consist of two hard-pine beams, with the struts coming between them. We will have two rods, instead of one, at R, coming down each side of the strut, and passing through an iron casting below the beams, forming supports for them. The height of truss from centre to centre of timbers we must limit to 18 inches, and we will space the trusses 8 feet on centres. Then the total floor-area supported by one girder equals 8 feet by 39 feet, equal to 312 square feet. The heaviest load to which the floor will be subjected will be the weight of students, for which 75 pounds per square foot will be ample allowance; and the weight of the floor itself will be about 25 pounds; so that the total weight of the floor and load will be 100 pounds per square foot. This makes the total weight liable to come on one girder 31,200 pounds. Then we find, Formula 14, = 11450 pounds. Tension in both rods R = 0.367 W The timber in the truss will be hard pine, and hence we must have 106800 or 107 square inches, area of cross-section in the strut, 1000 which is equivalent to a 9-inch by 12-inch timber. or, as that is not a merchantable size, we will use a 10-inch by 12-inch strut. The tie-beams will each have to carry one-half of 106000, or 53000 53000 2000 pounds; and the area of cross-section to resist this equals 27 inches, or 24 inches by 12 inches. The distributed load on one section of each tie-beam coming from the floor-joist equals 13 × 8 × 100 = 10400 pounds; and from Formula 23 we have 2 x W x L 2 × 10400 × 13 B 5 × D × A of each tie-beam must be 33 inches + 2 inches the tie-beams will be 6 inches by 12 inches. Each = 33 inches. 5 × 144 × 100 carry 5725 pounds, and their diameter will be nearly. Then the breadth 6 inches hence rod will have to Thus we have found, for the dimensions of the various pieces of the girder: Two tie-beams 6 inches by 12 inches; two rods at each joint, inch diameter, and strut-pieces 10 inches by 12 inches. CHAPTER XX. RIVETED PLATE-IRON GIRDERS. WHENEVER the load upon a girder or the span is too great to admit of using an iron beam, and the use of a trussed wooden girder is impracticable, we must employ a riveted iron-plate girder. Girders of this kind are quite commonly used at the present day; as they can easily be made of any strength, and adapted to any span. They are not generally used in buildings for a greater span than sixty feet. These girders are usually made either like Fig. 1 I Fig. 1. Fig. 2. or Fig. 2, in section, with vertical stiffeners riveted to the webplates every few feet along their length. The vertical plates, called "web-plates," are made of a single plate of wrought-iron, rarely less than one-fourth, or more than five-eighths, of an inch thick, and generally three-eighths of an inch thick. Under a distributed load, the web of three-eighths of an inch thick is generally sufficiently strong to resist the shearing-stress in the girder without buckling, provided that two vertical pieces of angle-iron are riveted to the web, near each end of the girder. These vertical pieces of angle-iron or T-iron, whichever is used, are called "stiffeners;" and when the girder is loaded at the centre, and sometimes when under a distributed load, it is necessary to use the stiffeners for the whole length of the girder, placing them a distance apart equal to the height of the girder. The web is only assumed to resist the shearing-stress in the girder. The top and bottom plates of the girder, which have to be proportioned to the loads, span, and height, are fastened to the web by means of angle-irons. It has been found, that in nearly all cases the best proportions for the angle-irons is 3 inches by 3 inches by 1⁄2 inch, which gives the sectional area of two angles five and a half square inches. The two angles and the plate taken together form the flange; the upper ones being called the "upper flange," and the lower ones the lower flange." RIVETS. - The rivets with which the plates and angle-irons are joined together should be three-fourths of an inch in diameter, unless the girder is light, when five-eighths of an inch may be sufficient. The spacing ought not to exceed six inches, and should be closer for heavy flanges; and in all cases it should not be more than three inches for a distance of eighteen inches or two feet from the end. Rivets should also not be spaced closer than two and a half times their diameter. Rules for the Strength of Riveted Girders. In calculating the strength of a riveted girder, it is customary to consider that the flanges resist the transverse strain in the girder, and that the web resists the shearing-strain. To calculate the strength of a riveted girder very accurately, we should allow for the rivet-holes in the flanges and angle-irons; but we can compute the strength of the girder with sufficient accuracy by taking the strength of the iron at ten thousand pounds per square inch, instead of twelve thousand pounds, which is used for rolled beams, and disregarding the rivet-holes. Proceeding on this consideration, we have the following rule for the strength of the girder : The height of the girder is measured in inches, and is the height of the web-plate, or the distance between the flange-plates. The web we may make either one-half or three-eighths of an inch thick; and, if the girder is loaded with a concentrated load at the centre or any other point, we should use vertical stiffeners the whole length of the girder, spaced the height of the girder apart. If the load is distributed, divide one-fourth of the whole load on the girder, in tons, by the vertical sectional area of the web-plate; and if the quotient thus obtained exceeds the figure given in the following table, under the number nearest that which would 1.4 × height of girder be obtained by the following expression, thickness of web then stiffening pieces will be required up to within one-eighth of the span from the middle of the girder. dx 1.4 t 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 3.08 2.84 2.61 2.39 2.18 1.99 1.82 1.66 1.52 1.40 1.28 1.17 1.08 1.00 0.92 EXAMPLE. A brick wall 20 feet in length, and weighing 40 tons, is to be supported by a riveted plate-girder with one web. The girder will be 24 inches high. What should be the area of each flange, and the thickness of the web? 3 × 40 × 20 Ans. Area of one flange = 10 × 24 = 10 square inches. Subtracting 5 square inches for the area of two 3-inch by 3-inch angle-irons, we have 5 square inches as the area of the plate. If we make the plate 8 inches wide, then it should be 5÷8, or § of an inch thick. The web we will make ₹ of an inch thick, and put two stiffeners at each end of the girder. To find if it will be necessary to use more stiffeners, we divide of 40 tons, equal to 10 tons, by the area of the vertical section of the web, which equals § of an inch × 24 inches = 9 square inches, and we obtain 1.11. The expression 1.4 × height of girder thickness of web , in this case, equals 89.6. The number near est this in the table is 90, and the figure under it is 1.08, which is a little less than 1.11; showing that we must use vertical stiffeners up to within 3 feet of the centre of the girder. These vertical stiffeners we will make of 23-inch by 24-inch angle-irons. From the formula for the area of flanges, the following table has been computed, which greatly facilitates the process of finding the necessary area of flanges for any given girder. Table of Co-efficient of Flanges for Riveted Girders. Co-efficient for determining the area required in flanges, allowing 10,000 pounds per square inch of cross-section fibre strain : RULE. - Multiply the load, in tons of 2000 pounds uniformily distributed, by the co-efficient, and divide by 1000 pounds. The quotient will be the gross area, in square inches, required for each flange. 23 30 750 643 563 500 409 33 825 707 619 550 495 34 450 375 346 321 800 281 265 250 775 664 581 517 465 423 388 358 332 310 291 274 258 800 686 600 533 480 436 400 369 343 320 300 282 267 450 413 381 354 330 309 291 275 464 425 392 364 340 319 300 283 477 438 404 375 386 350 328 309 292 360 338 318 300 396 370 347 326 308 407 380 356 335 317 390 366 344 325 EXAMPLE. Let us take the same girder that we have just computed. Here the span was 20 feet, and the depth of girder 24 inches. From the table we find the co-efficient to be 250; and multiplying this by the load, 40 tons, and dividing by 1000, we have 10 square inches as the area of one flange, being the same result as that obtained before. |