- Project Runeberg -  Machinists' and Draftsmen's Handbook /
305

(1910) Author: Peder Lobben - Tema: Mechanical Engineering
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MECHANICS. 305
Axle Friction.
The friction between bearings and shafts is frequently
called axle friction. This, of course, is sliding friction, but
owing to the fact that the surfaces in question are usually very
smooth and well lubricated, the coefficient of friction is smaller
than for ordinary slides.
Example 2.
A fly-wheel weighs 24,000 pounds, the diameter of the shaft
is 10 inches, and the coefficient of friction in the bearings is 0.08.
What force must be applied 20 inches from the center in order to
keep the wheel turning ?
Resistance due to friction — 24000 X 0.08 = 1920 pounds.
This resistance is acting at a radius of 5 inches, but the
force is acting at a radius of 20 inches ; therefore, the required
force necessary to overcome friction will be -.— ° x 5
=480 pounds.
20
^
How much power is absorbed by this frictional resistance if
the wheel is moving 72 revolutions per minute ?
Solution
:
The space moved through by the force is
72 x 20 X
J
x 3- 1416
= 753.984 feet, and 753.934 X 480 = 361,912.32 foot-pounds and
361912.32 _ 1Q Q7 horse.power>
33000

*


Horse=Power Absorbed by Friction in Bearings.
The horse-power absorbed by the friction in the bearings
for any shaft may be figured directly by the formula,
H-P = W X f X n X 3- 1416 x d
33000 X 12
This reduces to
:
HP = IV X / X n X d X 0.000008
H-P = Horse-power absorbed by friction.
W— Load on bearings in pounds.
d= Diameter of shaft in inches.
/= Coefficient of friction.
n = Number of revolutions per minute.
Calculating the previous example by this formula, we have :
H-P = 24000 X 0.08 X 72 X 10 X 0.000008 = 11.06 horse-
power, which is practically the same as figured before.

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