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fly-wheels. 357
Example.
A round rim of a fly-wheel is 4 inches in diameter and the
middle diameter of the wheel is 36 inches. What is the weight
of the rim ?
Solution:
ff=36X4X4X 0.64 = 369 pounds.
For a rim of rectangular section the weight will be
:
W= Width X thickness X D X 3.1416 X 0.26
W = Width X thickness X D X 0.816
Example.
The width of the rim is six inches, the thickness is two
inches, and the middle diameter of the rim is 48 inches. What
is the weight of the rim ?
Solution
:
J^=2X6X48X 0.816 = 470 pounds.
Centrifugal Force in Fly=Wheels and Pulleys.
Pulleys are not only liable to be broken by the stress due to
the action of the driving belt, but in fast-running pulleys and
fly-wheels the stress due to centrifugal force is far more dan-
gerous. This stress increases as the square of the velocity and
directly as the weight, therefore there is a limit to the velocity
at which fly-wheels and pulleys can be run with safety.
Generally speaking, increasing the thickness of the rim
does not increase its strength, because the total tensile strength,
the total weight of the rim, and, consequently, also the
centrifugal force, increase in the same proportion ; but it has
great influence upon the strength of the wheel to have the ma-
terial in the rim distributed to the best advantage. At the same
time it is very important to construct the rim and arms of such
proportions that the initial stress due to uneven cooling in the
foundry, is avoided.
The common formula is :
~ .., , , Mass X (velocity)2
Centrifugal force = ^ —
radius
Weight „ . . »XrX2?r
Mass = ——-2;
—
Velocity = —
32.2 50
Therefore,
r X 2 7T V
(
A
W 60
32.2 X r
W X «2
X r2
X 0.01096628
cf =
Cf = 32.2 X r
cf = IV X ft
1
X r X 0.00034
cf = Centrifugal force in pounds.
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