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SHAFTING. $6$
This calculation shows plainly how very necessary it is to
have bearings near the pulleys where shafts are loaded with
heavy pulleys and large belts. There is nothing more liable to
destroy a shaft than too much deflection, because the shaft is,
when running, continually bent back and forth, and at last it
must break. The fact must never be lost sight of that strength
and stiffness are two entirely different things and follow entirely
different laws; therefore, after calculations are made for
strength, the stiffness must also be investigated, as stiffness is
a very important property in shafting. The best way to over-
come too much transverse deflection is to shorten the distance
between the bearings. Of course, increasing the diameter of the
shaft will also overcome deflection, but shafting should never be
larger in diameter than necessary, because the first cost increases
with the weight, which increases as the square of the diameter,
and the frictional resistance will also increase with the increased
diameter ; consequently, also, the running expenses.
Torsional Strength of Shafting.
Shafting may be considered as a beam fastened at one end
and having a torsional load applied at the other end equal to
the pull of the belt on an arm of the same length as the radius of
the pulley. In Table No. 32, page 268, constant c is given as 580
pounds for wrought iron.
The formula for twisting stress, as explained under beams
( see page 267) is,
D*c n-J-p
P = " c
D
i
m = the length of the lever or arm in feet, and will here
be the radius of the pulley and be denoted by r. The length of
the shaft has no influence on its torsional strength, but only on
its angle of torsional deflection (see page 268). Using 10 as
factor of safety, the formula will be :
* 58
D = Diameter of shaft in inches.
r == Radius of pulley in feet.
W= Pull of belt in pounds.
58 = Constant, with 10 as factor of safety = Vw X 580,
taken from Table No. 32, page 268.
Frequently it is more convenient to calculate the torsional
strength of shafting according to the number of horse-power the
shaft is to transmit ( see page 317 ).
In the above formula, assume W to be 58 pounds, r to be
one foot, and D will be one inch. That is, a shaft one inch in
diameter is strong enough to resist, with 10 as factor of safety,
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