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248 STRENGTH OF MATERIALS.
Solution :
14400 X 10
^ 2X X 125
3
^ = V512
B = 8 inches in width.
H— 1% X 8" = 12 inches in thickness.
The weight of the beam itself is not considered in this
problem.
To Find the Size of a Beam to Carry a Given Load When
Also the Weight of the Beam is to be Considered.
Rule.
Calculate first the size of beam required to carry the load*
then figure what such a beam will weigh and add half of this
weight to the load, if the beam is fastened at one end and loaded
at the other, or supported under both ends and loaded at the
center, but add the whole weight of the beam to the weight of
the load if the load is distributed along the whole length of the
beam. Then figure the size of the required beam for this new
load.
Example.
Find width and thickness of a pitch pine beam to carry
2000 pounds, with 8 as factor of safety, and a span of 27 feet.
The beam is supported under both ends and loaded at the center ;
its own weight is also to be taken into consideration.
Solution
:
Find the constant for pitch pine in Table No. 30 to be 150,
and find the weight of pitch pine in Table No. 10 to be 50 pounds
per cubic foot. When the beam is supported under both ends
and loaded at the center it is four times as strong as if fastened
at one end and loaded at the other; therefore, constant 150 is
multiplied by 4. The load, 2000 pounds, multiplied by 8 as a
factor of safety, gives 16,000 pounds as breaking load of the
beam.
3
•
_ / 16000 X 27
> 2X X 150 X 4
3
B — \/ 320
B = 6.84" = width, and 1% X 6.84" = 10.26" = thickness.
The area is 6.84 X 10.26 == 70 square inches ;
the weight per
foot is 70 times 50 divided by 144, which equals 24.3 pounds, say 25
pounds. The weight of the beam is 25 X 27 = 675 pounds. This
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