ment, or anchor the rod by means of a hook, as explained in the following paragraph. The LENGTH OF EMBEDMENT to prevent slipping as computed by formula (16) is: Allowable bond stress, u = 80 lb per sq in (plain rounds or squares) 16 000 lb per sq in, l = 50 diameters of rod, or side of square bar in inches; 56 diameters of rod, or side of square bar, in inches; f. 18 000 lb per sq in, = Allowable bond-stress u = f. 16 000 lb per sq in, f. = = 18 000 lb per sq in, = = 100 lb per sq in (deformed bars); 40 diameters of rod, or side of square bar, in inches; 45 diameters of rod or side of square bar, in inches. The Joint Committee, 1924, requires that in continuous, restrained or cantilever beams, anchorage of the tensile negative reinforcement beyond the face of the support shall provide for the full maximum tension with bond stresses not greater than 0.04 f'c for plain bars and 0.05 f', for deformed bars. Such anchorage shall provide a length of bar not less than the depth of the beam. In the case of end supports which have a width less than three-fourths of the depth of the beam, the bars shall be bent down toward the support a distance not less than the effective depth of the beam. The portion of the bar so bent down shall be as near to the end of the beam as protective covering permits. In continuous or restrained beams, negative reinforcement shall be carried to or beyond the point of inflection. Not less than one-fourth of the area of the positive reinforcement shall extend into the support to provide an embedment of ten or more bar-diameters. In simple beams or freely supported end spans of continuous beams at least one-fourth of the area of the tensile reinforcement shall extend along the tension side of the beam and beyond the face of the support to provide an embedment of ten or more bar-diameters. Where increased shearing stresses are used, as noted on page 10, or increased bond stresses, as noted on page 61, SPECIAL ANCHORAGE of all reinforcement in addition to that required by the preceding paragraphs shall be provided as follows: In continuous and restrained beams, anchorage beyond points of inflection of one-third the area of the negative reinforcement and beyond the face of the support of one-third the area of the positive reinforcement, shall be provided to develop one-third of the maximum working stress in tension, with bond stresses not greater than those specified on page 10. At the edges of footings, anchorage for all the bars for one-third the maximum working stress in tension shall be provided within a region where the tension in the concrete, computed as an unreinforced beam, does not exceed 40 lb. per square inch. In simple beams or freely supported end spans of continuous beams at least one-half of the tensile reinforcement shall extend along the tension side of the beam to provide an anchorage beyond the face of the support for one-third of the maximum working stress in tension. The SPLICING OF REINFORCEMENT is ordinarily accomplished by a lap of sufficient length, as computed by Formula (16), to develop the required stress in the rod. In the case of the vertical reinforcement of columns, rods of 13 in or more in diameter were formerly milled at butts, threaded, and joined by couplings or enclosed by tight-fitting pipe-sleeves. The present practice, however, is to use lap-splices for all column reinforcements. The length of the splice is computed by the above formula and the minimum is usually set at 18 inches or 2 ft. When it is not possible to obtain the required anchorage by means of embedment, the customary practice is to HOOK THE END OF THE ROD. If the hook is made of semicircular form, and bent with a radius equal to at least four diameters of the rod and as detailed in Fig. 6, the full strength of the steel can be developed before the enclosed concrete is crushed. The use of such a hook is usually the best method of securing the ends of the raised rods at the beam terminations. Also, in the case of slab, or band reinforcement, terminat 8 Diameters Minimum (Radius of Bend 4 Diameters) ing in a spandrel beam or other similar member, the ends of the rods should be hooked in a like manner. The ANCHORAGE OF STIRRUPS is preferably made by semicircular hooks of radii at least equal to four diameters of the web-bar. As the tensile-stress at the termination of the stirrup is low, and there is little danger of crushing the enclosed concrete, a simple angular bend is often used as indicated in the typical beam designs. In order to justify the working stresses previously recommended, it is very important that STIRRUPS RE 4 Dia.Fig. 6. STRONGLY AND RIGIDLY ATTACHED TO THE MAIN LONGITUDINAL STEEL. This is best accomplished by passing the stirrups around the longitudinal reinforcement in either the top or bottom of the beam, and anchoring by a hook as near to the opposite face as fireproofing requirements permit. 8. Spacing of Bars in Beams and Girders. In order that the SHEARING VALUE OF THE CONCRETE between two adjacent rods, composing the main tensile reinforcement of a beam, or girder, should be sufficiently great to prevent its splitting before the adhesion or bond-strength is exceeded, the Joint Committee, 1924, requires that there should be a clear distance between rods lying in the same horizontal plane, equal to one and one-half times the diameter of a round rod or one and one-half times the diagonal of a square bar. If special anchorage is employed, as described on page 63, the clear spacing may be taken as equal to the diameter of a round bar, or to the diagonal of a square bar, but never less than 1 in or one and one-quarter times the maximum size of the coarse aggregate. The clear vertical distance between two layers of rods, or bars, may be taken as 1 in. CHAPTER V DESIGN OF BEAMS AND SLABS 1. Design-Procedure for Rectangular Beams. (1) The width of web, b, of the beam, or girder, is made large enough to accommodate the probable reinforcement (page 37), including the necessary protection (page 13); the weight of the beam is then computed on a basis of the assumed width and a depth d, estimated as equal to 1 in for each foot of clear span. (2) The maximum shear is determined; see page 50. (3) The maximum bending moments at the supports and near the mid-span and under the concentrations, for beams subjected to concentrated loads, are computed as described in Chapter III, Articles 1 and 2. M = (see Kb (4) The depth is computed by Formula (10), Chapter II, d table on page 20 for values of K), using the maximum bending moment as the value of M. (5) The area of the main tensile reinforcement is computed by Formula (11), A, = pbd, or by Formula (12), Chapter II, As = M f.jd = Values of p and j cor responding to the stresses employed are taken from the table on page 20. If the depth d has been determined as in step 4, the formula A, pbd gives a steel-area just sufficient to balance in tension the compressive value of the concrete. If considerations other than the compressive strength of the concrete determine the dimensions of the beam, the formula A, = M fsjd should be used. For all usual conditions encountered in typical construction, the value of j may be taken from the table on page 20, although not accurate except when the percentage of reinforcement employed is that which, with the permissible stresses, gives a balanced beam. For unusual conditions the exact value of j should be computed by the formula on page 18, using a trial area A,, found by taking j from the table on page 20. (6) The value of the maximum unit shearing stress is computed by Formula (1), Chapter IV, v = V/jbd. (7) The web-reinforcement is designed: By Formula (13), Chapter IV, x = (L/2) x (1 − ) By Formula (12), Chapter IV, s = (1⁄2)f.jdA./V (8) The bond-stress and anchorage are tested; By Formula (14), Chapter IV, u = V/Zojd By Formula (16), Chapter IV, 1 = (1⁄4)(f./u)i (9) The length of laps, location of hooks, etc., are noted. fc = 650 lb per sq in (over supports increased 15%) v', limited to 40 lb per sq in Span = = 20 ft between faces of supports 27 000 lb (exclusive of the weight of the beam) End-Conditions: Fully continuous (1) Assume b = 10 in. Then, estimating the effective depth d as 14 in for each foot of span, and 2 in of fireproofing below the center of the longitudinal reinforcement (11⁄2 in protection), the total depth is 27 in and the total uniformly distributed load is in which 150 is the weight, in pounds, of 1 cu ft of concrete. = = 16 315 lb = WL/12 = (32 625 × 20 × 12)/12 = 652 500 in-lb. (2) V 32 625/2 (3) By Formula M If this value is taken as the bending moment over the supports, and also at the mid-span, it is apparent that in rectangular-beam design the latter is the critical section as a 15% increase in the compressive stress is allowed over supports. (4) By Formula (10), Chapter II, d = 24.70.5+1.5 = 26.7 in. The value 27 in is used for the total depth. (5) By Formula (11), Chapter II, A, = pbd = 0.0077 X 10 × 24.7 = 1.90 sq in. Two 4-in squares and two 5%-in squares are used. One 34-in and one 5%-in bar are raised at the fifth-point of the clear span at an angle of 30% to the horizontal. (6) By Formula (1), Chapter IV, v = V/jbd 16 315/(0.875 X 10 X 25) = 75 lb = as the distance out from the supports to the section where the shear can be carried by the concrete alone. By Formula (12), Chapter IV, == S = (3⁄41⁄2)ƒ.jdA ̧/V = (3⁄41⁄2) × 16 000 × 0.875 × 25 × 0.22/16 315 7.05, or 7 in. By placing the first stirrup at a distance (1⁄2)s, 4 in from the face of support, omitting two stirrups through the distance covered by the double-bend rods, see page 56, and spacing equally at the minimum distance, there will be six -in round U-shaped stirrups at each end of the beam. (8) By Formula (14), Chapter IV, the bond on the negative reinforcement at the face of the support, u = V/Zojd = 16 315 [(2 × 2.5) + (2 × 3.0)] × 0.875 X 25 = 68 lb per sq in Bond in which 2.5 is the perimeter of the cross-section of the 5%-in square bars and 3.0 that of the 34-in square bars, there being one of each from each span. at the point of inflection, (9) The straight bars are carried to the center lines of columns and the double-bend bars are lapped to the fifth-points of the adjacent spans. It is generally desirable to hook the terminations of the bent bars. Hooks are not usually required at the terminations of the straight bars of continuous beams. 3. Load-Distribution in Slabs. In the case of square panels, with side supports of approximately equal rigidity, the design-load is equally distributed in both directions, and the reinforcement is the same each way. For oblong panels, however, a larger proportion of the load is carried on the shorter span, and when the length is equal to, or greater than, one-and-one-half times the width, the Joint Committee, 1916, recommends that the entire load be provided for by the steel spanning the shorter way. The following Formula (1) gives the proportion of the load carried on the shorter span, following this recommendation, for uniformly loaded panels of varying dimensions (see Table I): T= proportion of the load carried by the shorter span. Then r = (l/b) — 0.5 (1) Another method of distribution, based on the assumption that the deflections of strips crossing a panel parallel to the longer and shorter sides are proportional respectively to the fourth powers of the lengths of the strips, has also been widely used in practice and may be expressed by the formula |