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STRENGTH OF MATERIALS. 249
weight is distributed along the whole beam and, therefore, it does
not have any more effect than if half of it, or 337^ pounds,
was placed at the center, but as the beam is to be calculated with
8 as factor of safety, the weight allowed for the beam must be
337J£ X 8 = 2700 pounds. Thus, adding this weight to 16,000
pounds gives 18,700 pounds ; this new weight is used for calcu-
ing the size of the required beam.
;o<; ..:,
X 150 X 4
3
,
V 374
B = 7.2 inches = width, and 1% X 7.2" == 10.8 inches,
thickness.
This, of course is also a little too small, as only the weight
of a beam 6.84 inches by 10.25 inches is taken into account, but
if more exactness should be required the weight of this new
beam may be calculated and the whole figured over again,
and the result will be closer. This operation may be repeated
as many times as is wished, and the result will each time be
closer and closer, but never exact; but for all practical purposes
one calculation, as shown in this example, is sufficient.
Example 2.
Find width and thickness of a spruce beam to carry 4200
pounds distributed along its whole length. The span is 24 feet
;
use 10 as factor of safety, and also allow for the weight of beam.
The thickness of the beam is to be l}£ times its width.
Solution :
3
*=V-!
4200 X 24 X 10
2X X 125~X8
3
B = V 448
B = 7.65 inches, and £T= 11.48 inches.
„7
.
u ,, 7.65 X 11.48 X 24 X 32 4QO .
Weight of beam = —
—
= 468 pounds.
144
Adding ten times the weight of the beam to ten times the
weight to be supported, gives 46,680 pounds.
3
=4 46680 X 24
2% X 125 X 8
3
B = V 497.9
i? = 7.93 inches, and H =. \y
2 B= 11.9 inches, or prac-
tically, a beam 8 inches by 12 inches is required.
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