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Concrete Cutting Sawing New Ipswich NH New Hampshire

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From Equation 20, assuming s = 55,000, we have: M,=.012 >< 7 x 10 X 55,000 >< 8.25 = 372,900 inch-pounds. If the concrete beam were actually stressed with this moment, the total compression in the concrete would equal 372,900 -- 8.25 = 45,200 pounds. From Equation 13 we have: 45,200 =c'bkd=c' X 7 X .490 x 10,125-12 Solving for c', = 45,200 20.008 = 2,205 pounds, which is considerably more than that assumed 2,000. The practical interpretation of the above is that if the concrete beam is tested by Equation 22, indicating an ultimate moment of 330,137 inch-pounds, and the actual proposed loading, multiplied by its factor of safety, does not have a moment which exceeds this value, the compression in the concrete will not be more than 2,000 pounds per square inch, while the tension in the steel will be not greater than 47,640 pounds per square inch (ultimate value), which is safe but uneconomical. On the other hand, if Equation 20 were employed, indicating an ultimate moment of 372,900 pounds and the ultimate loading of the concrete beam seemed to require this moment, the steel would be all right, but the concrete would have an ultimate compression of 2,205 pounds, which would be dangerous for that grade of concrete. Therefore, as a general rule, whenever the percentage of steel has been assumed, both equations (20 and 22) should be tested. The lowest ultimate moment should be the limit which should not be exceeded by the ultimate moment of the actual loading, for the use of the higher value will mean an excessive stress in either the concrete or the steel. What will be the ultimate resisting moment of a 5-inch concrete slab made of a high quality of concrete (1:2:4), using the most economical percentage of steel? For this quality of concrete, r = 10; the ultimate compressive strength of the concrete is 2,700; and the ultimate tension in the steel is assumed at 55,000. Substituting these values in Equation 18, we find that the economical percentage of steel is 1.21. Interpolating this value of p in Table XIII, considering that r = 10, we have k = .424. Substituting this value of k in Equation 14, we find that x = .151 d. In the case of the 5-inch concrete slab, we shall assume that the center of gravity of the steel is placed 1 inch from the bottom of the concrete slab. Therefore d = 4 inches. For a concrete slab of indefinite width, we shall assume that b = 12 inches. Therefore our computed value for the ultimate resisting moment gives the moment of a strip of the concrete slab one foot wide, and the computed amount of the steel is the amount of steel per foot of width of the concrete slab. Substituting these various values in Equation 20, we find as the value of the ultimate resisting moment: M. = .0121 x 12 >< 4 x 55,000 >< .849 x 4 = 108,482 inch-pounds. The area of steel required for each foot of width is: A = .0121 >< 12 >< 4.5808 square inch. This equals .0484 square inch per inch of width. Since a'-inch square bar has an area of .25 square inch, we may provide the reinforcement by using -s-inch square bars spaced  = 5.17 inches, or, say, 5 inches. A very instructive comparison may be made by considering a 5-inch concrete slab with d = 4 inches, but made of 1: 3: 5 concrete. In this case we call r = 12; c = 2,000; and s (as before) = 55,000. By the same method as before, we obtain p = .0084; k = .395; and therefore x = .141 d. Substituting these values in Equation 20, we have: C = .0084 X 12) < 4 >< 55,000) < .859 x 4= 76,197 inch-pounds.

For example, a very, large amount of work is being done, using 1:3: 5 concrete. Sometimes concrete construction engineers will use the formula developed on the basis of 1:3:5 concrete, even when it is known that a richer mixture will be used.

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