values of the constant A are three times the values given in Table I., p. 374. For yellow pine, 300 pounds; for spruce, 210 pounds; and for white pine, 180 pounds, may be taken as reliable values for A. Continuous girder of Two equal spans, loaded equally at the centre of each span, 4 BX D2 X A L Breaking-weight = X (24) Continuous girder of THREE equal spans, loaded uniformly over each span, 5 BX D2 × A Breaking-weight = X L (25) Continuous girder of THREE equal spans, loaded equally at the centre of each span, STIFFNESS. The following formulas give the loads which the beams will support without deflecting more than one-thirtieth of an inch per foot of span. Continuous girder of Two equal spans, loaded uniformly over each span, Load on one span = BX D3 × e (27) Continuous girder of Two equal spans, loaded equally at centre of each span, 16 Bx D X e Load on one span = X 7 Continuous girder of THREE equal spans, loaded uniformly over each span, Load on one span = BX DX e (29) Continuous girder of THREE equal spans, loaded equally at the centre of each span, Load on one span 20 B x D3 X e (30) L2 The value of the constant e is obtained by dividing the modulus of elasticity by 12,960; and, for the three woods most commonly used as beams, the following values may be taken : Yellow pine, 137; white pine, 82; spruce, 100. For iron beams we may find the bending-moment by the formulas given, and, from tables giving the strength and sections of rolled beams, find the beam whose moment of inertia = bending-moment x depth of beam 2000 when the bending moment is in foot pounds. For example, we have a continuous 1-beam of three equal spans, loaded over each span, with 2000 pounds per foot, distributed. Each span being 10 feet, then, from formula 14a, we have 20,000 2000 = 10, and we must find a beam whose depth multiplied by ten will equal its moment of inertia. If we try a ten-inch beam, we should have 10 x 10 = 100; and we see from Tables, pp. 260-272, that no ten-inch beam has a moment of inertia as small as 100: so we will take a nine-inch beam. 9 × 10 =90, and the lightest nine-inch beam has a moment of inertia of 93: so we will use that beam. In the case of continuous I-beams of three equal spans, equally loaded with a distributed load, we may take four-fifths of the load on one span, and find the iron beam which would support that load if with only one span. EXAMPLE. - If we have an I-beam of three equal spans of 10 feet each loaded with 20,000 pounds over each span, what size beam should we use? Ans. of 20,000 = 16,000. The equivalent load for a span of one foot would be 16,000 × 10 = 160,000. From Tables, Chap. XIV., we find that the beam whose co-efficient is nearest to this is the nine-inch light beam, the same beam which we found to carry the same load in the preceding example. For beams of two equal spans loaded uniformly, the strength of the beam is the same as though the beam were not continuous. The formulas given for the re-actions of the supports and for the deflections of continuous girders with concentrated loads, were verified by the author by means of careful experiments on small steel bars. The other formulas have been verified by comparison with other authorities, where it was possible to do so; though one or two of the cases given, the author has never seen discussed in any work on the subject. CHAPTER XVIII. FLITCH PLATE GIRDERS. IN framing large buildings, it often occurs that the floors must be supported upon girders, which themselves rest upon columns; and it is required that the columns shall be spaced farther apart than would be allowable if wooden girders were used. In such cases the Flitch Plate girder may be used, oftentimes with advantage. A section and elevation of a Flitch Plate girder is shown in Fig. 1. IRON FLATE C Fig. 1. The different pieces are bolted together every two feet by threefourths-inch bolts, as shown in elevation. It has been found by practice that the thickness of the iron plate should be about onetwelfth of the whole thickness of the beam, or the thickness of the wood should be eleven times the thickness of the iron. As the elasticity of iron is so much greater than that of wood, we must proportion the load on the wood so that it shall bend the same amount as the iron plate: otherwise the whole strain might be thrown on the iron plate. The modulus of elasticity of wrought-iron is about thirteen times that of hard pine; or a beam of hard pine one inch wide would bend thirteen times as much as a plate of iron of the same size under the same load. Hence, if we want the hard-pine beam to bend the same as the iron plate, we must put only one-thirteenth as much load on it. If the wooden beam is eleven times as thick as the iron one, we should put eleven-thirteenths of its safe load on it, or, what amounts to the same thing, use a constant only eleventhirteenths of the strength of the wood. On this basis the following formulas have been made up for the strength of Flitch Plate girders, in which the thickness of the iron is one-twelfth of the breadth of the beam, approximately : As an example of the use of this kind of girder, we will take the case of a railway-station in which the second story is devoted to offices, and where we must use girders to support the second floor, of twenty-five feet span, and not less than twelve feet on centres, if we can avoid it. This would give us a floor area to be supported by the girder of 12 × 25 = 300 square feet; and, allowing 105 pounds per square foot as the weight of the superimposed load and of the floor itself, we have 31,500 pounds as the load to be supported by the girder. Now we find, by computation, that if we were to use a solid girder of hard pine, it would require a seventeen-inch by fourteen-inch beam. If we were to use an iron beam, we find that a fifteen-inch heavy iron beam would not have the requisite strength for this span, and that we should be obliged to use two twelve-inch beams. We will now see what size of Flitch Plate girder we would require, should we decide to use such a girder. We will assume the total breadth of both beams to be twelve inches, so that we can use two six-inch timbers, which we will have hard pine. The thickness of the iron will be one inch and one-eighth. Then, substituting in Formula 3, we have, Hence we shall require a twelve-inch by fourteen-inch girder. Now, for a comparison of the cost of the three girders we have considered in this example. The seventeen-inch by fourteen-inch hard-pine girder would contain 515 feet, board measure, which, at five cents a foot, would amount to $25.75. Two twelve-inch iron beams 25 feet 8 inches long will weigh 2083 pounds; and, at four cents a pound, they would cost $83.32. The Flitch-Plate girder would contain 364 feet, board measure, which would cost $18.20. The iron plate would weigh 13124 pounds, which would cost $52.50; making the total cost of the girder $70.70, or $13 less than the iron beams, and $45 more than the solid hard-pine beams. Flitch-Plate beams also possess the advantage that the wood almost entirely protects the iron; so that, in case of a fire, the heat would not probably affect the iron until the wooden beams were badly burned. |