10" : 40" :: 19 : Answer.
The answer in this case is 76, whereas the observed value is 68, or 8
lbs. less; this does not agree very well with the theory, but still
the difference, though 8 lbs., is only about 11 or 12 per cent. of the
whole, and we shall still retain the law, for certainly there is no
other that can express the result as well.
388. But the table will supply another verification. In experiment
No. 3 a 40" bar, 1" broad, and 0"·5 deep, broke with 38 lbs.; and in
experiment No. 7 a 20" bar of the same section broke with 74 lbs.; but
this is so nearly double the breaking weight of the 40" bar, as to be
an additional illustration of the law, that _for a given section the
breaking load varies inversely as the span_.
389. We next inquire as to the effect of the _breadth_ of the beam upon
its strength? For this purpose we compare experiments Nos. 3 and 4: we
there find that a bar 40" × 1" × 0"·5 is broken by a load of 38 lbs.,
while a bar just half the breadth is broken by 19 lbs. We might have
anticipated this result, for it is evident that the bar of No. 3 must
have the same strength as two bars similar to that of No. 4 placed side
by side.
390. This view is confirmed by a comparison of Nos. 7 and 8, where we
find that a 20" bar takes twice the load to break it that is required
for a bar of half its breadth. The law is not quite so well verified
by Nos. 5 and 6, for half the breaking weight of No. 5, namely 29·5
lbs., is more than 25, the observed breaking weight of No. 6: a similar
remark may be made about Nos. 9 and 10.
391. Supposing we had a beam of 40" span, 2" broad, and 0"·5 deep, we
can easily see that it is equivalent to two bars like that of No. 3
placed side by side; and we infer generally that the strength of a bar
is proportional to its breadth; or to speak-more definitely, _if two
beams have the same span and depth, the ratio of their breaking loads
is the same as the ratio of their breadths_.
392. We next examine the effect of the _depth_ of a beam upon its
strength. In experimenting upon a beam placed edgewise, a precaution
must be observed, which would not be necessary if the same beam were
to be broken flatwise. When the load is suspended, the beam, if merely
laid edgewise on the supports, would almost certainly turn over; it
is therefore necessary to place its extremities in recesses in the
supports, which will obviate the possibility of this occurrence; at the
same time the ends must not be prevented from bending upwards, for we
are at present discussing a beam free at each end, and the case where
the ends are not free will be subsequently considered.
393. Let us first compare together experiments Nos. 2 and 3; here
we have two bars of the same dimensions, the section in each being
1"·0 × 0"·5, but the first bar is broken edgewise, and the second
flatwise. The first breaks with 77 lbs., and the second with 38 lbs.;
hence the same bar is twice as strong placed edgewise as flatwise when
one dimension of the section is twice as great as the other. We may
generalize this law, and assert that the strength of a rectangular beam
broken edgewise is to _the strength of a beam of like span and section
broken flatwise, as the greater dimension of the section is to the
lesser dimension_.
394. The strength of a beam 40" × 0"·5 × 1" is four times as great as
the strength of 40" × 0"·5 × 0"·5, though the quantity of wood is only
twice as great in one as in the other. In general we may state that if
a beam were bisected by a longitudinal cut, the strength of the beam
would be halved when the cut was horizontal, and unaltered when the cut
was vertical; thus, for example, two beams of experiment No. 4, placed
one on the top of the other, would break with about 40 lbs., whereas if
the same rods were in one piece, the breaking load would be nearly 80
lbs.
395. This may be illustrated in a different manner. I have here two
beams of 40" × 1" × 0"·5 superposed; they form one beam, equivalent to
that of No. 1 in bulk, but I find that they break with 80 lbs., thus
showing that the two are only twice as strong as one.
396. I take two similar bars, and, instead of laying them loosely
one on the other, I unite them tightly with iron clamps like those
represented in Fig. 56. I now find that the bars thus fastened together
require 104 lbs. for fracture. We can readily understand this increase
of strength. As soon as the bars begin to bend under the action of the
weight, the surfaces which are in contact move slightly one upon the
other in order to accommodate themselves to the change of form. By
clamping I greatly impede this motion hence the beams deflect less, and
require a greater load before they collapse; the case is therefore to
some extent approximated to the state of things when the two rods form
one solid piece, in which case a load of 152 lbs. would be required to
produce fracture.
397. We shall be able by a little consideration to understand the
reason why a bar is stronger edgewise than flatwise. Suppose I try to
break a bar across my knee by pulling the ends held one in each hand,
what is it that resists the breaking? It is chiefly the tenacity of
the fibres on the convex surface of the bar. If the bar be edgewise,
these fibres are further away from my knee and therefore resist with a
greater moment than when the bar is flatwise: nor is the case different
when the bar is supported at each end, and the load placed in the
centre; for then the reactions of the supports correspond to the forces
with which I pulled the ends of the bar.