W

the weights W, are 1, 2, 3, 4 also, and

becomes,,3,4 =

b

1.0 in all cases, showing that when the loads are proportional to the breadths the deflections are constant. But R, or 8× W, will not be constant, but become 1 x 1 = 1:1 x 2 = 2: 1 x 3 = 3 and 1 x 4 = 4, &c., showing that R is simply and directly proportional to b.

Then, for the influence of the length: the same Rule (659) shows that d = L3 x W; then with lengths 1, 2, 3, 4, W, or the dead loads, will obviously be in the simple inverse ratio, 1,,,, &c.: hence the deflections, or d

=

=

4: 33 ×

=

=

L3 x W, become 9: and 43 ×

=

=

16, and 1:

=

13 × 1 = 1 : 2 R, or the power to resist Impact, or 8 × W, becomes 1 x 1 4 × 1/2 = 2: 9 × 3 3 and 16 x 1 4, which are in the simple direct ratio of the lengths. This is remarkable, being precisely the reverse of the effect of length on the dead load, where of course the strength is inversely as the length.

We thus find that the power of Rectangular Beams in resisting Impact, or R, is simply and directly proportional to the length multiplied by the depth, and by the breadth. Putting these results into the form of Rules, we have for Rectangular beams:

breadth in

=

In which d = the depth of the Beam in inches: b inches: L= length between supports in feet: w falling weight, say in lbs.: h height fallen by w in inches, and R = the Specific Resilience of the Material, as given by cols. 5 or 6 in Table 67; cols. 9, 12 in Table 64, &c.

=

These Rules do not allow for the resistance from the Inertia of the Beam or of any dead load it may bear before receiving the force of the blow, and by which the results may be modified very considerably.

(781.) "Effect of the Inertia of the Beam."-Mr. Hodgkinson has shown by his experiments that in resisting impact, the

power of a heavy beam is to that of a light one as the inertia of the beam plus the falling weight is to the falling

In which, the letters have the same signification as in (780), except I, which is the Inertia of the beam and the load upon it. The inertia of a beam uniform in section from end to end, supported at the ends, and struck in the centre, may be taken at half the weight between supports. To this has to be added the whole central load (if any), or if otherwise distributed, it must be reduced to an equivalent central load.

(786.) The application of these rules may be illustrated by an example:-say we have a beam of English Oak 12 inches deep, 6 inches wide, 20 feet long between supports, and we require the height from which a weight of 5 cwt., or 560 lbs., must fall to strain the beam to th of the breaking strain. First we have to find the inertia of the beam from its weight: we have 1 x x 20 = 10 cubic feet, and by col. 2 of :—we 1 Table 150, 48·4 × 10 484 lbs. for the weight, the inertia is therefore 242 lbs., and taking the value of r at 2.04 from col. 6 of Table 67, we get by Rule (782):

=

h

12 × 6 × 20 × 2·04 × (242 + 560)

=

= 7.51 inches fall.

5602

To find the height of fall to break the beam with the same

falling weight, we obtain the value of R from col. 5 of the same Table = 78.4, and h becomes

12 × 6 × 20 × 78.4 × (242 + 560)

5602

= 288.7 inches,

or 24 feet: hence the ratio of the breaking and safe heights is 288 77.51 = 38.4 to 1, as in col. 9: see (825).

1120

(787.) To vary the illustration, say that the beam was loaded with a central dead weight of 10 cwt., or 1120 lbs., adding which to the inertia of the beam itself we obtain I +242 1362 lbs., and for th of the breaking strain, h becomes

12 × 6 × 20 × 2·04 × (1362 + 560)

5602

=

= 18 inches fall,

whereas when unloaded the fall was 7.51 inches only (800).

If the extra load of 10 cwt. had been equally distributed all over the length of the beam, it would have been equivalent to a central load of 5 cwt., or 560 lbs., hence I would be 560 + 242 802 lbs., and h becomes

=

12 × 6 × 20 × 2·04 × (802 + 560) = 12.75 inches fall.

5602

(788.) The three rules in (783), &c., by which d, b, and d × b are respectively determined, are difficult in application, because the inertia of the beam depends on the dimensions which are unknown, but we can assume dimensions and solve the question by repeated approximations to any desirable degree of accuracy. Say we require the dimensions of a beam of Elm 15 feet long to bear safely a weight of 8 cwt., or 896 lbs., falling 12 inches. We will assume the dimensions at 12 inches square, hence the beam contains 15 cubic feet, weighs by Table 150, 36.65 × 15 = 550 lbs., its inertia 550÷2 275 lbs.; the value of r from col. 6 of Table 67 is 2.13, and the rule 8962 × 12 (785) will now give d × b

=

=

= 257,

15 × 2·13 × (275 + 896) and 257 = 16 inches square, instead of 12 inches as we assumed. But 16 inches would be too much, because the weight, and thereby the inertia, would be greater than we

assumed. Assuming 15 inches square, or 1·25 feet, as a second approximation, the weight comes out 1.25 × 1·25 × 15 × 36 15 847 lbs., hence the inertia

=

becomes

8962 × 12

=

=

424 lbs., and d × b

228, and 228 = 15.1

15 × 2·13 × (424 + 896) inches square, agreeing sufficiently closely with the sizes we assumed. By (821) it is shown that the beam might be of any dimensions at pleasure, so long as d x b = 228; thus it might be 14 x 16.3 = 228, or 12 x 19 228, &c., &c., and it would be unimportant whether it was struck by the falling weight in the direction of its greatest or least dimension, as shown in (824).

=

It will be observed that these Rules apply only to beams of rectangular sections; the following apply to beams of all sections, but are not so facile in application :—

In which W = a given Statical weight on the beam, in pounds. F = Flexure or deflection produced by W, in inches.

h

=

=

height fallen by w, in inches.

f flexure in inches produced by the impulse of w. Inertia of the beam and its load, as in (785), &c.

I

=

(791.) These rules connect together the laws of the Statical and Dynamic forces, enabling us to reason from one to the other: thus from the known strength and stiffness of a beam of any form, material, mode of fixing, &c., under a statical load or dead weight, we may calculate the effect of an impulsive strain upon it.

As an illustration of the application of these rules, we may take Mr. Bevan's experiments in Table 120:-they were made on a beam whose inertia or half-weight between supports was 63.5 lbs., a deflection of 1 inch was produced by a dead weight of 148 lbs., and the falling weight was 28 lbs. The deflections

produced by various heights of fall are given in col. 2 and as calculated by the rule in col. 3. Thus, with 12 inches fall,

fo=

12 1 x 282 × 2

(118x (63-5 +28))

periment it was 1.25 inch.

1.18 inch deflection: by ex

TABLE 120.-Of IMPACT on BEAM of WOOD, 18 feet long, &c.

(792.) Again:-say that we have a beam in which a dead weight of 400 lbs. produces a deflection of 1.2 inch, the inertia or half-weight between supports being 600 lbs., and the falling weight 200 lbs.; we require the fall to produce a deflection of say 2 inches. Then by Rule (789):

h =

2.5o × 400 × (600 + 200)

1.2 × 2002 × 2

= 20.8 inches fall.

It will be observed that in both these examples, the dimensions of the beam are not given, nor are they required by the rules in (789), &c.

(793.) But these rules are based on the supposition that the elasticity of the deflecting beam is perfect, or that the deflections are strictly and simply proportional to the loads even up to the breaking point; this is far from the truth with cast iron (688), and is not strictly true perhaps with any material, although sufficiently so for practical purposes, with wrought iron and steel up to the "limit of Elasticity" (692). This will be seen by comparing Tables 121 and 122, for the rule in (789) shows that, other things being the same, the fall h should vary as f2, and a comparison of cols. 2 and 3 of Table 121 shows that