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Technical Section 


 1.5 Determination of Bearing Size

(a) Basic Definitions

In the course of many years of experience with ball bearings and extensive testing, it has been found that the prediction of the load capacity of a ball bearing is a statistical event related to the fatigue life of the bearing. This makes the sizing of ball bearings more difficult than that of many other machine elements.

A basic phenomenon in ball bearings is that ball bearing life has been found to be inversely proportional to the cube of the bearing load. This means that when the load is doubled, the life expectancy of the bearing is reduced by a factor of eight. This phenomenon has been studied extensively and has led to the adoption of an industry-wide national standard for rating ball bearings pioneered by the American Bearing Manufacturers Association (formerly Anti-Friction Bearing Manufacturers Association, Inc.), 1200 19th Street, N.W., Suite 300, Washington, D.C. 20036-2433.

The following represents a summary of the load rating of ball bearings of less than one inch in diameter, according to ANSI-AFBMA Standard 9-1978: "Load Rating and Fatigue Life for Ball Bearings" – reprinted with the permission of the American National Standards Institute, Inc., 11 West 42nd Street, 13th Floor, New York, N.Y. 10036.

Ball bearings were formerly rated on the basis of the compressive stress in the most heavily loaded ball. Except for static loads, experience has shown that the actual cause of failure is fatigue. Fatigue characteristics are thus used for load rating and are dependent to a large extent on experimental results.

The life of a ball bearing is the life in hours at some known speed, or the number of revolutions, that the bearing will attain before the first evidence of fatigue appears on any of the moving elements. Experience has shown that the life of an individual ball bearing cannot be precisely predicted. Fatigue characteristics are thus used for load ratings.

Even if ball bearings are properly mounted, adequately lubricated, protected from foreign matter, and are not subject to extreme operating conditions, they can ultimately fatigue. Under ideal conditions, the repeated stresses developed in the contact areas between the balls and the raceways eventually can result in fatigue of the material which manifests itself as spalling of the load carrying surfaces. In most applications, the fatigue life is the maximum useful life of a bearing. This fatigue is the criterion of life used as the basis for the first part of this standard.

The material in the standard which follows assumes bearings having nontruncated contact area, hardened good quality steel as the bearing material, adequate lubrication, proper ring support and alignment, nominal internal clearances, and adequate groove radii. In addition, certain high-speed effects such as ball centrifugal forces and gyroscopic moments are not considered.

The following nomenclature and definitions are used in life testing of bearings. A multitude of identical bearings are tested under same conditions:

RATING LIFE is the life at which 10 percent of bearings have failed and 90 percent of them are still good. This value is designated as L10 and is expressed in millions of revolutions.

LIFE of an individual ball bearing is the number of revolutions (or hours at some given constant speed) designated as L which the bearing runs before the first evidence of fatigue develops in the material of either ring (or washer) or of any of the rolling elements.

MEDIAN LIFE is the life at which 50 percent of bearings failed and 50 percent are still good. It is designated as L50, which is generally not more than five times the RATING LIFE, L10.

BASIC LOAD RATING "C" for a radial or angular contact ball bearing is the calculated, constant, radial load which a group of apparently identical bearings with stationary outer ring can theoretically endure for a RATING LIFE of one million revolutions of the inner ring. For a thrust ball bearing, it is the calculated, constant, centric, thrust load which a group of apparently identical bearings can theoretically endure for a RATING LIFE of one million revolutions of one of the bearing washers. The basic load rating is a reference value only of the base value of one million revolutions RATING LIFE having been chosen for ease of calculation. Since applied loading as great as the basic load rating tends to cause local plastic deformation of the rolling surfaces, it is not anticipated that such heavy loading would normally be applied.

 

 

(b) Determination of Basic Load Rating

The basic load rating C for a rating life of one million revolutions for radial and angular contact ball bearings, except filling slot bearings, with balls not larger than 1 in. diameter, is given by the equation:

s14df5.gif (2028 bytes)

where:

  • i = number of rows of balls in the bearing

  • a = nominal angle of contact (angle between line of action of ball load and plane perpendicular to bearing axis)

  • Z = number of balls per row

  • D = ball diameter

  • fc = a constant from Table 1-2, as determined by the value of (D cos a)/dm

  • dm = pitch diameter of ball races

NOTE: For balls larger than 1 inch diameter, the exponent for D is 1.4.

To get a better feel for the meaning of one million revolutions, it is attained in 8 hrs at a speed of 2,084 rpm. Most ball bearings, however, may have intended life many times exceeding one million revolutions.

In the above formula, dm represents the pitch diameter of the ball races. It can be expressed as follows:

figure8t.gif (4298 bytes)s14df6.gif (1974 bytes)

A and B are dimensions as shown. However, assuming that inner ring and outer ring wall thicknesses are the same, A becomes outside diameter, and B the bore of the bearing.

Values of fc are shown in Table 1-2
for different values of (D cos a)/dm.

 

RATING LIFE L10 in millions of revolutions for a ball bearing application can be calculated from:

s14df7.gif (1928 bytes)

 

where:

C = the basic load rating as previously defined and
P = the load.

(c) Illustrative Examples

Example 1

Consider an ABEC 3 single row, radial ball bearing having 10 balls of 1/16" diameter, 0.300" inner race diameter and 0.452" outer race diameter in a single shield configuration.

a = 0° (radial bearing)

Z = 10 (number of balls)

D = 1/16" (ball diameter) and

s14df8.gif (2285 bytes)

Therefore,s14df9.gif (2067 bytes)

From Table 1-2 this value yields (from third column) a value of fc=4530. Substituting these values in Equation (5) for C, we obtain:

s14df10.gif (2023 bytes)

This means that for a load of P = 143 lbs, the rating life of this ball bearing will be one million revolutions and 90% of a group of such ball bearings will be expected to complete or exceed this value.


Table 1-2* Values of fc
table12.gif (16214 bytes)

NOTES:

  1. a. When calculating the basic load rating for a unit consisting of two similar, single row, radial contact ball bearings, in a duplex mounting, the pair is considered as one, double row, radial contact ball bearing.
    b. When calculating the basic load rating for a unit consisting of two, similar, single row, angular contact ball bearings in a duplex mounting, "Face-to-Face" or "Back-to-Back", the pair is considered as one, double row, angular contact ball bearing.
    c. When calculating the basic load rating for a unit consisting of two or more similar, single angular contact ball bearings mounted "in Tandem", properly manufactured and mounted for equal load distribution, the rating of the combination is the number of bearings to the 0.7 power times the rating of a single row ball bearing. If the unit may be treated as a number of individually interchangeable single row bearings, this footnote (1) c. does not apply.

  2. Use to obtain C in newtons when D is given in mm.

  3. Use to obtain C in pounds when D is given in inches.

* Reprinted by permission of the American National Standards Institute, 11 West 42nd Street, 13th Floor, New York, NY 10036. (from ANSI-AFBMA Std. 9-1978)


Suppose now it is desired to determine the "L" life of this bearing when operating at 200 rpm and a load of 50 lbs, the life being evaluated in hours of operation.

Let the life in hours be denoted by L, and let N denote the rpm of the bearing. We then have:

s14df11.gif (2115 bytes)

Substituting N = 200, P = 50 and C = 143 into Equation (8), we obtain L = 1949 hours.

NOTE: L10 is bearing life in millions of revolutions; L is bearing life in hours.

A table showing required life at constant operating speed has been given by N. Chironis ("Today’s Ball Bearings", Product Engineering, December 12, 1960, pp. 63-77, table on p. 68). This table is reproduced below with the permission of McGraw-Hill Book Company, New York, N.Y.

table13.gif (15491 bytes)

In order to provide data for larger size bearings as well as additional examples, Table 1-4 is given.

Table 1-4 Dimensions and Basic Load Ratings for Conrad-Type
Single-Row Radial Ball Bearings

table14.gif (27247 bytes)

 

 

Example 2:

Find the value of C for a 207 radial bearing.

Solution:

s14df12.gif (3731 bytes)

From Equation (5) for C:

C = 4550 x 4.327 x 0.2258 = 4440 lbs,  load for 1 million revolutions with 90 percent probability that it will be attained or exceeded.

(d) Relationship between Load and Number of Revolutions

In some cases, it is needed to determine the new value of the permitted loading when the number of revolutions N is changed.

Experimentally, it was proven that:

s14df13.gif (1952 bytes)

where N is number of revolutions and P is radial load.

Furthermore, it was established that

 

s14df14.gif (2105 bytes)

or subsequently:s14df15.gif (1945 bytes)

It has to be made clear that C is the basic load rating in lbs. for a rating life of 1 million revolutions, and this fact establishes the above relationship.

If a bearing has a rating life expressed in number of revolutions designated by N, the life of the bearing expressed in hours, designated by L, can be found from:

N = 60 n L

where n is the actual speed in rpm of the bearing.

Example 3

For Example 2 where we found C = 4440 lbs., find the radial load P1 for a rating life of 500 hours, at 1500 rpm.

s14df16.gif (1983 bytes)

Apply: C = 4.440 lbs., n = 1,500 rpm, and L = 500 hrs

s14df17.gif (2404 bytes)

(e) Combined Axial and Radial Loads

This condition is dealt with by ANSI-AFBMA Standard 9-1978 which defines the combined load to be expressed as:

P = C1 (X • i • Fr + Y • Fa) (11)

table15.gif (3719 bytes)

where value C1 is a service factor which is shown in Table 1-5.
In the above equation:

  • i = race rotation factor equal 1 for inner ring rotation, 1.2 for outer ring rotation.

  • Fr and Fa are radial and axial components, respectively, of the load.

  • X and Y are factors to be used as shown in Table 1-6.

NOTE: Y is the axial or thrust factor determined from the value of

s14df19.gif (1742 bytes)

Table 1-6 Values of X and Y
table16.gif (25410 bytes)

(1) Two similar, single row, angular contact ball bearings mounted "Face-to-face" or "Back-to-back" are considered as one, double row, angular contact bearing.

(2) Values of X, Y and e for a load or contact angle other than shown in Table 5-5 are obtained by linear interpolation.

(3) Values of X, Y and e shown in Table 5-5 do not apply to filling slot bearings for applications in which ball-raceway contact areas project substantially into the filling slot under load.

(4) For single row bearings, when Fa/Fr ­ e, use X = 1 and Y = 0.

* Reprinted by permission of the American National Standards Institute, 11 West 42nd Street, 13th Floor, New York, NY 10036 (force ANSI-AFBMA Std. 9-1978).

 

Example 4

For a bearing dealt with in Example 2, assume that it carries a combined load of 400 lbs radially and 300 lbs axially at 1200 rpm. The outer ring rotates, and the bearing is subjected to moderate shock. Find the rating life of this bearing in hours.

s14dp21.gif (5113 bytes)

NOTE: The impact load on a bearing should not exceed the static capacity as given by Table 1-4 or the race may be damaged by Brinelling from the balls. This load may be exceeded somewhat if the bearing is rotating and the duration of the load is sufficient for the bearing to make one or more complete revolutions while the load is acting.

Example 5

What change in the loading of a ball bearing will cause the expected life to be doubled?

Solution:

Let N1 and P1 be the original life and load for the bearing. Let N2 and P2 be the new life and load.

Then: N2 = 2N1

By Equation (9):

s14dp20.gif (2640 bytes)

Hence a reduction of the load to 79 percent of its original value will cause a doubling of the expected life of a ball bearing.

(f) Variable Loading of Bearings

Ball bearings frequently operate under conditions of variable load and speed. Design calculations should take into account all portions of the work cycle and should not be based solely on the most severe operating conditions. The work cycle should be divided into a number of portions in each of which the speed and load can be considered as constant.

Suppose P1, P2,... are the loads on the bearing for successive intervals of the work cycle. Let N1 be the life of the bearing, in revolutions, if operated exclusively at the constant load P1. Let there be N1’ applications of load P1. Then N1’/N1 represents the proportion of the life consumed in this portion of the cycle.

Let N2 be the life of the bearing, in revolutions, if operated exclusively at load P2. Let there be N2’ applications of load P2. Then N2’/N2 represents the proportion of the life consumed by load P2.

A corresponding statement can be made for each portion of the work cycle. The sum of these proportions represents the total life of the bearing or unity. Then:

s14f1.gif (2211 bytes)

Let Nc be the life of the bearing under the combined loading. Let N1’ = a1Nc where a1 represents the proportion of the total life, consumed under load P1. In a similar way, N2’ = a2 Nc, N3’ = a3 Nc, and so on. Substitution in Equation (12) yields:

s14f2.gif (2075 bytes)

Using Equation (10):

s14f3.gif (2286 bytes)

Combining these last two equations we can obtain:

s14f4.gif (3386 bytes)

From previous definition of a it is obvious that a1 + a2 + ... must equal unity. The application of this equation will be demonstrated by the following examples.

Example 6

A ball bearing is to operate on the following work cycle:

  • Radial load of 1400 lbs at 200 rpm for 25% of the time

  • Radial load of 2000 lbs at 500 rpm for 20% of the time

  • Radial load of 800 lbs at 400 rpm for 55% of the time

Total rpm is to be 1100.

Additional conditions:

The inner ring rotates; loads are steady. Find the minimum value of the basic rating load C for a suitable bearing for this application if the required life is 7 years at 4 hours per day.

Since both the load as well as the speed for the particular load varies, we have to establish the actual work cycle per minute.

table16atbot.gif (4087 bytes)

The following table should be constructed:

s14f5.gif (2356 bytes)

A working year is assumed to consist of 250 days.

Total life duration of the bearing expressed in hours will become 7 x 250 x 4 = 7000 hours, whereas this expressed in number of revolutions becomes:

s14f6.gif (2314 bytes)

Inputing this data in the formula (13), previously derived in 1.5 (f):

s14f7.gif (4759 bytes)

In order to choose the appropriate bearing, we refer to Table 1-4 from which we find that a bearing such as No. 308 should be satisfactory, keeping in mind there is but a 90 percent probability that the required life will be attained or exceeded.

Example 7

A 306 radial ball bearing with inner ring rotation has a 10-sec work cycle as follows:

s14f8st.gif (3527 bytes)

Find the rating life of this bearing in hours and in years of 250 working days of 2 hours each.

Solution:

Since the bearing chosen is No. 306, from Table 1-4:

Z = 8, D = 0.5 and i = 1.

s14f9.gif (2171 bytes)

From Table 1-6 for this value of 200, a value for Y will be 1.45 and X will be 0.56.

From Equation (11) and Table 1-5, for the combined axial and radial loads with light shock and 2-second duration:

qbc-053.gif (2531 bytes)

Since P2 is a pure radial load:

P2 = Fr = 600 lbs

The number of revolutions for the 2-second time duration will be:

whereas for the 8-second time duration will be:

The combined total number of revolutions in 10 seconds is:

then,

From formula (13)

Using C = 5120 in Table 1-4 for bearing No. 306:

This is the number of revolutions the bearing will endure. The total number of revolutions during the 10-second operation was established as being 190. Therefore, the number of revolutions per minute will be:

qbc-054.gif (1965 bytes)

From Equation (8):

This expressed in years of operation will become

(g) Static Loading of Bearings

Up to this point we have been dealing with dynamic loading of bearings. This is the condition when there is relative motion between the rings of the bearings and the balls that are rotating. If this is not the case, as a result of static concentrated loads of the balls against the races, the depressions of the balls into the races will gradually enlarge, and permanent indentations will remain. The static capacity is ordinarily defined as the maximum allowable static load that does not impair the running characteristics of the bearing to make it unusable.

This permanent deformation under the balls is known as Brinnelling and takes place at moderate to high loads. The magnitude of the permissible load is found by methods given in the standards. Calculations for the bearings of Table 1-4 have been made and are shown in the column headed Pst.

When very smooth and quiet operation is required, the loading should be no more than about one-half the static capacity.

Back and forth rotation of the shaft through small angles can cause early failure of bearings unless the load is very light. Lubrication is difficult because the oil or grease may not be replenished back of a ball or roller before the motion is reversed.

(h) Effect of Increased Confidence Levels

When a bearing is installed there is no way of knowing whether it is one of the 90 percent that are good or one of the 10 percent that will not attain the rating life. In other words, one can have but 90 percent confidence that the bearing will achieve or exceed its rating life, usually designated L10.

In some cases a greater degree of reliability is required. The expected life will of course be reduced as the reliability is made higher. Let an adjusting factor a1 be taken such that life Ln is equal to a1 L10. Factors a1 for different values of the reliability are given in Table 1-7. Life L10 is the rating life.


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