads, and generally can carry higher loads than ball bearings due to greater contact area. Taper roller bearings are used, for example, as the wheel bearings of most cars, trucks, buses, and so on. The downsides to this bearing is that due to manufacturing complexities, tapered roller bearings are usually more expensive than ball bearings; and additionally under heavy loads the tapered roller is like a wedge and bearing loads tend to try to eject the roller; the force from the collar which keeps the roller in the bearing adds to bearing friction compared to ball bearings.

 
Spherical roller bearings
Spherical roller bearings use rollers that are thicker in the middle and thinner at the ends; the race is shaped to match. Spherical roller bearings can thus adjust to support misaligned loads. However, spherical rollers are difficult to produce and thus expensive, and the bearings have higher friction than a comparable ball bearing since different parts of the spherical rollers run at different speeds on the rounded race and thus there are opposing forces along the bearing/race contact.
 
Thrust bearing
An axial load is supported by this type, typically to support a vertical shaft against gravitational loads. Spherical, conical or cylindrical rollers are used; and non rolling element bearings such as hydrostatic or magnetic bearings see some use where particularly heavy loads or low friction is needed.



Sizing a Rolling Bearing - Examples

Problem 1.

Determine a suitable deep groove ball bearing with 65mm bore to carry radial load of 5.15kN at 1000rpm for a nominal life of 20000 hours.
 
Solution. (Using the ISO L10 method - Outlined on page 8)
Data:
    bore = 65mm
    N = 1000 RPM
    h or Lh= 20000 hours
    Fr = 5.15kN

1. Determine the number of hours. (It tells you...  h = 20000)

2. Convert to design life in millions of revs;

       
3.  Determine radial load, Fr, and axial load, Fa, for the bearing.
        Fr=5.15kN
        Fa=0kN

4.  Calculate Fa/Fr.
        Fa/Fr=0

5.  Select bearing from tables p 26,27 for a shaft size of d = 65mm:
        Range for dynamic load rating C: 11900 to 119000  
        Range for static load rating Co: 9650 to 78000

6.  Ratio Fa/Co = 0.
        Now read graph on p22;  (Note: The vertical axis is Fa/Co)
        If Fa/Co = 0  then e =  0.2 (estimated from graph)

7.   Equivalent dynamic bearing load P:
        Since Fa/Fr  =  0, then  Fa/Fr  < e, so P = Fr + Y*F =   Fr + 0 = 5.15 kN

8.   Bearing life equation:
        L10 = (C / P ) ^3
        Re-arrange to solve for C...
       

9.  Select bearing from catalogue with C greater than 54.7kN..(See p27).
        6213 bearing:, C=55.9kN, Co=34000
        ID=65mm, OD=140mm, Breadth = 33mm

10.  Calculate equivalent static bearing load, Po.
        Po=0.6*Fr+0.5*Fa  so  Po = Fr + 0 = 5.15kN
        So Po=5.15kN is much less than Co=34kN.  So bearing is OK.

11.  Check minimum radial load on bearing. (There is none- OK)
 
12.  Check maximum speeds.
        Limiting speeds: 5300rpm with grease, 6300rpm with oil. This is fine.


 
Problem 2.
A deep groove ball bearing is required to carry a radial load of 2.18kN and a thrust load of 450N at a speed of 1600rpm.  Select a suitable bearing for a working life of 10000 hours.

 
Solution. (Using the ISO L10 method - Outlined on page 8)
Data:
    Fr = 2.18kN
    Fa = 450N
    RPM = 1600
    Lh = 10000
    ID = 28mm (minimum shaft size)

1.  10000 hours
2.   Determine the required basic rating life, L10, of the bearing in millions of revolutions.       
   
 
3.  Determine radial load, Fr, and axial load, Fa, for the bearing.
        Fr = 2.18kN
        Fa = 450N

4.  Calculate Fa/Fr.
        Fa/Fr = 450 / 2180 = 0.206

5.  Select bearing from p26.
        D=28mm does not exist, so go up to 30mm
        Range of load rating: From (C=4.49kN, Co=2.9kN) to (C=43.6kN, Co=23.6kN)        

6.  Use ratio to get e from graph on p22.
        F
a/Co = 0.450/2.9 = 0.155 (min size)
        Fa/Co = 0.450/23.6 = 0.019 (max size)
        Now read graph on p22;  (Note: The vertical axis is Fa/Co)
        If Fa/Co = 0.155  then e =  0.325 (from graph)
        If Fa/Co = 0.019  then e =  0.22 (estimated from graph)

7.   Equivalent dynamic bearing load P:
        In all bearing cases above, Fa/Fr < e, so P = Fr = 2.18 kN

8.   Bearing life equation:
        L10 = (C / P ) ^3
        Re-arrange to solve for C...
        

9.  Select bearing from catalogue with C greater than 21.5kN..(See p26).
        6306 bearing: C=28100,  Co=14600
        ID=30mm,  OD=62mm, B=19mm

10.  Calculate equivalent static bearing load, Po.
        Po=0.6*Fr+0.5*Fa   =  0.6*2.18 + 0.5*0.45 = 1.533kN
        So Po=1.533kN is much less than Co=14.6kN.  So bearing is OK.

11.  Check minimum radial load on bearing.
        Is
Fr>0.01C?   Where Fr=2.18, 0.01C =  0.01*28.1 =  0.281
        Yes!  (Bearing axial loading is OK)

12.  Check maximum speeds.
        Limiting speeds: 9000rpm with grease, 11000rpm with oil. This is fine.
 

Problem 3.
A deep groove ball bearing with minimum bore of 60mm is required to carry a radial load of 4kN and a thrust load of 2.2kN at 1000rpm.  Select a suitable bearing for a working life of 10000 hours.

 
Solution.
Data:
    bore*60mm
    Fr=4kN
    Fa=2.2kN
    RPM=1000
    Lh=10000 hours

1.  Determine the required basic rating life, L10, of the bearing in millions of revolutions.


 

2.  Determine radial load, Fr, and axial load, Fa, for the bearing.

    Fr=4kN
    Fa=2.2N

3.  Calculate Fa/Fr.

    Fa/Fr=2200/4000=0.55

4.  Estimate factors X and Y.

Fa/Fr*0.44 estimate X=0.56, Y=1.5

5. Calculate the Equivalent dynamic bearing load, P.

    P=X*Fr+Y*Fa=(0.56*4)+(1.5*2.2)
    P=5.54kN

6.  Calculate the Basic dynamic load rating, C.

 
7.  Select bearing from catalogue with C greater than step 6.

6212 bearing:
    C=47500
    Co=28000
    ID=60mm
    OD=110mm

8. Calculate Fa/Co.

Fa/Co=2200/28000=0.079

9.  Check X and Y.

e = 0.28
so Fa/Fr greater than e
so Xactual=0.56,Yactual=1.6

Repeat step 5 to step 8

5. Calculate the Equivalent dynamic bearing load, P.

    P=X*Fr+Y*Fa=(0.56*4)+(1.6*2.2)
    P=5.76kN

6.  Calculate the Basic dynamic load rating, C.

 
 
7.  Select bearing from catalogue with C greater than step 6.

we require next largest bearing

6312 bearing:
    C=81900
    Co=48000
    ID=60mm
    OD=130mm

10.  Calculate Equivalent static bearing load, Po.

Po=0.6*Fr+0.5*Fa
Po=3.5kN
but, Fr=4kN
therefore use Po=4kN
 
bearing is OK.

11.  Check maximum speeds.

Limiting speeds:
    5000rpm with grease
    6000rpm with oil.

 

Problem 4.
The vertical shaft shown is supported by two deep groove ball bearings.  Bearing A at the top is allowed to float whereas bearing B at the bottom carries all the axial thrust.  The shaft is driven by vee belts through a pulley.  The sum of the belt tensions is 5.4kN and the estimated weight of the pulley assembly is 900N.  The shaft has a minimum diameter, based on strength, of 24mm and rotates at 300rpm.  Use the SKF catalogue for selection of two bearings based on a life of 9000 hours and a shock factor of 2 for vee belt forces.

problem 4 

Solution:

Data:
    F1+F2=5.4kN
    FaA=0
    FaB=900N
    RPM=300
    Shock factor=2
    Lh=9000 hours
    IDmin=24mm


1.  Determine the required basic rating life, L10, of the bearing in millions of revolutions.



2.  Determine radial load, Fr, and axial load, Fa, for the bearings.

Shock factor, f=2, meaning all radial loads must be doubled.

Using moments,





 
3.  Calculate Fa/Fr.

Bearing A:  Fa/Fr=0

Bearing B:  Fa/Fr=900/3600=0.25



4.  Estimate factors X and Y.

Bearing A:  Fa/Fr=0, X=1, Y=0

Bearing B:  Fa/Fr<0.44, X=1, Y=0


5. Calculate the Equivalent dynamic bearing load, P.

Bearing A:  P=Fr=7.2kN

Bearing B:  P=Fr=3.6kN    

 
6.  Calculate the Basic dynamic load rating, C.

Bearing A
 

Bearing B


7.  Select bearings from catalogue with C greater than step 6.

Bearing A
  Bearing B
6305 bearing 6205 bearing
ID=25mm ID=25mm
OD=62mm OD=52mm
C=22500 C=14000
Co=11400 Co=6950


8. Calculate Fa/Co.

Bearing A
Fa/Co=0

Bearing B
    Fa/Co=900/6950=0.1295



9.  Check X and Y.

Bearing B

e = 0.31
e greater than Fa/Fr
therefore X=1,Y=0 is used.

10.  Calculate Equivalent static bearing load, Po.

Bearing A

Po=Fr=7.2kN

Bearing B

Po=0.6*Fr+0.5*Fa
Po=0.6*3.6+0.5*0.9
Po=2.61kN
but,
    Fr=3.6kN
therefore Po=Fr=3.6kN

in both cases Po>Co

11.  Check maximum speeds.

Limiting speeds:
    
Bearing A
    11000rpm with grease
    14000rpm with oil.

Bearing B
    12000rpm with grease
    15000rpm with oil.


Assignments

1. FLYWHEEL BEARING ASSEMBLY. A flywheel is held by 2 deep groove ball bearings.



(a) Draw a FBD, determine the loads on the shaft and complete the statics (reactions)
(b) Determine the minimum size of the shaft (using simple beam bending equations) . Note: This is not a complete shaft design according to Shaft Design Standards - we will do this later. Assume a simple working stress of 100MPa.
(c) Determine the loads on the bearings
(d) Select bearings. Assume 25000 hour life.
(e) Explain the bearing arrangement in terms of assembly issues, load sharing, radial and axial loads, preload and temperature changes causing expansion, maintenance.