1-5/8 | 2 |1-1/16
75 | 12 | 10 | 3 | 6x9 | 3 | 6x11| 2 |6x8 | 2 |6x6 | 1 |6x6 | 2 |1-7/8 | 2 |1-3/16
100 | 15 | 11 | 3 | 6x10| 3 | 6x12| 2 |8x9 | 2 |6x8 | 1 |6x8 | 2 |2-3/16| 2 |1-5/16
125 | 18 | 12 | 4 | 6x10| 4 | 6x13| 2 |8x10| 2 |6x9 | 1 |6x9 | 2 |2-5/8 | 2 |1-3/8
150 | 21 | 13 | 4 | 8x10| 4 | 8x14| 3 |9x10| 3 |6x9 | 2 |6x9 | 3 |2-3/8 | 3 |1-3/16
175 | 24 | 14 | 4 |10x12| 4 |10x15| 3 |9x11| 3 |8x8 | 2 |8x8 | 3 |2-5/8 | 3 |1-1/4
200 | 27 | 15 | 4 |12x12| 4 |12x16| 3 |9x12| 3 |8x10| 2 |8x10| 3 |2-7/8 | 3 |1-3/8

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, sho