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The load is uniformly distributed; and the moment: Mx 37,350 X 240 1,120,500 inch-pounds. We shall assume that the concrete beam is to have a depth ci to the reinforcement, of 22 inches, and shall utilize Equation 39 to obtain an approximate value for the area. Substituting the known quantities in Equation 39, we have: 1,120,500 = A >< 16,000 x (22-1.67) - A = 3.44 square inches. For T-concrete concrete beams with very wide concrete slabs and great depth of concrete 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 concrete slab; x = - kd = 1.21, and therefore (ci - x) = 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 afforded by six f-inch round bars (6 X .60 = 3.60) with considerable to spare.

From Table XVIII we find that six 1-inch bars (either square or round), if placed in one row, would require a concrete 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 concrete 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 concrete 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 H 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 concrete beams was considered as 20 feet, which is the distance from center to center of the concrete columns (or of the girders). This is certainly more nearly correct than to use the net span between the concrete columns (or girders), which is yet unknown, since neither the concrete 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 concrete beam is computed above as 37,350 pounds. The load on the girder is therefore the equivalent of this load concentrated at the center, or of double the load (74,700 pounds) uniformly distributed. Assuming for a trial value that the girder will be 12 inches by 22 inches below the concrete 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 distributed over the girder. Its moment in the center therefore equals X 79,100 X 192 = 1,898,400 inch-pounds. The width of the concrete slab in this case is almost indefinite, being twenty feet, or forty-eight times the thickness of the concrete slab. We shall therefore assume that the compression is confined to a width of fifteen times the concrete 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, k = .207, and lcd = 5.175. This shows that the neutral axis is below the concrete slab, and that it belongs to Case 1, Article 286. Checking the computation of lcd from Equation 34, we compute lcd = 5.18, which is probably the more correct value because computed more directly.

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