The load on a beam is that on an area of S feet by 20 feet, and equals S X 20 X 210 = 33,600 pounds for live and dead load. As a rough trial value, we shall assume that the beam will be 12 inches wide and 15 inches deep below the slab, or a volume of 1 X 1.25 X 20 cubic feet = 25 cubic feet, which will weigh 3,750 pounds. Adding this, we have 37,350 pounds as the total live and dead load carried by each beam. The load is uniformly distributed ; and the moment: We shall assume that the beam is to have a depth d to the reinforce ment, of 22 inches, and shall utilize Equation 39 to obtain an approxi mate value for the area. Substituting the known quantities in Equation 39, we have: For T-beams with very wide slabs and great depth of beam, the percentage of steel is always very small. In this case, p = 3.44 (96 X 22) = .00163. Such a value is beyond the range of those given in Table XV, and therefore we must compute the value of k from Equation 27; and we find that k = .165; kd = 3.63, which shows that the neutral axis is within the slab; x = a kd = 1.21, and therefore (d — r) = 20.79. Substituting these values in the upper part of Equation 29 in order to find the value of c, we find that c = 309 pounds per square inch. Substituting the known values in the second half of Equation 29, in order to obtain a more precise value of s, we find that s = 15,737 pounds per square inch.
The required area (3.44 square inches) of the bars will be af forded by six Zr-inch round bars (6 X .60 = 3.60) with considerable to spare. From Table XVIII we find that six ;-inch bars (either square or round), if placed in one row, would require a beam 14.72 inches wide. This is undesirably wide, and so we shall use four bars in the lower row, and two above, and make the beam 11 inches wide. This will add nearly an inch to the depth, and the total depth will be 22 + 3, or 25 inches. The concrete below the slab is therefore 11 inches wide by 20 inches deep, instead of 12 inches wide by 15 inches deep, as assumed when computing the dead load. The section of 220 square inches will therefore weigh more than the suggested section of 180 square inches; but the difference in dead load weight is so small that it is unnecessary to alter the calculations, especially /since the unit-stresses in the concrete and steel are both lower than the working limits. It should also be noted that the span of these beams was considered as 20 feet, which is the distance from center to center of the columns (or of the girders). This is certainly more nearly correct than to use the net span between the columns (or girders), which is yet unknown, since neither the columns nor the girders are yet designed. There is probably some margin of safety in using the span as 20 feet.
The load on one beam is computed above as 37,350 pounds. The load on the girder is therefore the equivalent of this load con centrated at the center, or of double the load (74,700 pounds) uni formly distributed. Assuming for a trial value that the girder will be 12 inches by 22 inches below the slab, its weight for sixteen feet will be 4,392, or say 4,400 pounds. This gives a total of 79,100 pounds as the equivalent total live and dead load uniformly dis tributed over the girder. Its moment in the center therefore equals X 79,100 X 192 = 1,898,400 inch-pounds.
The width of the slab in this case is almost indefinite, being twenty feet, or forty-eight times the thickness of the slab. We shall therefore assume that the compression is confined to a width of fifteen times the slab thickness, or that b' = 75 inches. Assume for a trial value that d = 25 inches; then from Equation 39, if s = 16,000, we find that A = 5.08 square inches.. Then p = .0027; and, from Equation 27, le = .207, and led = 5.175. This shows that the neutral axis is below the slab, and that it belongs to Case 1, Article 286. Checking the computation of kd from Equation 34, we com pute kd = 5.18, which is probably the more correct value because computed more directly. The discrepancy is due to the dropping of decimals during the computations. From Equation 33, we com pute that x = 1.72; then (d — x) = 23.28. Substituting the value of the moment and of the dimensions in the upper part of Equation 35, we compute c to be 420 pounds per square inch. Simi larly, making substitutions in the lower part of Equation 35, using the more precise value of (d — x) for the lever-arm of the steel, we find s = 16,052 pounds per square inch. The student should verify in detail all these computations.
The total required area- of 5.08 square inches may be divided into, say, S round bars inch in diameter. These would have an area of 4.81 square inches. The discrepancy is about five per cent. These bars, placed in two rows, would require that the beam should be at least 10.78 inches wide. We shall call it 11 inches. The total depth of the beam will be three inches greater than d, or 2S inches. This means 23 inches below the slab, and the area of concrete below the slab is therefore 11 X 23 = 253 square inches, rather than 12 X 22 = 264 square inches, as assumed for trial.
Shear. The shearing stresses between the rib and slab of the girder are of special importance in this case. The quantity of Article 293 equals the total compression in the concrete, which equals the total tension in the steel, which equals, in this case, 16,032 X 5.08 = S1,544 pounds. This equals 3 bzl, in which b = 11, 1 = 16 (feet), and z is to be determined.