As the cast-iron arch of a bowstring girder serves only to resist compression, its place can as well be filled by an arch of brick, footed on a pair of cast-iron skew-backs; and these held in position by a pair of wrought-iron tie-rods, as shown in Fig. 49. This system of construction is preferable to the bowstring girder, in that the tie-rods are not liable to injury7 by "shrinking in," and the cost is less. From Transverse Strains, Art. 596, we have -
an equation in which D is the diameter in inches of each of the two tie-rods of the brick arch; U is the load in pounds uniformly distributed over the arch; / is the span of the arch in feet; and d, in inches, is its versed sine, or its height measured from the centre of the tie-rod to the centre of the
. thickness or height of the arch at middle.
This equation may be put in words as follows: Rule LXV. - Multiply the weight by the length; divide the product by 9425 times the depth, and the square root of the quotient will be the diameter of each rod.
Example. - What should be the diameter of each of the pair of tie-rods required to sustain a brick arch 20 feet span from centres, with a versed sine or height at middle of 30 inches, to carry a brick wall 12 inches thick and 30 feet high, weighing 100 pounds per cubic foot? The load upon this arch will be for so much of the wall as will occur over the opening, which will be about one foot less than the span of the arch, or 20 - 1 = 19 feet. Therefore, the load will equal 19 x 30X 1 x 100 = 57,000. pounds; and hence, U = 57000, l = 20, d = 30, and, by the rule -
or, the diameter of each rod is required to be 2 inches.