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364 SHAFTING.
a torsional load of 58 pounds acting on an arm one foot long.
Assuming this 58 pounds to act on the rim of a pulley of one
foot radius, two feet in diameter, and making one revolution per
minute, it will transmit power at a rate of 58 X 6f = 364f foot-
pounds per minute ; but one horse-power is 33,000 foot-pounds
per minute, and if the shaft should transmit one horse-power it
, 33000 , . . „
must make og, 4 = 90.52 revolutions per minute. Hence the
practical formulas for torsional strength of shafting:
V-
rr DB
Xn „ _ H X 90
ri = n = —
—
90 Z> 3
D = Diameter of shaft in inches.
H= Number of horse-power transmitted by the shaft.
n = Number of revolutions made by the shaft per minute.
90 = Constant, using 10 as factor of safety, and assuming
the torsional strength to be as given in Table No. 32.
Torsional Deflection in Shafting.
In constructing different kinds of machinery it is frequently
necessary to consider the torsional deflection. The formula
for torsional deflection for wrought iron ( see page 271) will be :
e _ 0.00914 X m L P
jyi This will transpose to
s _ 48 HL
S = Deflection in degrees.
H= Number of horse-power transmitted.
, , , ,
0.00914 X 33000 _ A„
48 = Constant ;
calculated thus, —9 v s 1416
L = Length of shaft in feet between the force and the
resistance.
n =. Number of revolutions made by the shaft per minute.
D = Diameter of shaft in inches.
Example.
How many degrees is the deflection of a shaft two inches
in diameter, 50 feet long, making 300 revolutions per minute and
transmitting 15 horse-power, applied at one end and taken off at
the other ?
Solution
:
48 HL = 48 X 15 X 50 _
6 ~ ~VD^ 300 X 2* ~ i/2 de-rees’
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