294. Numerical Illustration. It is required to test the beam which was computed in Example 1 of Article 291. Here the total compressive stress in the flange = cbkd = 1 X 432 X 96 X344 = 71,332 pounds. But this compressive stress measures the shearing stress between the flange and the rib. This beam requires six '-inch bars for the reinforcement. We shall assume that the rib is to be 11 inches wide, and that four of the bars are placed in the bottom row, and two bars about 2 inches above them. The effect of this will be to deepen the beam slightly, since d measures the depth of the beam to the center of the reinforcement, and, as already computed numerically in Article 290, the center of gravity of this combination will be - of an inch above the center of gravity of the lower row of bars. Substituting in Equation 40 the values = 71,332, b = 11, and L = 20, we find, for the unit-value of z, 10S pounds per square inch. This shows that the assumed dimensions of the beam are satisfactory in this respect, since the true shearing stress permissible in concrete is higher than this.
But the beam must be tested also for its ability to withstand shear in vertical planes along the sides of the rib. Since the slab in this case is 5 inches thick and we can count on both surfaces to with stand the shear, we have a width of 10 inches to withstand the shear, as compared with the 11 inches on the underside of the slab. The unit-shear would therefore be of the unit-shear on the under side of the slab, and would equal 119 pounds per square inch. Even this would not be unsafe, but the danger of failure in this respect is usually guarded against by the fact that the slab almost invariably contains bars which are inserted to reinforce the slab, and which have such an area that they will effectively prevent any shearing in this way.
Testing Example 2 similarly, we may find the total compression C from Equation 32, and that it equals As = 16,000 X 3.37 = 54,000 pounds. The steel reinforcement is six i-inch bars, and by Table XVIII we find that if placed side by side, the beam must be 13.19 inches in width, or, in round numbers, 13f inches. Substituting these values in Equation 37, we find, for the value of z, 45 pounds per square inch. Such a value is of course perfectly safe. The shear along the sides of the beam will be considerably greater, since the slab is only four inches thick, and twice the thickness is but S inches; therefore the maximum unit-shear along the sides will equal 45 times the ratio of 13.25 to S, or 75 pounds per square inch. Even this would be perfectly safe, to say nothing of the additional shearing strength afforded by the slab bars.
295. Shear in a T-Beam. The shear here referred to is the shear of the beam as a whole on any vertical section. It does not refer to the shearing stresses between the slab and the rib.
The theoretical computation of the shear of a T-heam is a very complicated problem. Fortunately it is unnecessary to attempt to solve it exactly. The shearing resistance is certainly far greater in the case of a T-beam than in the case of a plain beam of the same width and total depth and loaded with the same total load. There fore, if the shearing strength is sufficient, according to the rule, for a plain beam, it is certainly sufficient for the T-beam. In the first
example of Article 291, the total load on the beam is 30,000 pounds. Therefore the maximum shear V at the end of the beam, is 15,000 pounds. In this particular case, d — x = 12.25. For this beam, d= 13.75 inches, and b = 11 inches. Substituting these values in Equa tion 31, we have: Although this is probably a very safe stress for direct shearing, it is more than double the allowable direct tension due to the diagonal stresses; and therefore ample reinforcement must be provided. If only two of the -;-inch bars are turned at an angle of 45° at the end, these two bars will have an area of 1.54 square inches, and will have a working tensile strength (at the unit-stress of 16,000 pounds) of 24,640 pounds. This is more than the total vertical shear at the ends of the beam; and we may therefore consider that the beam is protected against this form of failure.
296. Numerical Illustration of Slab, Beam, and Girder Con= struction. 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 S feet by 20 feet. We may therefore consider that girder d supports the load of b (on a floor area S ft. by 20 ft.) as a concentrated load in the center. Or, we may consider that, ignoring the beams, the girder supports a uniformly dis tributed 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 = .35S; 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 S feet long will carry a total load of 1,650 pounds, and its moment will be X 1,6S0 X 96 = 20,160 inch-pounds. Using the first half of Equation 29, we can substitute the known values, and say that: 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 S 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 = pled = .01 X 1 X 4.21 = .0421 square inch. Using 1-inch round bars whose area equals .1963 square inch, the required spacing of the bars will be .1963 ÷ .0421 = 4.66 inches. Practically this would be called 4b inches.