. Iron truss bridges for railroads. Methods of calculating strains, with a comparison of the most prominent truss bridges, and new formulas for bridge computations; . ns are equal to each other, and their sum is equal tothe entire horizontal component that originally required transmission to the left, and itneutralizes the horizontal force developed on the right. On the Bottom Chord we see thatthe greatest tension is equal to either horizontal component of the weight diminished byone of the top compressions {ebr=em), and that the tensions diminish toward the furtherabutment, and at the foot of
. Iron truss bridges for railroads. Methods of calculating strains, with a comparison of the most prominent truss bridges, and new formulas for bridge computations; . ns are equal to each other, and their sum is equal tothe entire horizontal component that originally required transmission to the left, and itneutralizes the horizontal force developed on the right. On the Bottom Chord we see thatthe greatest tension is equal to either horizontal component of the weight diminished byone of the top compressions {ebr=em), and that the tensions diminish toward the furtherabutment, and at the foot of the last Tie become 0, thereby showing that all the hori-zontal forces developed in the process of transmitting the portions of the weight to eachabutment, neutralize each other through the horizontal members of the truss, leaving nounneutralized force to cause motion. We also see that the weights are transmitted onlyby the vertical or inclined parts, the horizontal parts being only used for the necessary 22 IRON TRUSS BRIDGES FOR RAILROADS. decompositions required to pass the weights from incUned to vertical members, andvice versa. Fia. 14. ^ i G ^ c . r ^ C. There is a class of so-called trusses, distinct from all others, that are more properly-trussed girders. Figure 14 represents one of the class which we will have to discuss indetail further on. It is manifest that, there being but two members meeting at A, thecomponent parts of the weight must pass up those two. As they make equal angleswith each other, the weight is halved. So far 0 has received but half the weight, whenthe relative distances from A to JB and C show that it should receive three-fourths. Thecomponent at JE is resolved vertically and horizontally. The vertical component trans-mitted to J^is resolved into Ft, that goes to increase the forces at C; and into F^i, thatgoes to B. An examination of the figure will show that, as before, the horizontal forcesdeveloped in the Top Chord are equal and opposite, whil
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Keywords: ., bookcentury1800, bookdecade1870, bookpublishernewyo, bookyear1870
