Let us suppose it to be 12' long, 5" broad, and 7" deep. This is five
times as strong as a beam 1" broad and 7" deep for we may conceive the
original beam to consist of 5 of these beams placed side by side (Art
391); the beam 1" broad and 7" deep, is 7 times as strong as a beam 7"
broad, 1" deep (Art. 393). Hence the original beam must be 35 times as
strong as a beam 7" broad, 1" deep; but the beam 7" broad and 1" deep
is seven times stronger than a beam the section of which is 1" × 1",
hence the original beam is 245 times as strong as a beam 12' long and
1" × 1" in section; of which we can calculate the strength, by Art.
388, from the proportion—
144" : 40" :: 152 : Answer.
The answer is 42·2 lbs., and thus the breaking load of the original
beam is about 10,300 lbs.
399. It will be useful to deduce the _general_ expression for the
breaking load of a beam _l"_ span, _b"_ broad, and _d"_ deep, supported
freely at the ends and laden in the centre.
Let us suppose a bar _l"_ long, and 1" × 1" in section. The breaking
load is found by the proportion—
_l_ : 40 :: 152 : Answer;
and the result obtained is 6080/_l_. A beam which is _d"_ broad,
_l"_ span, and 1" deep, would be just as strong as _d_ of the beams
_l"_ × 1" × 1" placed side by side; of which the collective strength
would be—
6080
———————— × _d_.
_l_
If such a beam, instead of resting flatwise, were placed edgewise,
its strength would be increased in the ratio of its depth to its
breadth—that is, it would be increased _d_-fold—and would therefore
amount to
6080
————————— × _d_².
_l_
We thus learn the strength of a beam 1" broad, _d"_ deep, and _l"_
span. The strength of _b_ of these beams placed side by side, would be
the same as the strength of one beam _b"_ broad, _d"_ deep, and _l"_
span, and thus we finally obtain
6080
———————— _d_² × _b_.
_l_
Since _b d_ is the area of the section, we can express this result
conveniently by saying that the breaking load in lbs. of a rectangular
pine beam is equal to
area of section × depth;
6080 × —————————————————————————
span
the depth and span being expressed in inches linear measure, and the
section in square inches.
400. In order to test this formula, we have calculated from it the
breaking loads of all the ten beams given in Table XXIV. and the
results are given in the sixth column. The difference between the
amount calculated and the observed mean breaking weight is shown in the
last column.
401. Thus, for example, in experiment No. 7 the span is 20", breadth,
1", depth 0"·5; the formula gives, since the area is 0"·5,
0·5 × 0·5
_P_ = 6080 —————————— = 76
20
This agrees sufficiently with 74 lbs., the mean of two observed values.
402. Except in experiments Nos. 5 and 10, the differences are very
small, and even in these two cases the differences are not sufficient
to make us doubt that we have discovered the correct expression for the
load generally sufficient to produce fracture.
403. We have already pointed out that a beam begins to sustain
permanent injury when it is subjected to a load greater than half that
which would break it (Art. 368), and we may infer that it is not in
general prudent to load a beam which is part of a permanent structure
with more than about a third or a fourth of the breaking weight. Hence
if we wanted to calculate a fair working load in lbs. for a beam of
pine, we might obtain it from the formula.
area of section × depth.
1500 × —————————————————————————
span
Probably a smaller coefficient than 1500 would often be used by the
cautious builder, especially when the beam was liable to sudden blows
or shocks. The coefficient obtained from small selected rods such as
we have used would also be greater than that found from large beams in
which imperfections are inevitable.
404. Had we adopted any other kind of wood we should have found a
similar formula for the breaking weight, but with a different numerical
coefficient. For example, had the beams been made of oak the number
6080 must be replaced by a larger figure.