The first and third methods are approximate, the second is exact for the particular engine and train loads assumed.

The advocates of the concentrated-load system maintain that a uniform load, or equal loads concentrated at the panel-points, does not represent what actually occurs when two coupled engines followed by a train pass over the bridge, and the same applies to a less extent with an engine excess.

The advocates of the uniform load contend that the concentrated-load system is unnecessarily complicated, as it is possible to assume such uniformly distributed loads as shall cover the maxima stresses which actually occur without producing errors which are excessive on the side of safety or danger. Moreover, that the actual wheel loads and uniform loads used in calculating the stresses may be considerably modified by the use of heavier engines and trains, or different wheel spacings, in the future.

With regard to the uniform-load system. Sir B. Baker 1 estimates the equivalent uniform load for heavily engined lines as follows:

The loads given in the foregoing table are not now sufficient to provide for the traffic on heavily engined lines, although they were ample at the time the table was compiled.

The weight of engines and the trains hauled by them have been considerably increased during the last ten years; the use of heavy gradients of 1 in 40 in America and the colonies necessitates the use of heavy engines. Figs. 324 and 325 illustrate two types of consolidation goods engines in use on the New South Wales Government Railways, where grades as steep as 1 in 30 occur. Fig. 329 shows one of Mr. Cooper's wheelload diagrams known as Class A, which has been very largely

"Short-Span Railway Bridges."

used in calculating the stresses on American bridges; much heavier engines are, however, in use in America.

Consolidation engines on New South Wales Government Railways.

Fig. 326 represents a diagram of half-engine loads, which will be made use of in connection with an example of the concentrated-load system.

In estimating the equivalent uniform load per foot run for the purpose of calculating the maxima stresses in a structure, due consideration should be given to the probable increase in the engine and train loads which may be required to pass over the structure in the future.

Half-engine loads used in tabulations for maxima stresses in example on the concentrated-load system.

34.24

-18.08 ★- 3.75-★-4·3- ★-4·3-★ - 7:09 - ★-4·53-★ -3.07- ★-4·83- ★

Ө

FIG. 326.

The same uniform load will not generally be suitable for both trusses and plate web girders, being slightly greater in the former than in the latter. The uniform load producing maxima bending moments is generally less than that required to produce the maxima shearing stresses.

This diagram might be still further simplified by using the nearest foot in all cases, or by using the first place of decimals only in the spacing of the driving wheels.

бобы

Mr. Theodore Cooper, in the following table, gives the equivalent uniform loads for the 101 American tons consolidation engine (Fig. 327), which is not very different from the New South Wales consolidation engine (Fig. 325); also for a still heavier engine known as the Lehigh Heavy Grade engine (Fig. 328).

8-1

543

+58 +4·54571

DOO

4

30,000

30,000

FIG. 327.

Lehigh Heavy Grade engine.

54:3

-3·7--48 4'

D O O O O O

18,000

FIG. 328.

TABLE LIV.

Two 101 American tons consolidations, followed by 3000 lbs.

Mr. Cooper specifies for the extra heavy Class A service:-

Two engines like Fig. 327 followed by a train of 3000 pounds per foot run, or 80,000 pounds equally distributed on two pairs of drivers spaced 7 feet centres, as shown in Fig. 329.

40,000

FIG. 329.

40,000

Professor Waddell has prepared the following table of

equivalent uniformly distributed live loads which produce practically the same bending moments in plate web girders as the concentrated wheel-load diagram shown in Fig. 330:

He further recommends that, in order to produce the maximum shearing stress, an amount be added to the total

distributed load divided by two, which is given by the follow

ing formula :

W = A + Bl

where W the additional load,

the span; A and B are constants which, for the engine loads shown in Fig. 330, are 8000 and 100 respectively.

Other constants may be determined for different engine loads. Some engineers use two different sets of tables, one for maxima moments, the other for maxima shears.

In order to find the maxima reactions of longitudinal girders upon the cross-girders, Professor Waddell gives the following simple rule, which may be used instead of the method explained in the design for a deck of a railway bridge, Chapter XIV., Fig. 320. Multiply the uniformly distributed live load per foot run for a span of two panel lengths by the length of one panel.1 Thus for a 20-foot panel the maximum cross-girder reaction. is

1 Trans. of the Amer. Soc. of C.E., February and March, 1892.

or 18.58 tons at each point of attachment. With the halfengine loads shown in Fig. 326, the equivalent distributed load would be 4700 lbs. per foot for a span of 40 feet, and the reaction is

or 21 tons at each point of attachment, agreeing very well with the reaction calculated from the wheel concentrations, Fig. 320. The rule is, however, strictly true, and is proved thus—

AR

In Fig. 331 we have R 2Wx

=

FIG. 331.

In a girder of length l the central bending moment is—

Therefore, as the rule is true for a single load, it is true for any summation of loads.

In simple Pratt trusses, with panel length of 25 feet, Professor Waddell gives the following equivalent distributed loads to cover the stresses produced by the engine diagram, Fig. 330.

In using this table instead of the actual concentrations for bending moments, the greatest error on the side of safety occurs in the 100-feet truss, and is 477 per cent. The greatest error on the side of danger is 2.21 per cent. In the shearing stresses, the loads given in the table produce stresses the error of which