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252 STRENGTH OF MATERIALS.
constant in Table No. 30, and the cube root of this quotient is
the diameter of the beam.
Example.
A round spruce beam is fastened into a wall, and is to
carry 1200 pounds on the free end projecting 4 feet from the
wall, with 8 as a factor of safety, the weight of the beam not to
be considered. Find diameter of beam.
Solution
:
D -
-<-1200 X 4 X 8
0.6 X 125
D–
3
=V-
3
38400
75
£> = \/~512~
D = 8 inches diameter.
Load Concentrated at Any Point, Not at the Center of
a Beam.
If a beam is supported at both ends and loaded anywhere
between the supports but not at the center (see Fig. 25), it will
carry more load than if it was loaded at the center. With
regard to breaking, the carrying capacity is inversely as the
square of half the beam to the product of the short and the long
ends between the load and the support. For instance, a beam
10 feet long is of such size that when it is supported under
both ends and loaded at the center it will carry 1400 pounds.
How many pounds will the same beam carry if loaded 3 feet
from one end and 7 feet from the other ?
Solution
:
x _ 1400 X 52
X =
7X3
1400 X 25
21
X= 1666% pounds.
If weight of beam is also in
eluded in its center-breaking-load
the formula will be :
FIG. 25.
b F-«H«—F
-h
a X b
P = Breaking load (including
weight of beam) if applied at the center in pounds.
F= Half the length of the span.
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