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The discrepancy is due to the dropping of decimals during the computations. From Equation 33, we compute that x = 1.72; then (d - x) = 23.28. Substituting the value of the moment and of the dimensions in the upper part of Equation 35, we compute c to be 420 pounds per square inch. Similarly, making substitutions in the lower part of Equation 35, using the more precise value of (ci - x) for the lever-arm of the steel, we find s = 16,052 pounds per square inch. The student should verify in detail all these computations. The total required area of 5.08 square inches may be divided into; say, 8 round bars - 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 concrete beam should be at least 10.78 inches wide. We shall call it 11 inches. The total depth of the concrete beam will be three inches greater than ci, OT 28 inches. This means 23 inches below the concrete slab, and the area of concrete below the concrete slab is therefore 11 X 23 = 253 square inches, rather than 12 X 22 square inches, as assumed for trial. The shearing stresses between the rib and concrete slab of the girder are of special importance in this case.

The quantity 5h 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, 1 = 16 (feet), and z is to be determined: z = 81,544 -- (3 x 11 >< 16) 154 pounds per square inch. This measures the maximum shearing stress under the concrete 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 concrete beam (where this maximum shear .occurs) nearly to the top of the concrete slab. The vertical planes on each side of the rib have a combined width of 10 inches, and therefore the unit-stress is >< 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 concrete slab, say, one foot wide, which is reinforced by concrete 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 concrete slab bars, bars shall be placed across the girder and through the top of the concrete 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--inch bars, 5 feet long, spaced 12 inches apart. The shear of the girder, taken as a whole, should be computed as for simple concrete 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 concrete beams enter the girders, there is a tendency for the concrete beams to tear their way out through the girder. The total load on the girder by the two concrete beams on each side, is of course equal to the total load on one concrete beam, and equals 37,350 pounds. Some of the reinforcing bars of the concrete beam will be bent up diagonally so that they enter the girder near its open end. We therefore have 2 X 22 > (11 = 484 square inches, the area to be sheared out. Dividing this figure into 37,350 will give 77 pounds per square inch.

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