In which 3.25 is the effective depth in inches to the upper layer of steel. Use % in round rods, 51⁄2 in on centers. As the above result is based on a value of j applying only to a balanced beam, which is not the condition existing in the longer span, it is slightly too conservative and may be corrected by finding the actual value of j corresponding to the approximate steel area as computed. The difference, however, is usually negligible. Assuming that the negative resisting moments over continuous supports are to be made equal to the positive resisting moments at the centers of the spans, the reinforcement is arranged so as to give equivalent areas at both sections. Testing the bond stress on the reinforcement at the face of the supports, we have, for the shorter span, by Formula (14), Chapter IV, In which 787 is one-half the total load carried on the shorter span; 3, the number of rods in the top of the slab over the supports, per foot of slab-width (4 in on centers); 1.18 the perimeter of the cross-section of a 3%-in round rod, in inches; 0.875, the value of j; and 3.75, the actual depth d of the slab, in inches, as designed. At the point of inflection, since alternate rods are raised at the per sq in, which value is acceptable for plain rods. The steel must be arranged so as to give the same sectional area at the top of the slab over the supports as at the mid-span. All splices lap 20 in (50 diameters of a rod), and the rods of the end spans are anchored into the spandrel beams or other supporting members. Example 2. One-Way Reinforcement. Panel-dimensions: 6 ft 0 in X 18 ft 0 in, between faces of supports Finish: Monolithic End-conditions: Fully continuous Assume a slab, 4 in in total thickness, and a unit dead load of 48 lb per sq ft Total unit load = 200 + 48 248 lb per sq ft. As the length of this panel is more than one and one-half times the width, the entire load is provided for by the steel spanning the shorter dimension. Total thickness of slab = 2.63 +1.0+ 0.25 = 3.88, or approximately 4 in in which 1 in is the allowance for protection, and 0.25 in the approximate halfdiameter of each rod. Steel required for the shorter span, by Formula (17), Chapter II, Round rods, 3%-in in diameter, and spaced 53⁄4 in on centers, are therefore used. For the longer span, three %-in round rods per panel are placed transversely to the main reinforcement and spaced evenly from girder to girder. There are provided, also, for each panel, four 3-in round rods, of a length equal to the girder-width plus 30 in, and placed with a uniform spacing parallel with the longer rods across the top of the girder, as shown in Fig. 3. Fig. 3. Typical Concrete Slab with One-Way Reinforcement Testing the bond-stress on the reinforcement adjacent to the point of inflection, for a section of slab 1 ft 0 in wide, by Formula (14), Chapter IV, In this equation 744 is one-half the total load in pounds on a strip 1 ft wide; 1.05, the number of rods in the bottom of the slab, per foot of width (111⁄2 in on centers); 1.23, the perimeter of the cross-section of a 38-in round deformed rod in inches; 0.875 the value of j; and 2.75, the actual depth d, of the slab, in inches, as designed. The bond-stress being very high, a deformed rod should be used in both directions for the sake of uniformity, and when not continuous the rods lying in the bottom of the slab should be spliced. All splices lap 15 in (40 diameters of the rod), and the rods of the end spans are anchored into the spandrel beams or other supporting members. The steel must be arranged so as to give the same sectional area at the top of the slab over the supports as at the midspan. 7. Design-Loads on T Beams and T Girders. Having determined the arrangement of the slab-reinforcement, and knowing the total unit live and dead load per square foot, the design-load carried by a T beam is found by multiplying this unit load by the floor-area which the beam supports, and adding to this product any other incidental loads, and the weight of the stem of the beam. In the case of slabs with one-way reinforcement the floor-area supported is equal to the product of the length of the beam by the sum of the two half slabspans on either side. When a slab has a two-way reinforcement the loads carried to the supporting beams are not uniformly distributed, the loading being a maximum at the middle of the span and diminishing toward the ends. The bending moments, for a simply-supported beam, may be closely approximated, by assuming that the loading varies as the ordinates of a triangle, with its apex over the middle of the span.* Let W2 = load per square foot of slab carried to longer beam; load per square foot of slab carried to shorter beam; 1 longer dimension of slab in feet (length of longer beam); = shorter dimension of slab in feet (length of shorter beam). For semicontinuous or fully continuous beams these results are multiplied by 4% and 2% respectively. The loads carried by girders are usually a combination of concentrated loads due to beam-reactions, and the uniformly distributed weight of the girder itself, plus a small portion of the floor-slab whose weight is transmitted directly to the girder. Weights of building materials and live loads are given in the tables on pages 7 and 8. 8. Effective Width of Flange. For purposes of design in computing the sectional area of concrete available to resist the compressive stresses, the width of slab which may be considered as acting as the flange of a T beam is limited by the Joint Committee, 1924, as follows: (1) It shall not exceed one-fourth of the span-length of the beam. Compare Formulas for the Design of Rectangular Floor-Slabs and the Supporting Girders, by H. M. Westergaard, Proc. Amer. Concrete Inst., 1926, page 16. (2) Its overhanging width, on either side of the web, shall not exceed eight * times the thickness of the slab, nor one-half the clear distance to the next beam. In order that these assumptions may be justified it is necessary to provide sufficient steel, in the form of stirrups or bent rods of the main longitudinal reinforcement, to bond the flange to the web properly and to provide adequate shear resistance at the junction of the beam with the slab, and the latter must be built integrally with the beam. If a flange occurs on one side only, as in the case of a spandrel-beam, the flange width for design purposes should not exceed one-tenth of the span length of the beam, and the overhanging width from the face of the web should not be more than six times the thickness of the slab, nor one-half the clear distance to the next beam. Sometimes it is convenient to use a T section in order to increase the concretearea of a beam, even when there is no slab. In such cases the width of the flange b should not be over four times the width of the web b'; and the thickness of flange t, not less than one-half the width of the web. 9. Working Formulas for T Beams. In the design of T beams and T girders, two cases are generally considered; first, that in which the neutral surface falls either within the flange or is at the bottom of it, and secondly, that in which the neutral surface falls within the web. In the first case, which occurs only when the slab is very thick in proportion to the depth of the beam, the bending-moment computations are the same as for rectangular members, the width of the flange instead of the width of the web being substituted for b, and the steel-ratio p being based upon the total area bd. In the second case, which is the one ordinarily encountered, formulas are developed upon the same fundamental assumptions, and in the same manner, as for rectangular beams. Fig. 4 illustrates the relations of the different stresses. Neglecting the amount of compression resisted by the web, let t = total slab-thickness on either side of the web, in inches; 2 = distance from the compression-surface of the beam to the center of The remaining notation is the same as for rectangular beams, and is given in Chapter II. * Many codes including that of New York City limit this dimension to six times the thickness of the slab. Center of Compression Center of fc = fsk/[n(1 − k)] (See Formula (3), Chapter II) (See Formula (12), Chapter II) (9) (10) (11) jd Trace of Neutral Axis Flange Web Fig. 4. Stress-Diagram and Cross-Section of Reinforced-Concrete T Beam Formulas (12) and (13) giving approximate results, but erring slightly on the safe side, are deduced by substituting t for jd, and (1⁄2fc) for Se (1-2) in Formulas (9) and (11), as follows: 2kd These are the formulas generally used in practice. The equations for SHEAR, WEB-REINFORCEMENT, and BOND-STRESS, are the same as for rectangular beams. In regard to the compression in the web, the following formulas may be used for large beams, in which the web forms a considerable part of the compressive |