Trusses of 300 ft. span, 30 ft. rise, and 26 ft. wide, without any
arch, but it has a wrought iron lower chord, and is only proportioned
for a moving load of 1000 lbs. per ft. run. [Vide Vose on R.R.
construction.]

In order to ensure uniformity in strength in the chords--but one joint
should be allowed in a panel--and that should come at the centre of
the panel length--but in long spans this cannot always be done.


=Web Members.= We will now proceed to calculate the web members of a
Howe Truss of the foregoing dimensions, when subjected to the strains
above mentioned.

=Braces.= The end braces must evidently support the whole weight of
the bridge and load, which for one end of one truss will be 134400
lbs., and as these braces are in pairs,--67200 lbs. will be the strain
vertically on the stick--but as this stick is a diagonal--whose
vertical is 15 ft., and horizontal 10 ft., we shall have for its
length 18 ft. in round numbers, whence the strain along the diagonal
will be found from the proportion 15 : 18 :: 67200 : 80640 lbs.,
whence we have an area of 80 inches required for compression, or a
stick of 8" x 10". Now, to ascertain if this is stiff enough for
flexure, we will substitute these values in the equation

2240 bd³
W = --------, and we have
L²

[TeX: $W = \frac{2240 \times bd^3}{L^2}$]