there are more timbers required.

Mr. G.L. Vose, in his admirable work on R.R. Construction, observes
very truly that "The braces, at the end of a long span, may be nearer
the vertical than those near the centre, as they have more work to do.
If the end panel be made twice as high as long, and the centre panel
square, the intermediates varying as their distance from the end, a
good architectural effect is produced."

Now it is necessary for us to have some data from which to determine
the approximate weight of the bridge, and also its load. These can be
found by comparing weights of bridges in common use, as obtained from
reports. In a small bridge of short span, the weight of the structure
itself may be entirely neglected, because of. the very small
proportion the strains caused by it bear to those due to the
load;--but, in long spans, the weight becomes a very important element
in the calculations for strength and safety--inasmuch as it may exceed
the weight of the load.

In all Bridges of 120 ft. span, about 1/3 of a ton, per foot run, will
be the weight of each truss for a single track, including floor
timbers--transverse bracing, &c. If the bridge were loaded with
Locomotives only, the greatest load would be, on the whole bridge--160
tons = 1.33 tons per ft. run of the bridge or .666 tons per ft. run of
each truss. Now if we make the rise of the bridge 15 ft., and divide
the span into 12 panels of 10 ft. each, we shall have for total weight
of bridge and load 240 tons, or for a single truss 10 tons to each
panel.

=Lower Chords.= Now to find the tension on the Lower Chords,

W x S
T = ----- and supplying values, we have
8 h

[TeX: $T = \frac{W \times S}{8 h}$]

240 x 120
T = --------- = 240 tons, or 537600 lbs.,
8 x 15

[TeX: $T = \frac{240 \times 120}{8 \times 15 = 240$}

for the two Lower Chords, and 1/2 of this, or 268800 lbs. for one
chord. The Tensional Strength of timber for safety may be taken at
2000 lbs. per square inch of section, and hence the area of timber
required to sustain the above strain will be

268800
------ = 134.4 sq. inches.
2000

[TeX: $\frac{268800}{2000} = 134.4$]

But this chord has also to sustain the transverse strains arising from
the weights passing over it, and, as in the case of a Locomotive, the
weight of 20 tons on 2 pair of drivers, (or 10 tons for one truss,)
may be concentrated on the middle point of a panel--the chord