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Concrete Cutting Sawing Sharon NH New Hampshire

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For convenience in future computations, we shall consider L to represent the length of the concrete beam, measured in feet. All other dimensions are measured in inches. Therefore the total shearing force along the lower side of the flange will be: Sh=zxbXLX12=3bzL. There is also a possibility that a concrete beam may fail in case the flange (or the concrete slab) is too thin; but the concrete slab is always reinforced by bars which are transverse to the concrete beam, and the concrete slab will be placed on both sides of the concrete beam, giving two shearing surfaces. It is required to test the concrete beam which was computed in Example 1 of Article 291. Here the total compressive stress in the flange = - cbkd = X 432 >< 96 X344 = 71,332 pounds. But this compressive stress measures the shearing stress Sh between the flange and the rib. This concrete beam requires six j--inch bars for the reinforcement. We shall assume that the rib is to be 11 inches wide, and that four of the bars are placed in the bottom row, and two bars about 2 inches above them. The effect of this will be to deepen the concrete beam slightly, since ci measures the depth of the concrete beam to the center of the reinforcement, and., as already computed numerically in Article 290, the center of gravity of this combination will be of an inch above the center of gravity of the lower row of bars.

Substituting in Equation 40 the values Sh = 71,332, b = 1.1, and L = 20, we find, for the unit-value of z, 108 pounds per square inch. This shows that the assumed dimensions of the concrete beam are satisfactory in this respect, since the true shearing stress permissible in concrete is higher than this. But the concrete beam must be tested also for its ability to withstand shear in vertical planes along the sides of the rib. Since the concrete slab in this case is 5 inches thick and we can count on both surfaces to withstand the shear, we have a width of 10 inches to withstand the shear, as compared with the 11 inches on the underside of the concrete slab. The unit-shear would therefore be - of the unit-shear on the underside of the concrete slab, and would equal 119 pounds per square inch. Even this would not be unsafe, but the danger of failure in this respect is usually guarded against by the fact that the concrete slab almost invariably contains bars which are inserted to reinforce the concrete slab, and which have such an area that they will effectively prevent any shearing in this way. Testing Example 2 similarly, we may find the total compression C from Equation 32, and that it equals As = 16,000 X 3.37 = 54,000 pounds. The steel reinforcement is six *-inch bars, and by Table XVIII we find that if placed side by side, the concrete beam must be 13.19 inches in width, or, in round numbers, 1314 inches. Substituting these values in Equation 37, we find, for the value of z, 45 pounds per square inch. Such a value is of course perfectly safe. The shear along the sides of the concrete beam will be considerably greater, since the concrete slab is only four inches thick, and twice the thickness is but 8 inches; therefore the maximum unit-shear along the sides will equal 45 times the ratio of 13.25 to 8, or 75 pounds per square inch.

Even this would be perfectly safe, to say nothing of the additional shearing strength afforded by the concrete slab bars. The shear here referred to is the shear of the concrete beam as a whole on any vertical section. It does not refer to the shearing stresses between the concrete slab and the rib. The theoretical computation of the shear of a T-concrete beam is a very complicated problem. Fortunately it is unnecessary to attempt to solve it exactly. The shearing resistance is certainly far greater in the case of a T-concrete beam than in the case of a plain concrete beam of the same width and total depth and loaded with the same total load.

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