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STRENGTH OF GEAR TEETH 4 II
Solution: Using the formula,
W= sXpX/X y.
and inserting the values corresponding to the problem, we have,
W = 4000 X 2 X 5 X 0,114 = 456o pounds.
A force of 4560 pounds acting at a speed of 600 feet will give
4560 X 600
33000
= 83 horse P°wer-
For bevel gears Mr. Lewis gives the following formulas,
referring to Fig. 16:
7)3 -73
W= sXpX/X yX
3D2
X (D—d)
W = Safe load in pounds.
5 = Allowable stress in the
material which is de-
pending upon the speed
See table No. 55.
D = large diameter of gear.
d = small diameter of gear.
/ = Face of bevel gear in inches.
P = Circular pitch at the large
diameter.
n=Actual number of teeth.
iV=Formative number of teeth
which can be calculated by
multiplying the secant of angle
a (See Fig. 16) by the actual
number of teeth thus:
N = n X secant a.
The formative number of teeth is also the same number
of teeth as would correspond to the radius R.
R = Back cone radius.
(See explanation on page 399.)
y = A factor, depending upon the formative number N
and also upon the shape of the teeth, obtained from Table No.
To illustrate the use of the formula, Mr. Lewis gives the
Find the working strength of a pair of cast iron miter gears
of 50 teeth, 2 inch pitch, 5 inch face at 120 revolutions per minute.
Radial flank system.
In this case D = 31.8 inches, d = 24.8 inches. The angle
a is 45 degrees, therefore secant a is 1.4; the formative number
N will be 1.4 X 50 = 70. The corresponding value of y is
found in table No. 54 to be 0.071.
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