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362 SHAFTING.
Constant C may be calculated from experiments by the
formula,
c _ S D*
Z 3
W
S = Deflection in inches noted in the specimen, when
supported under both ends and loaded transversely
at the middle between supports.
D = Diameter of specimen in inches.
L = Distance between supports of specimen in feet.
W= Experimental load in pounds.
Example.
A round specimen placed in a testing machine, supported
under both ends and loaded at the middle with 2000 pounds,
deflects 0.1 inch. The diameter of the specimen is two inches
and the distance between supports is three feet. Calculate
constant C for this kind of material.
Solution
:
0.1 X 2 4
C =
3 3
X 2000
C = ±-u
= 0.0000296 inch.
54000
Thus, the deflection for this kind of material is 0.0000296
inch per pound of load, applied at the middle, between supports,
for a round bar one inch in diameter and one foot between
supports.
Allowable Deflection in Shafts.
The distance between the hangers must always be deter-
mined with due consideration to the allowable transverse deflec-
tion in the shafting, especially when the shaft is loaded with
large pulleys and heavy belts, remembering that the deflection
increases directly with the transverse load and with the cube of
the length between the bearings, (see page 254). The allowable
transverse deflection in shafting ought not to exceed 0.006 to
0.008 inch per foot of span ( see page 266). A beam of wrought
iron one foot long and one inch square, when supported under
both ends and loaded at the middle, will deflect 0.0000156 inch
per pound of load, (see Table No. 31, page 259), and a round beam
deflects 1.7 times as much as a square beam, when the diameter
and side are equal. A round shaft, one inch in diameter and
one foot long, when loaded at the middle with 144 pounds will,
therefore, deflect 144 X 1.7 X 0.0000156 = 0.00382 inch.
Thus, this load does not give more than an allowable de-
flection. But, suppose the distance between bearings is doubled
and the load decreased one-half; the ultimate strength of the
shaft will be the same, but the deflection will be 72 X 1.7 X 23
X
0.0000156 = 0.01528 == 0.0764 inch per foot.
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