Assume a floor construction as outlined in skeleton form in Fig. 107. The columns are spaced 16 feet by 20 feet. Girders which support the alternate rows of beams, connect the columns in the 16-foot direction. The live load on the floor is 150 pounds per square foot. The concrete is to be a 1:2:4 mixture, with r= 10, and c = 600. Required the proper dimensions for the girders, beams, and slab.

The load on the girders may be computed in either one of two ways, both of which give the same results. We must consider that each beam supports an area of 8 feet by 20 feet. We may therefore consider that girder d supports the load of b (on a floor area 8 ft. by 20 ft.) as a concentrated load in the center. Or, we may consider that, ignoring the beams, the girder supports a uniformly distributed load on an area 16 ft. by 20 ft, The moment in either case is the same. Assume that we shall use a 1 per cent reinforcement in the slab. Then, from Table XV, with r = 10, and p = .01, we find that k = .358; then x = .119 d, or (d - x) = .881 d. As a trial, we estimate that a 5-inch slab (or d = 4) will carry the load. This will weigh 60 pounds per square foot, and make a total live and dead load of 210 pounds per square foot. A strip one foot wide and 8 feet long will carry a total load of 1,680 pounds, and its moment will be 1/8 X 1,680 X 96 = 20,160 inch-pounds. Using the first half of Equation 29, we can substitute the known values, and say that:

20,160 = 1/2 X 600 X 12 X .358 d X .881 d

= 1,135 d2 d2 = 17.76 d = 4.21

In this case the span of the slab is considered as the distance from center to center of the beams. This is evidently more exact than to use the net span (which equals eight feet, less the still unknown width of beam), since the true span is the distance between the centers of pressure on the two beams. It is probable that the true span (really indeterminable) will be somewhat less than 8 feet, which would probably justify using the round value of d = 4 inches, and the slab thickness as 5 inches, as first assumed. The area of the steel per inch of width of the slab = pbd = .01 X 1 X 4.21 = .0421 square inch. Using 1/2-inch round bars whose area equals .1963 square inch, the required spacing of the bars will be .1963 .0421 = 4.06 inches. Practically this would be called 4 5/8 inches.

Fig. 107. Skeleton Outline of Floor Panel.

Fig. 107. Skeleton Outline of Floor-Panel.

The load on a beam is that on an area of 8 feet by 20 feet, and equals 8 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:

M = - 1/8 X 37,350 X 240 = 1,120,500 inch-pounds.

We shall assume that the beam is to have a depth d 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 X 16,000 X (22-1.67) A = 3.44 square inches.

For T-bearns 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 = 1/3 kd = 1.21, and therefore (d - 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 7/8-inch round bars (6 X .60 = 3.60) with considerable to spare. From Table XVIII we find that six 7/8-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 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 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 1/8 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, k = .207, and kd = 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 compute 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 S3, we compute 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. Similarly, 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.