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Concrete Cutting Sawing South Hampton NH New Hampshire

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Therefore, if the shearing strength is sufficient, according to the rule, for a plain concrete beam, it is certainly sufficient for the T-concrete beam. In the first example of Article 291, the total load on the concrete beam is 30,000 pounds. Therefore the maximum shear V at the end of the concrete beam is 15,000 pounds. In this particular case, d — x = 12.25. For this concrete beam, d= 13.75 inches, and b = 11 inches. Substituting these values in Equation 31, we have: V+15,000 and V=IIX 12.25 111 pounds per square inch. Although this is probably a very safe stress for direct shearing, it is more than double the allowable direct tension due to the diagonal stresses; and therefore ample reinforcement must be provided. If only two of the i-inch bars are turned at an angle of 450 at the end, these two bars will have an area of 1.54 square inches, and will have a working tensile strength (at the unit-stress of 16,000 pounds) of 24,640 pounds. This is more than the total vertical shear at the ends of the concrete beam; and we may therefore consider that the concrete beam is protected against this form of failure. Assume a concrete floor construction as outlined in skeleton form in Fig. 107.

The concrete columns are spaced 16 feet by 20 feet. Girders which support the alternate rows of concrete beams connect the concrete columns in the 16-foot direction. The live load on the concrete floor is 150 pounds per square foot. The concrete is to be a 1:2:4 mixtures, with r = 10 and c'= 600 require the proper dimensions for the girders, concrete beams, and concrete slab. The load on the girders may be computed in either one of two ways, both of which give the same results. We must consider that each concrete beam supports an area of 8 feet by 20 feet. We may therefore consider that girder d supports the load of b (on a concrete floor area 8 ft. by 20 ft.) as a concentrated load in the center. Or, we may consider that, ignoring the concrete beams, the concrete beam a girder supports a uniformly distributed load on an area 16 ft. by 20 ft. The moment in either case we estimate that a 5-inch concrete slab (or d = 4) will carry the load. This will weigh 60 pounds per square foot, and make a total live and dead load of 210 pounds per square foot. A strip one foot wide and 8 feet long will carry a total load of 1,680 pounds, and its moment will be - x 1,680 X 96 = 20,160 inch-pounds. Using the first half of Equation 29, we can substitute the known values, and say that: 20,160 = X 600 X 12 X .358d x .881 d=1,135d? d2==17.76 and d = 4.21. In this case the span of the concrete slab is considered as the distance from center to center of the concrete beams.

This is evidently more exact than to use the net span (which equals eight feet, less the still unknown width of concrete beam), since the true span is the distance between the centers of pressure on the two concrete beams. It is probable that the true span (really indeterminable) will be somewhat less than 8 feet, which would probably justify using the round value of d = 4 inches, and the concrete slab thickness as 5 inches, as first assumed. The area of the steel per inch of width of the concrete slab = pbd = .01 X 1 X 4.21 = .0421 square inch. Using k-inch round bars whose area equals .1963 square inch; the required spacing of the bars will be .1963 — .0421 = 4.66 inches. Practically this would be called 4- inches. The load on a concrete beam is that on an area of 8 feet by 20 feet, and equals 8 x 20 X 210 = 33,600 pounds for live and dead load. As a rough trial value, we shall assume that the concrete beam will be 12 inches wide and 15 inches deep below the concrete slab, or a volume of 1 X 1.25 >< 20 cubic feet = 25 cubic feet, which will weigh 3,750 pounds. Adding this, we have 37,350 pounds as the total live and dead load carried by each concrete beam.

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