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2l8 STRENGTH OF MATERIALS.
=v
p xf
6- X 0.7854
» = t24000 X 10
55000 X 0.7854
P — Vl^Q
D = 2.358, or nearly 2^ inches diameter.
To Find the Diameter of a Bolt to Resist a Given Load.
Rule.
Multiply pull in pounds by the factor of safety. Multiply
the ultimate tensile strength of the material by 0.7854 ; divide
this first product by the last and extract the square root from
the quotient which will then be diameter of bolt at the bottom
of the thread.
D = {~~ ~’
P =
f=
S X 0.7854
D2
X S X 0.7854
f
B2
X S X 0.7854
P
D = Diameter of bolt or screw in the bottom of the thread.
P = Load or pull in pounds.
f= Factor of safety.
•S" = Ultimate tensile strength per square inch.
0.7854 is constant = —
4
Note.—Bolts are frequently exposed to a considerable
amount of initial stress, due to the tightening of nuts, which
must always be allowed for when deciding upon the load to be
considered when calculating their diameter.
Example.
Find diameter of a bolt to sustain a load of 4,450 pounds,
taking 10 as factor of safety and ultimate tensile strength of the
iron to be 50,000 pounds per square inch.
Solution
:
, 4450 X 10
50000 X 0.7854
D = 1.064" in the bottom of thread ; thus, a l^" screw,
standard thread, which is lj1
^ inches in diameter at the bottom
of thread, will be the bolt to use.
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