These bridges are subject to a compressive strain along the top string or chord, and to a tensile strain along the bottom string.

The strains are equal when the strings are parallel: they are greatest in the middle of the span, and diminish towards the ends.

The method of finding these strains, on the principle of the lever, was explained in Sect. I., and is given by the following rule, when the load is uniformly distributed over the roadway.

363. To find the Strain in lbs. at the Centre of the Top and Bottom Strings.

Rule. - Multiply the span in feet by the weight in lbs. on each truss, including its own weight, and divide by 8 times the depth of the girder or truss in feet.

For every 1000 lbs. of strain thus found allow in the upper string 1 square inch area of cross section, and for every 700 lbs. allow 1 square inch in the bottom string; a greater quantity of timber being required in the bottom than in the top string to compensate for the weakness caused by joints when the pieces are subject to a tensile strain.

Example. - Find the horizontal strain in lbs. on the strings or chords of an ordinary Howe truss bridge having a span of 100 feet and a depth of 12 feet, the gross distributed load on each truss being 75 tons, or 168,000 lbs. 100x 168000/8 x 12 = 175000 lbs., the strain at the centre of the top and bottom strings, which, divided by 1000 for the top and 700 for the bottom, give

175 Squre inches and 250 square inches respectively for the sectional area of the strings in each truss, and by making the depth 10 inches, the width must be 17 1/2 inches in the top and 25 inches in the bottom. It is a good practical rule (and one which is observed in Howe's bridges) to make the upper chord or string consist of three, and the lower of four timbers to each truss; a joint will then occur in each panel, and the pieces should be sufficiently long to extend over four panels. With this arrangement three of the timbers must be supposed to sustain the whole strain, since that which contains the joint is not capable of opposing any resistance.*

364. In a solid beam resting on two supports, the strain at the ends of the strings is nothing, and it increases uniformly towards the centre; but in a bridge truss of a single span there will be a horizontal strain at the end of the brace nearest the abutment, which will equal the weight on the brace multiplied by the co-tangent of the inclination of the brace, which, if 45°, the horizontal strain will equal the vertical weight at the end. If the angle with the horizontal is greater than 45°, (which is generally the case) the horizontal strain will be less than the weight, and consequently it will be safe in practice to assume the horizontal strain at the end of the string, or, more correctly, at the end of the first brace, as equal to the vertical force acting on that brace. This vertical force is equal to one-half of the whole weight on the truss, or in the example given in Art. 363 it will be 84,000 lbs., and the cross section to resist it will be 84 inches.

Having determined the cross section of the strings at the centre and end, a uniform increase between these points will fulfil all the necessary conditions.

365. Another circumstance must be taken into consideration in determining the size of the strings that carry the load directly, and which produces a cross strain upon the portion that lies between any two posts. It is that of a beam supported at the ends and uniformly loaded, or loaded in the middle, as the case may be. The cross section to support this load should be calculated by the rules for the stiffness of beams, given in Section II. In a well-proportioned truss the

* Haupt's ' General T eory of Bridge Construction.,' dimensions calculated to fulfil these last conditions are usually much less than those required in the former case.

366. When the load on a truss is confined to the centre instead of being uniformly distributed, the strain on the strings at the centre will be doubled.