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Concrete Cutting Sawing Newmarket NH New Hampshire

Concrete Cutting Sawing Newmarket NH New Hampshire

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Whenever the percentage of steel is greater than the economical percentage, as is usual, then the upper of the two formulae of Equation 29 should be used. When in doubt, both should be tested, and that one giving the lower moment should be used. Using p = .0075, and r = 12, we have k = .343; x = .114d; and (ci - x) = .886 ci. Then, since p is greater than the economical value, we use the upper formula of Equation 29, and have: M=76 bd2. What is the working moment for a concrete slab with 5-inch thickness to the steel, the concrete having the properties described above? Calling b = 12 inches, M 76 X 12 >< 25 = 22,800 inch-pounds, the concrete slab having a span of 8 feet is to support a load of 150 pounds per square foot. The concrete is to be as described above, and the percentage of steel is to be 0.75. What is the required thickness it of the steel? A strip 12 inches wide has an area of 8 square feet, and carries a load of 1,200 pounds. The moment = - Wi = >< 1,200 X 06=14,400 inch-pounds. For a strip 12 inches wide, b = 12 inches, and Al = 76 X 12 X d2 & 912d2 = 14,400; d2 = 15.79; ci = 3.97 inches—or, say, 4 inches. The necessity of very frequently computing the required thickness of concrete slabs, renders very useful a table such as is shown in Table XVI, which has been worked out on the basis of 1:3:5 concrete, and computed by solving Equation 23 for various thicknesses d, and for various spans L varying by single feet. It should be noted that the loads as given are ultimate loads per square foot, and that they therefore include the weight of the concrete slab itself, which must be multiplied by its factor of safety, which is usually considered as 2. For example, in the numerical case of Article 269, we computed that there would be a total load of 700 pounds on a span of 6 feet. In the column headed 6, we find 794 on the same line as the value of 3.0 in the column ci. This shows that 3.0 is somewhat excessive for the value of ci. We computed its precise value to be 2.82. On the same line, we find under "Spacing of Bars," that *-inch square bars spaced 5 inches will be sufficient. In the above more precise calculation, we found that the bars could be spaced 6 inches apart, as was to be expected, since the computed ultimate load is considerably less than the nearest value found in the table. What is the ultimate load that will be carried by a 5-inch concrete slab on a span of 10 feet, using 1:3:5 concrete? The 5 inches here represents the total thickness, and we shall assume that the effective thickness (d) is 1 inch less. Therefore d = 4 inches. On the line opposite d = 4 in Table xvi and under the column L = 10, we have 508, which gives the ultimate load per square foot. A 5-inch concrete slab will weigh approximately 60 pounds per square foot, allowing 12 pounds per square foot per inch of thickness. Using a factor of 2, we have 120 pounds, which, subtracting from 508 leaves 388 pounds; dividing this by 4, we have 97 pounds per square foot as the allowable working load. Such a load is heavier than that required for residences or apartment houses. It would do for an office building. The concrete floor of a factory is to be loaded with a live load of 300 pounds per square foot, the concrete slab to be supported on concrete beams spaced 8 feet apart. What must be the thickness of the concrete floor-concrete slab? With 1,200 pounds per square foot ultimate load for the live load alone, we notice in Table XVI, under L = 8, that 1,241 is opposite to d = 5. This shows that it would require a concrete slab nearly 6 inches thick to support the live load alone. We shall therefore add another half-inch as an estimated allowance for the weight of the concrete slab; and, assuming that a 6k-inch concrete slab having a weight of 78 pounds per square foot will do the work, we multiply 300 by 4, and 78 by 2, and have 1,356 pounds per square foot as the ultimate load to be carried.

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