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STRENGTH OF MATERIALS. 269
Example 1.
A wrought iron shaft is eight inches in diameter, and the
force acts upon a lever two feet long. How much force must
be applied in order to twist off or to destroy the shaft?
Solution
:
8*^5S0_ = 512 X 560
=148?4S0 p0unds.
2 2
Example 2.
A force of 870 pounds is acting with a leverage of four feet
in twisting a wrought iron shaft. What must be the diameter
of the shaft in order to resist the twisting stress, with 10 as a
factor of safety ?
Solution :
3
D = \l P m X 10
^ 580
3
D — V 00 = 3.914, or, practically, a 4-inch shaft.
Note.—Ten is used as a multiplier of the twisting moment,
P m, because 10 is the factor of safety. Constant 580 is taken
from Table No. 32.
Example 3.
A round bar of cast-iron four inches in diameter is to be
twisted off by a force of 3200 pounds. How long a leverage is
necessary ? (c for cast-iron, in Table No. 32, is 450).
Solution
:
... D* c
P
43
X 450 _ 64 X 450 —- 9 feet long.
3200 3200
Example 4.
Experiments are made upon a cast-iron round bar 2 inches
in diameter with a leverage of h% feet ; the bar is twisted off at
a force of 832 pounds. Calculate constant c, or the force in
pounds if acting with a leverage of one foot, which will break a
round bar of the same material one inch in diameter.
Solution
:
c
__ P m
, = 832X5X_ 4368 = ^ ds
23
8
P
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