This section is from the book "Elementary Principles Carpentry", by Thomas Tredgold. Also available from Amazon: Elementary Principles Of Carpentry.

Fur the end braces, multiply that portion of the weight in lbs. supported by the pier ( = one-half the gross weight in a uniformly loaded girder) by the length of the brace in feet. The product divided by the vertical height of the brace or distance between the top and bottom strings will give the strain on the brace in lbs. This again divided by 1000 will give the sectional area required for the timber.

Example. - Find the strain and size of the timber for the end braces of a simple truss girder 100 feet span, 10 feet in depth between the strings, the panels 9 feet wide, and the gross distributed load on the girder 168,000 lbs., the length of the brace being 13.45 feet.

84000 x 13.45/10 = 112980 lbs., the pressure transmitted through the brace, and allowing 1000 lbs. to the square inch, the sectional area required is 113 square inches nearly.

368. With a uniform load the strains on the braces diminish from the ends to the middle of the girder until they become nothing.

In practice, as there never is a brace exactly in the middle, those on each side of it must sustain the portion of the load borne by the centre panel, i. e. one-half to each brace increased by the length of the brace to the vertical height as for the end ones. Taking the width of the panel at 9 feet, or 1/11 th of the span nearly, the load on each panel of the girder in the last example will be 15,273 lbs., the half of which multiplied by 13.45, the length of the brace, and divided by 10, the vertical depth, gives 20,542 lbs. nearly for the strain, and allowing 1000 lbs. to the square inch, as before, we obtain a sectional area of 20 1/2 square inches for the brace nearest to the middle of the girder. The intermediate braces can easily be proportioned between these extremes. By the same rules the strains on the braces can be obtained when they intersect one another as in the common lattice, the only difference being that the strain or pressure is divided among a greater number.

369. Where the unsupported length of the brace is considerable in proportion to the diameter, the 1000 lbs. per square inch will not be a sufficient allowance, and the rules for long pillars, as given in Sect. II., must be applied. The braces are seldom so long as to necessitate this; they are almost always in practice supported at one or more points, which reduces the length.

370. When the truss is loaded in the middle only, the braces will all be strained alike, and will equal one-half the lead increased in proportion to the length of the brace divided by the vertical height as for a uniform load.

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