A = .0121 X 12 X 4 = .5SOS square inch.
This equals .0484 square inch per inch of width. Since a 1-inch square bar has an area of .25 square inch, we may provide the rein forcemeat by using 4--inch square bars spaced 0484 = 5.17 inches, or, say, 51 inches.
Example 3. A very instructive comparison may be made by considering a 5-inch slab with d = 4 inches, but made of 1:3:5 con crete. 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 Equa tion 20, we have: The area of steel per foot of width is: This would require square bars spaced 7.33 inches. Although the amount of steel required in this slab is considerably less than was required in the previous case, the ultimate moment of the slab is also very much less. In fact the reduction of strength is very nearly in proportion to the reduction in the amount of steel. There fore, it must be observed that, although the percentage of steel used with high-grade concrete is considerably higher, the thickness of the concrete will be considerably less; and in spite of the fact that the percentage of steel may be higher, its absolute amount for a slab of equal strength may be approximately the same.
Example 4. Another instructive principle may be learned by determining the required thickness of a slab made of 1:3:5 concrete, which shall have the same ultimate strength as the high-grade con crete mentioned in example 2. In other words, its ultimate moment per foot of width must equal 108,482 inch-pounds. The values of r, c, and s are the same as in example 3, and therefore the value of p must be the same as in example 3; therefore p = .00S4. Since r and p are the same as in example 3, k again equals .395, and there fore x = .141 d. We therefore have from Equation 20: = 10S,482 = .0084 X 12 X d X 55,000 X .S59 X d.
Solving this equation for d, we find = 22.7S; and d = 4.77. The area of the steel A = p b d = .0084 X 12 X 4.77 = .481. This is considerably less than the area of steel per foot of width as computed in example 2 for a slab of equal strength. On the other hand, the slab of 1:3:5 concrete will require about 15 per cent more concrete. It will also weigh about 10 pounds per square foot more than the thinner slab, which will reduce by that amount the permissible live load. The determination of the relative economy of the two kinds of concrete will therefore depend somewhat on the relative price of the concrete and the steel. The difference in the total cost of the
two methods is usually not large; and abnormal variation in the price of cement or steel may be sufficient to turn the scale one way or the other.
Determination of Values for Frequent Use. The above methods of calculation may be somewhat simplified by the determi nation, once for all, of constants which are in frequent use. For example, a very large amount of work is being done, using 1:3 : 5 concrete. Sometimes engineers will use the formube developed on the basis of 1:3:5 concrete, even when it is known that a richer mixture, will be used. Although 'such a practice is not economical, the error is on the side of safety; and it makes some allowance for the fact that a mixture which is nominally richer may not have any greater strength than the values used for the 1:3:5 mixture, on account of defective workmanship or inferior cement or sand: Some of the constants for use with 1:3: 5 mixture and 1:2:4 mixture will now be worked out.
For the 1:3:5 mixture, r = 12; c = 2,000; and we shall assume s = 55,000. On the basis of such values, the economical percentaye of steel is .S4 per cent. Under these conditions, 1, will always he .305; and x will equal .141 d. Therefore the term (d — x) will always equal .559 d, or, say, .86 d, which is close enough for a working value. Since the above values for c and s represent the ultimate values, the resulting moment is the ultimate moment, which we shall call .1/„. Therefore, for 1:3:5 concrete, we have the constant values: = Xbd x 55,000 x .se/ = 397 b d' (23) .t = .00S-1 li d ld—x) = .S6d - Similarly we can compute a corresponding value for 1:2:4 concrete, using the values previously allowed for this grade: Numerical Example. A flooring with a live-load capacity of 150 pounds per square foot, is to be constructed on I-beams spaced 6 feet center to center, using 1:3: 5 concrete. What thickness of slab will be required, and how much steel must he used? Answer. Using the approximate estimate, based on experience, that such a slab will weigh about 50 pounds per square foot, we can compute the ultimate load by multiplying the live load, 150, by four, and the dead load, 50, by two, and obtain a total ultimate load of 700 pounds per square foot. A strip 1 foot wide and 6 feet long (between the beams) will therefore carry a total load of 700 x 6 = 4,200 pounds. Considering this as a simple beam, we have: mo o