Handbook of Railroad Construction; For the use of American engineers. / Containing the necessary rules, tables, and formulæ for the location, construction, equipment, and management of railroads, as built in the United States.

by George L. Vose

Fig. 103 A.

That the resultant of the forces acting upon the top of the tower may be vertical, the angles G C A, and A C H, fig. 103, must be equal; if not, the masonry must be so arranged as to cause the resultant to pass through the centre of gravity. When more than one span is used, and the openings are unequal, that the intermediate pier or piers shall not be pulled over, the cable of the largest, and consequently heaviest span, must have a greater inclination from the horizontal than that of the shorter span; the product of the tensions by their respective inclinations must be equal. Mr. Roebling’s plan in connecting several spans, is to attach the cables of adjacent spans to a pendulum upon the pier, by which arrangement the difference in tension upon the different cables swings the pendulums, without racking the masonry.

Problem 4.

228. Given the weight per foot of bridge and load, to find the tension at the lowest point of the curve. The formula for the minimum tension, that at the vertex F of the curve, is

T = ph²
2f;

where p is the weight per foot of bridge and load, h the half distance between the points of suspension, and f the versed-sine. Thus the span being one thousand feet, the versed-sine eighty feet, and the load per lineal foot six thousand lbs., the formula becomes

T = 6000 × 500²
160 = 9375000 lbs. or 4185 tons.

The maximum tension is at the points of support, and is expressed by the formula

T = ph
2f[h² + 4f²]^½: