The necessity of very frequently computing the required thickness of 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 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 d. This shows that 3.0 is somewhat excessive for the value of d. We computed its precise value to be 2.82. On the same line, we find under "Spacing of Bars," that 3/8 -inch square bars spaced 5 1/2 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.

Example 1. What is the ultimate load that will be carried by a 5-inch slab on a span of 10 feet, using 1:3:5 concrete?

Answer. 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 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.

Example 2. The floor of a factory is to be loaded with a live load of 300 pounds per square foot, the slab to be supported on beams spaced 8 feet apart. What must be the thickness of the floor-slab?

Answer. 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 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 slab; and, assuming that a 6 1/2-inch 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. Under L = 8, in Table XVI, we find that 1,356 comes between 1,241 and 1,501, showing that a slab with an effective thickness d of about 5 1/4 inches will have this ultimate carrying capacity. The total thickness of the slab should therefore be about 6 1/4 inches. The table also shows that 1/2-inch bars spaced about 5 3/4 inches apart will serve for the reinforcement. We might also provide the reinforcement by 3/8-inch square bars spaced a little over 3 inches apart; but it would probably be better policy to use the half-inch bars, especially since the 3/8-inch bars will cost somewhat more per pound.