.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, and supplying values, we have = 240 tons, or 537600 lbs., for the two Lower Chords, and ½ 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 ÷ 2000 = 134.4 sq. inches. 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 must be so proportioned as to safely bear, as a horizontal beam, this weight. Suppose we take three sticks of 8" x 12", to form the chord (the greater dimension being the depth,) we shall have 3 x 8" x 12" = 288 square inches area of section, and

we shall have after deducting allowances (288-128) 160 square inches area, giving an excess over 134.4, the area demanded, sufficient to cover allowances for any accidental strain.
 

Upper Chords. The upper chords are compressed as forcibly as the lower ones suffer tension—owing to the action and reaction of the diagonals. In this case the compression is 268800 lbs., and as 1 square inch of section will safely bear 1000 lbs., we have for the area required, 268800 ÷ 1000 = 268.8 square inches,—three pieces 8" x 11" will give 264 square inches and this area will require no reduction, as the whole chord presses together when properly framed and is not weakened by splicing. So far, the calculations made would apply to either of the three Bridges mentioned, as well as to a Warren Truss. But now, to obtain the dimensions of the web members, so called, of the Truss, it is necessary to decide upon the specific variety. The form of Bridge in more general use in the United States is called the Howe Truss, from its inventor, and in spans of 150 feet, and under, is very reliable; for spans exceeding 150 ft. it should be strengthened either by Arch Braces or by the addition of Arches, as the heavy strains from the weight of bridge and load bearing on the feet of the braces near the abutments, tend to cripple and distort the truss by sagging, although the Baltimore Bridge Co. have built a Wooden Howe Bridge of two 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 , and we have , or reducing, W = 55308 lbs. Now, these proportions will give ample strength for both flexure and compression, for if we block the two sticks composing the end brace together, and firmly connect them by bolts, we shall have a built beam of 24" x 10"—whence = 165925 lbs., and as 134400 lbs. was all that the conditions demand, we really have an excess of strength. The next set of braces supports the weight of the rectangle included between the upper ends of the braces and the two chords, and the dimensions of the sticks are calculated in the same manner. We find, as we approach the centre of the bridge, that the strains on the braces become less, and consequently their scantling should be reduced, but in ordinary practice this is seldom done.
 

Rods. The next thing is to ascertain the dimensions of the various tie rods. It is evident that the same weight comes upon the first set of rods, as on the first set of braces—which will give for the rods at one end of one truss, 134400 lbs.; and as there are two of these rods, each will sustain a strain of 67200 lbs.—and, at 15,000 lbs. per square inch, will have an area of 4.48 sq. inches, and, by Vose's Tables, must have a diameter of 2½ inches. The sizes of the rods in each set will decrease towards the centre of the bridge as the weight becomes less.