The total required area of 5.08 square inches may be divided into, say, 8 round bars 7/8 inch in diameter. These would have an area of 4.81 square inches. The discrepancy is about five per cent. These bars, placed in two rows, would require that the beam should be at least 10.78 inches-wide. We shall call it 11 inches. The total depth of the beam will be three inches greater than d, or 28 inches. This means 23 inches below the slab, and the area of concrete below the slab is therefore 11 X 23 = 253 square inches, rather than 12 X 22 = 264 square inches, as assumed for trial.

Shear. The shearing stresses between the rib and slab of the girder are of special importance in this case. The quantity Sh of Article 293 equals the total compression in the concrete, which equals the total tension in the steel, which equals, in this case, 16,052 X 5.08 = 81,544 pounds. This equals 3 bzl, in which b = 11, l = 16 (feet), and z is to be determined.

z = 81,544 ÷ (3x ll x 16) = 154 pounds per square inch.

This measures the maximum shearing stress under the slab, and is almost safe, even without the assistance furnished by the stirrups and the bars, which would come up diagonally through the ends of the beam (where this maximum shear occurs) nearly to the top of the slab. The vertical planes on each side of the rib have a combined width of 10 inches, and therefore the unit-stress is 11/10 x 154 = 169 pounds per square inch. This is a case of true shear, and a 1:2:4 concrete should stand such a stress with a large factor of safety. But there are still other shearing stresses in these vertical planes. Considering a strip of the slab, say, one foot wide, which is reinforced by slab bars that are parallel to the girder, the elasticity of such a strip (if disconnected from the girder) would cause it to sag in the center. This must be prevented by the shearing strength of the concrete in the vertical plane along each edge of the girder rib. On account of the combined shearing stresses along these planes, it is usual to specify that when girders are parallel with the slab bars, bars shall be placed across the girder and through the top of the slab for the special purpose of resisting these shearing stresses. Some of the stresses are indefinite, and therefore no precise rules can be computed for the amount of the reinforcement. But since the amount required is evidently very small, no great percentage of accuracy is important. A recent specification on this point required 3/8-inch bars, 5 feet long, spaced 12 inches apart.

The shear of the girder, taken as a whole, should be computed as for simple beams, as already discussed in Article 295; and stirrups should be used, as described in Article 279.

Another special form of shear must be considered in this problem. Where the beams enter the girders, there is a tendency for the beams to tear their way out through the girder. The total load on the girder by the two beams on each side, is of course equal to the total load on one beam, and equals 37,350 pounds. Some of the reinforcing bars of the beam will be bent up diagonally so that they enter the girder near its top, and therefore the beam could not tear out without shearing through the girder from near its top or for a depth of, say, 22 inches (3 inches less than d). We therefore have 2 X 22 X 11 = 484 square inches, the area to be sheared out. Dividing this into 37,350 gives 77 pounds per square inch. Although this is probably a safe shearing stress, many engineers would consider it advisable to use special V-shaped stirrups (see a, Fig. 108) to strengthen the beam against such stress. If the angle of these stirrups with the vertical is, say, 45°, then the stress in the bars on each side will be .707 of the total load, assuming that these bars were to take all the stress. This would mean that these bars would have a stress of about 26,406 pounds, and at 16,000 pounds per square inch would require a total area of 1.65 square inches. Three 3/4-inch bars would therefore more than provide the necessary area, even assuming that these stirrups took the entire load, and disregarding the stirrups such as would ordinarily be placed in the beam, and also disregarding the shearing strength of the concrete. If, therefore, these stirrups are made of Fig. 108. Detail of Reinforcement at Junction of Beam and Girder.

1/2-inch bars instead of 3/4-inch bars, the shearing stresses in the concrete due to the beam will be amply provided for. A complete detailed drawing will show all of the bars required for a panel between four of these columns. The student should study this drawing (see Fig. 109) in connection with the foregoing demonstrations of the dimensions of the bars and of the concrete.