38. As an illustration of the nature of the “tie” and “strut,” and also
for the purpose of giving a useful example of the decomposition of
forces, I use the apparatus of Fig. 17 (see next page).
It represents the principle of the framework in the common lifting
crane, and has numerous applications in practical mechanics. A rod of
wood B C 3' 6" long and 1" × 1" section is capable of turning round
its support at the bottom B by means of a joint or hinge: this rod is
called the “jib”; it is held at its upper end by a tie A C 3' long,
which is attached to the support above the joint. A B is one foot long.
From the point C a wire descends, having a hook at the end on which
a weight can be hung. The tie is attached to the spring balance, the
index of which shows the strain. The spring balance is secured by a
wire-strainer, by turning the nut of which the length of the wire can
be shortened or lengthened as occasion requires. This is necessary,
because when different weights are suspended from the hook the spring
is stretched more or less, and the screw is then employed to keep the
entire length of the tie at 3'. The remainder of the tie consists of
copper wire.
39. Suppose a weight of 20 lbs. be suspended from the hook W, it
endeavours to pull the top of the jib downwards; but the tie holds it
back, consequently the tie is put into a state of tension, as indeed
its name signifies, and the magnitude of that tension is shown to be
60 lbs. by the spring balance. Here we find again what we have already
so often referred to; namely, one force developing another force that
is greater than itself, for the strain along the tie is three times as
great as the strain in the vertical wire by which it was produced.
[Illustration: FIG. 17.]
40. What is the condition of the jib? It is evidently being pushed
downwards on its joint at B; it is therefore in a state of compression;
it is a strut. This will be evident if we think for a moment how
absurd it would be to endeavour to replace the jib by a string or
chain: the whole arrangement would collapse. The weight of 20 lbs. is
therefore decomposed by this contrivance into two other forces, one of
which is resisted by a tie and the other by a strut.
[Illustration: FIG. 18.]
41. We have no means of showing the magnitude of the strain along the
strut, but we shall prove that it can be computed by means of the
parallelogram of force; this will also explain how it is that the tie
is strained by a force three times that of the weight which is used.
Through C (Fig. 18) draw C P parallel to the tie A B, and P Q parallel
to the strut C B then B P is the diagonal of the parallelogram whose
sides are each equal to B C and B Q. If therefore we consider the force
of 20 lbs. to be represented by B P, the two forces into which it is
decomposed will be shown by B Q and B C; but A B is equal to B Q, since
each of them is equal to C P; also B P is equal to A C. Hence the
weight of 20 lbs. being represented by A C, the strain along the tie
will be represented by the length A B, and that along the strut by the
length B C. Remembering that A B is 3' long, C B 3' 6", and A C 1', it
follows that the strain along the tie is 60 lbs., and along the strut
70 lbs., when the weight of 20 lbs. is suspended from the hook.
42. In every other case the strains along the tie and strut can
be determined, when the suspended weight is known, by their
proportionality to the sides of the triangle formed by the tie, the
jib, and the upright post, respectively.
43. In this contrivance you will recognize, no doubt, the framework of
the common lifting crane, but that very essential portion of the crane
which provides for the raising and lowering is not shown here. To this
we shall return again in a subsequent lecture (Art. 332). You will of
course understand that the tie rod we have been considering is entirely
different from the chain for raising the load.
44. It is easy to see of what importance to the engineer the
information acquired by means of the decomposition of forces may
become. Thus in the simple case with which we are at present engaged,
suppose an engineer were required to erect a frame which was to sustain
a weight of 10 tons, let us see how he would be enabled to determine
the strength of the tie and jib. It is of importance in designing
any structure not to make any part unnecessarily strong, as doing so
involves a waste of valuable material, but it is of still more vital
importance to make every part strong enough to avoid the risk of
accident, not only under ordinary circumstances, but also under the
exceptionally great shocks and strains to which every machine is liable.
45. According to the numerical proportions we have employed for
illustration, the strain along the tie rod would be 30 tons when the
load was 10 tons, and therefore the tie must at least be strong enough
to bear a pull of 30 tons; but it is customary, in good engineering
practice, to make the machine of about ten times the strength that
would just be sufficient to sustain the ordinary load. Hence the crank
must be so strong that the tie would not break with a tension less
than 300 tons, which would be produced when the crane was lifting 100
tons. So great a margin of safety is necessary on account of the jerks
and other occasional great strains that arise in the raising and the
lowering of heavy weights. For a crane intended to raise 10 tons, the
engineer must therefore design a tie rod which not less than 300 tons
would tear asunder. It has been proved by actual trial that a rod of
wrought iron of average quality, one square inch in section, can just
withstand a pull of twenty tons. Hence fifteen such rods, or one rod
the section of which was equal to fifteen square inches, would be just
able to resist 300 tons; and this is therefore the proper area of
section for the tie rod of the crane we have been considering.
46. In the same way we ascertain the actual thrust down the jib; it
amounts to 35 tons, and the jib should be ten times as strong as a
strut which would collapse under a strain of 35 tons.
47. It is easy to see from the figure that the tie rod is pulling
the upright, and tending, in fact, to make it snap off near B. It is
therefore necessary that the upright support A B (Fig. 17) be secured
very firmly.
LECTURE III.
_PARALLEL FORCES._
Introduction.—Pressure of a Loaded Beam on its
Supports.—Equilibrium of a Bar supported on
a Knife-edge.—The Composition of Parallel
Forces.—Parallel Forces acting in opposite
directions.—The Couple.—The Weighing Scales.