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STRENGTH OF MATERIALS. 243
FIG 21
To Find the Transverse Strength of Beams when Their
Section is Not Uniform Throughout the
Whole Length.
Example.
A beam made of wrought iron is fastened at one end and
loaded at the other (as shown in Fig. 21). The largest part
is 5 inches in diameter and the smallest part is 4 inches in
diameter. Where will it break ?
and what is the breaking load ?
Note. —Naturally, the beam
will break at either A or B\
therefore, calculate first the break-
ing load of a round beam of
wrought iron 4>^ feet long and 5
inches in diameter, next a beam 3
3 feet long (the distance from P to
i?).and 4 inches in diameter.
Solving for strength at A :
Z>3 0.6 C
P =
P =
L
53
X 0.6 X 600
4^
P = 10,000 pounds.
Solving for strength at B :
p __ 43
X 0.6 X 600
P = 64 X
P = 7680 pounds.
Thus, the weakest point of the beam is at B, where its cal-
culated breaking load is only 7860 pounds, while the calculated
breaking load at A is 10,000 pounds.
When a beam is not of uniform section throughout its
whole length and is supported under both ends and loaded
somewhere between the supports, calculate first the reaction
on each support ; then consider the beam as if it was fastened
by the load, and consider the reaction at each support as a load
at the free end of a beam of length and section equal to the
length and section between its load and support.
Example.
The largest diameter of a round cast-iron shaft is 3 inches
and the smallest diameter is 2 inches. The length, mode of
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