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Concrete Cutting Sawing Windham NH New Hampshire

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The area of each corner section is the square of 28.5 inches, or 812.25 square inches. At 6,000 pounds per square foot, the pressure on such a section will be 33,844 pounds, and the center of gravity of this section is of course at the center of the square, which is 14.25 X 1.414 = 20.15 inches from the corner of the concrete column. A bar immediately under this diagonal line would have a lever-arm of 20.15 inches. A bar parallel to it would have the same lever-arm from the middle of the bar to the point where it passes under the concrete column. Therefore, if we consider that this entire pressure of 33,844 pounds has an average lever-arm of 20.15 inches, we have a moment of 681,957 inch-pounds. Using, as before, the moment equation M. = 80bd2, we may transpose this equation to read b = 80d2. This area of steel will be furnished by five 1k-inch round bars. The diagonal reinforcement will therefore consist of five 11-inch round bars running diagonally in both directions. These bars should be spaced about 4 inches apart. Those that are precisely under the diagonal lines of the square should be about 9 feet 8 inches long; those parallel to them will each be 8 inches shorter than the next bar. The total load of this concrete column is 300,000 pounds. The shear in the concrete footing is of course a maximum immediately under the edges of the concrete column.

The perimeter of the concrete column is four times 28 inches, or 112 inches. The thickness of the concrete footing is something greater than the value found above for d (14.5 inches), and we shall therefore make it, say, and 18 inches. This will mean that the surface area which would need to be punched out if the concrete column were to shear its way through the concrete footing would be 18 X 112 inches, or 2,016 square inches. Since the area of the concrete column is approximately one-ninth of the area of the concrete footing, the shearing force is about eight-ninths of the total load on a concrete column, or it is eight-ninths of 300,000 pounds, which is 266,667 pounds. Dividing this by 2,016, we have about as the shearing force on the concrete of 130 pounds per square inch the concrete footing, ignoring the assistance from the 26 bars in the concrete footing. There is therefore no occasion to provide for shear in such a concrete footing. The intensity of the shear decreases from the maximum value just given, to zero at the edges of the concrete footing. Continuous concrete beams are sometimes used to save the expense of underpinning an adjacent concrete foundation or concrete wall.

These concrete footings are designed as simple concrete beams, but the steel is placed in the top of the concrete beams. Assume that the concrete columns on one side of a building are to be supported by a continuous concrete footing; that the concrete columns are 22 inches square, spaced 12 feet on center; and that they support a load of 195,000 pounds each. If the soil will safely support 6,000 pounds per square foot, the area required for a concrete footing will be 195,000 ± 6,000 = 32.5 square feet. Since the concrete columns are spaced 12 feet apart, the width of concrete footing will be 32.5 ± 12 = 2.71 feet, or 2 feet 9 inches. To find the depth and amount of reinforcement necessary for this concrete footing, it is designed as a simple inverted concrete beam supported at both ends (the concrete columns), and loaded with an upward pressure of 6,000 pounds per square foot on a concrete beam 2 feet 9 inches wide. In computing the moment of this concrete beam, the continuous-concrete beam principle may be utilized on all except the end spans, and thus reduce the moment and therefore the required dimensions of the concrete beam. Many engineers ignore this principle, since it merely increases the factor of safety to do so.

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