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From the Rolls-Royce experimental archive: a quarter of a million communications from Rolls-Royce, 1906 to 1960's. Documents from the Sir Henry Royce Memorial Foundation (SHRMF).
Technical paper on the design, load distribution, and stress analysis of ball and roller bearings.

Identifier  ExFiles\Box 20\7\  Scan006
Date  23th August 1911
  
August 23, 1911
THE MOTOR TRADER
COM
457

The Elements of Ball and Roller Bearing Design.

By Arnold C. Koenig.

Distribution of Load Stresses.

In any ball bearing that is under load only the balls in one-half of the circle can be (theoretically) engaged at any one time, therefore, the entire load is assumed to be sustained by one-half of the total number of balls in the bearing.

Assuming that the total load “W” that is sustained by the bearing is so distributed along the projection of its horizontal diameter that the force of downward pressure at all points of contact between the outer race and the balls (as A, B, C, etc.) is of equal magnitude, and that this downward force at each point of application (as B, C, D, etc.) is the resultant of two forces which act upon the same points in radial and tangential directions with reference to the circumference of the ball race.

8.—Illustrating Mode of Determining Load Stresses.

Let b be the continuation of the line P* to the horizontal diameter and r the radius corresponding to the direction of force P¹.

Then, under the rule of similar triangles:

P : r :: P¹ : b

but, b = cos φ, in terms of r, therefore if r = 1
then,
P¹ = P cos φ

When the ball is directly under the load, as at A.{Mr Adams}
Then angle φ = 0 deg., therefore cos φ = 1
and, P¹ = P = maximum pressure on a single ball.

The total load on the bearing
W=P (1+2 cos φ + 2 cos 2 φ + 2 cos 3 φ . . . . + 2 cos n φ)

The sum of the cosines of the angles of all the balls in one-half of the bearing will therefore equal the load-factor K, which, multiplied by the safe load capacity of one ball, gives the maximum safe load which may be placed upon the bearing.

Thus, let K = load-factor taken from Table I. (appended).
W = total load on the bearing.
P = maximum safe capacity of single ball from Table II. (appended).
Then—
W/K = the greatest load which comes upon a single ball.
and P x K = total safe load capacity of the bearing.

Table I.—Showing Values of K (Load Factor).
For bearings with 6 up to 26 balls.

Number of Balls. | K.{Mr Kilner}
6 | 2.00000
7 | 2.30411
8 | 2.61312
9 | 2.92371
10 | 3.23601
11 | 3.54357
12 | 3.86371

Number of Balls. | K.{Mr Kilner}
13 | 4.17942
14 | 4.49393
15 | 4.80914
16 | 5.12584
17 | 5.44240
18 | 5.75712
19 | 5.99273

Number of Balls. | K.{Mr Kilner}
20 | 6.31374
21 | 6.63460
22 | 6.95310
23 | 7.27547
24 | 7.59577
25 | 7.91742
26 | 8.23568

A Deduction.—The reactions which result from the variable peining force of the balls rolling under variable loads and speeds defy definite analysis, and destroy the accuracy of all calculations which are based upon other conditions alone, so that any assumption is justified which indicates a distribution of load more nearly

* Digest of a paper read at the recent Dayton meeting of Society of Automobile Engineers, U.S.A.
* Concluded from page 497 ante.

in accord with the practical conditions indicated. The writer’s experiments with ball bearings indicate that the stresses and reactions which occur in the loaded half of a ball race transmit stresses and reactions of much greater magnitude around the entire ring than is ordinarily assumed to be the case. The assumption of load distribution here given is therefore based merely upon its mathematical simplicity, and the fact that it indicates larger values for the radial pressures at the low points in the semi-circle (such as C and D in Fig. 8) than would be indicated by assuming either a uniform or a concentrated load upon a beam having a length equal to the diameter of the enclosing circle, and calculating the pressures of reactions at intermediate points in the usual way.

Table II.—Showing Physical Properties of Steel Balls.

DIAMETER OF BALL D.{John DeLooze - Company Secretary}
Inches. | Mm. | d² Inches. | CRUSHING STRENGTH BETWEEN | SAFE LOAD PER BALL FOR
| | | Ball on Ball | Plates | 1/4" Grooves | [Bearing Type 1] | [Bearing Type 2] | [Bearing Type 3]
| | | 60000 d² | 80000 d² | 100000 d² | Lbs. | Lbs. | Lbs.
1/4" .25 | 6.25 | .0625 | 3,750 | 5,000 | 6,250 | 125 | 83.3 | 42
5/16" .2812 | 7.144 | .0791 | 4,746 | 6,328 | 7,910 | 158 | 105.3 | 53
3/8" .3750 | 9.525 | .1406 | 8,436 | 11,248 | 14,063 | 280 | 187.2 | 93
7/16" .4375 | 11.112 | .1914 | 11,484 | 15,312 | 19,140 | 382 | 255.2 | 128
1/2" .5000 | 12.700 | .2500 | 15,000 | 20,000 | 25,000 | 500 | 333.3 | 166
9/16" .5625 | 14.287 | .3164 | 18,984 | 25,312 | 31,640 | 632 | 421.2 | 210
5/8" .6250 | 15.875 | .3906 | 23,436 | 31,248 | 39,062 | 781 | 520.8 | 260
11/16" .6875 | 17.462 | .4726 | 28,356 | 37,808 | 47,265 | 945 | 630.0 | 315
3/4" .7500 | 19.050 | .5625 | 33,750 | 45,000 | 56,250 | 1125 | 750.0 | 375
13/16" .8125 | 20.637 | .6602 | 39,612 | 52,816 | 66,015 | 1320 | 880.2 | 440
7/8" .8750 | 22.225 | .7656 | 45,936 | 61,248 | 76,560 | 1531 | 1020.8 | 510
15/16" .9375 | 23.813 | .8789 | 52,734 | 70,312 | 87,890 | 1758 | 1171.9 | 586
1" 1.0000 | 25.400 | 1.0000 | 60,000 | 80,000 | 100,000 | 2000 | 1333.3 | 666
1 1/8" 1.1250 | 28.575 | 1.2656 | 75,936 | 101,248 | 126,560 | 2531 | 1687.5 | 844
1 1/4" 1.2500 | 31.750 | 1.5625 | 93,750 | 125,000 | 156,250 | 3125 | 2083.3 | 1042
1 1/2" 1.500 | 38.100 | 2.2500 | 135,000 | 180,000 | 225,000 | 4500 | 3000.0 | 1500

NOTE.—In determining the load capacity of bearings of the four-point contact type, with angular race surfaces, the maximum working load for a single ball may be determined by multiplying the tabular capacity of the ball for that type, as given in Table II. above, by the cosine of the angle of inclination of the race surfaces from the horizontal or by the sine of half the angle included between them.

Formula for Determining Diameter of Enclosing Circle.
To determine the diameter of the enclosing circle of any ring of balls:
Let n = number of balls,
d = diameter of ball in inches.
D¹ = diameter of enclosing circle in inches.
φ = 360 degrees ÷ n.
D¹¹ = diameter of circle through ball centres.
x = clearance between balls, or space occupied by any form of separating device used.

Then D¹ = d + (d + x) / sin ½ φ

In the full type of bearing take x = 0.005 inch or more. If separating devices are used, then x = diameter, or thickness of one separating unit.

Also D¹ = K (d+x) + d
In which the factor K is taken from Table I (super).

D¹¹ = D¹ - d

x = 2 sin ½ φ ( (D¹-d)/2 ) - d
x = sin ½ φ (D¹-d) - d
Or, x = sin ½ φ (D¹¹-d)

Analysis of Load Stresses.
The various stresses to which the outer race of a ball bearing is subjected by the pressure moment of the load are;
(1) Direct compression (elastic deformation) at the points of contact with the balls.
(2) The peining force which results from the rolling motion which occurs simultaneously with the application of compressive
  
  


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