In the case of restrained beams, or cantilevers, when maximum tension occurs near the support, careful attention must he given to the anchorage of the bars. Bats used for diagonal tension reinforce ment, either vertical stirrups or inclined bars, have maximum tension at the neutral axis, and must have a sufficient embedment on the com pression side of the neutral axis to resist the maximum tension in the steel.
Lateral Spacing of Steel.—The horizontal tension rods in a rein forced concrete beam must be so spaced as to leave a sufficient area of concrete between them to carry the shear communicated to the con crete by the portion of the bars below the minimum section of con crete. This would require that for circular bars the horizontal sec tion between rods be capable of carrying a shearing stress equal to the bond stress on the lower half of the bars. If be the clear spacing between the bars and i the diameter of the bar, for the round bar For the values of unit stress recommended by the Joint Com mittee (v=6 per cent and u=4 per cent of the ultimate compressive iu strength), v=-!,u, and for round bars, 1.05i.
For square bars with sides vertical, or and for square bars with diagonals vertical, 3i.
For deformed bars these values would be increased in the ratio of 5 to 4.
The Joint Committee recommends 1 that: The lateral spacing of parallel bars should not be less than three diameters from center to center, nor should the distance from the side of the beam to the center of the nearest bar be less than two diameters. The clear spacing between two layers of bars should he not less than 1 inch. The use of more than two layers is not recommended, unless the layers are tied together by adequate metal con nections, particularly at and near points where bars are bent up or bent down. Where more than one layer is used at least all bars above the lower layer should be bent up and anchored beyond the edge of the support.
110. Design of Beams.—The methods of applying formulas and tables in the design of rectangular beams is illustrated in the following examples: (7) Design a rectangular beam to have a span of 25 feet and carry a uniform load of 600 pounds per linear foot, in addition to its own weight, using working stresses recommended by the Joint Com mittee for concrete of 2000 lb./in 2 compressive strength.
Solution.—From Table VII, for n=15, f = 16,000 and f = 650, we find R= 10S, p = .0078, j= .874.
Assume weight of beam = 300 pounds per linear foot.
Then .1I = 8 = (600-I-300) 12 = 843750 in.-lb.
= R= S43750 and for b= 12, d=25.5, for d=23.6 Taking and total depth, weight of beam= 14 X 25 . 5 X = 372 pounds per linear foot. The
assumed load is too small. Assume weight of beam =400 pounds per linear foot.
elf b in.-lb., and S 108 1 For b= 14, d= 14 =24.9. Using b and d = 25, make h= 27.
Then weight of beam= 27X14X150 - 141 =394 pounds per linear foot, which agrees with the assumption.
Horizontal steel, A = pdb = .0078 X 14 X 25 = 2.73 From Table X, seven b-inch square bars give A = 2.73 five }-inch square bars give A =2.81 six 1-inch round bars give A = 2.65 Seven i-inch square bars, spaced is inch c. to c. or six }-inch round bars, spaced 21 inches c. to c. might be placed in the width of 14 inches, meeting the requirement of spacing 3 diameters c. to c. We will use five 1-inch square bars, spaced 22 inches c. to c. and 2 inches from side of beam.
The section is sufficient for shear, and no diagonal tension reinforce ment is necessary.
Bond Stress, Table X, for five 1-inch bars, .720=5 X3.00 =15 and (19) U n) 15 =31.91b. in?, which is less than the allowable stress.
S. A simple beam of 10-foot span to center of bearings, is to carry a load of 400 pounds per linear foot. Design the beam, assuming n =15. safe value of unit shear= 120 and for diagonal tension on concrete=40 Solution.—Assuming the weight of beam as 65 pounds per linear foot, the total load is (400+65)10=4650 pounds.
The four "1-inch square bars will fit in the width of beam with proper spacing, but we will use three *-inch bars.
If the concrete extend 1 inches below the center of the steel, 1i = d+ 1 z 1 inches, and the weight of beam is 5 X11.5 X 150/144 =60 lb./ft.; this is a little less than our assumed weight.
Maximum Shear, This is less than 120 in 2, and the dimensions of the beaus are sufficient.
Reinforcement for diagonal tension is needed beyond the point _ where v=40 or 40l =3.7 feet. Reinforcement is required to 5-3.7=1.3 foot=15.6 inches from the support.
Vertical Stirrups. — If we assume s=-111=5 inches for the stirrup next the support, we have (13).
For U-shaped stirrup, the section of rod required Avill be one-half of this, or 0.045 one-quarter in round bars are sufficient, and three stirrups may be used, spaced 3, 8, and 13 inches from the middle of support.
Bond Stress.—For the horizontal steel, Table X, .o=3 X1.75 =5.25 and (19) 'it= buffo= 5X54:525=51.4 in?, which is less than the allowable bond stress, and no anchoring is necessary.
15000 For the vertical stirrups (21) _ 4X S0 X4 =11.8 inches, or the stirrups need 11.8 inches above the neutral axis for anchor age; they must therefore have hooked ends.