Both of these tables were calculated for a single Railroad track, and would answer equally well for a double Highway Bridge.  In the bridge according to Trautwine’s Table, each lower chord is supposed to have a piece of plank, half as thick as one of the chord pieces, and as long as three panels, firmly bolted on each of its sides, in the middle of its length.

* * * * *

=PRATT’S BRIDGE.=

This is opposite in arrangement of parts to a Howe Bridge, as the diagonals are rods, and sustain tension, and the verticals are posts, and suffer compression: 

Example.—­Span = 100 feet. 
Rise = 12 "
Panel = 10 "
Weight per lineal ft. = 3000 lbs.

The tension on the lower, or compression on the upper chord, will be

300000 x 100
------------ = 3333333 lbs.
96

[TeX:  $\frac{300000 \times 100}{96} = 3333333$]

The dimensions of the chord and splicing would be found in the same manner as for a Howe Truss.

=Suspension Rods.= Fig. 1, Pl.  III., represents an elevation of a Pratt Bridge.  Now, it is evident that the first sets of rods must support the weight of the whole bridge and its load, which we have found to be 300000 lbs.  Each truss will have to sustain 150,000 lbs., and each end set of rods 75,000 lbs.  Now, if there are two rods in each set,—­each rod will have to bear a strain of 37500 lbs., and this will have an increase due to its inclination, so that the strain on it must be found by the following proportion: 

    Height :  diagonal ::  W :  W’ or

    12 :  15.8 ::  37500 :  49375 lbs.

Referring to the Table for bolts, we find that 2-1/8 gives a strength a little in excess, and will be the proper size.  The next set of rods bear the weight of the whole load, less that due to the two end panels, and so on.  Fig. 2, Pl.  III, shows the manner of applying the rods.  The bevel block should be so fitted to the chord that it will not have a crushing action.

=Counters.= Top and bottom chords are always used in this bridge, and consequently the counter rods have only to sustain the movable load on one panel.  The weight of the moving load cannot be more than 2000 lbs. per lineal foot which, for a panel of 10 ft., gives 20000 lbs., or 10,000 lbs. for each set, and if we have two rods in a set, the strain on each rod will be 5000 lbs., increasing this for inclination, we shall have,

  12 :  15.8 ::  5000 :  6585 lbs.,

requiring a rod of 3/4 of an inch diameter.  The posts in this bridge correspond to the braces of the Howe Truss, but being vertical, are not so large.

Subjoined are two Tables, one by Prof.  G.L.  Vose, and one by Mr. Trautwine, giving principal dimensions for bridges of different spans of the Pratt type of Truss.

TABLE OF DIMENSIONS OF A PRATT TRUSS.