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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 mixtures, 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 mixtures 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 percentage of steel is .84 per cent. Under these conditions, k will always be .395; and x will equal .141 d. Therefore the term (d - x) will always equal .859 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 310. Therefore, for 1:3:5 concrete, we have the constant values: M0 = .0084 x b d >< 55,000 >< .86d similarly we can compute a corresponding value for 1:2:4 concrete, using the values previously allowed for this grade: M0=565 bd7. Concrete flooring with a live-load capacity of 150 pounds per square foot, is to be constructed on 1-concrete beams spaced 6 feet center to center, using 1:3:5 concrete. What thickness of concrete slab will be required,-and how much steel must be used? Using the approximate estimate, based on experience, that such a concrete 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 concrete beams) will therefore carry a total load of 700 >< 6 = 4,200 pounds. Considering this as a simple concrete beam, we have: Placing this numerical value of M. = 397 b d2, as in Equation 23, we have 37,800 = 397 b d2. In this case, b = 12 inches. Substituting this value of b, we solve for d2, and obtain d2 = 7.93, and ci = 2.82 inches. Allowing an extra inch below the steel, this will allow us to use a 4-inch I concrete slab. Theoretically we could make it a little less. Practically this figure should be chosen. The required steel, from Equation 23, equals .0084 bd. Taking b = 1, we have the required steel per inch of width of the concrete slab = .0084 X 2.82 = .0237 square inch. If we use --inch square bars which have a cross-sectional area .25 of .25 square inch, we may space the bars = 10 inches. This reinforcement could also be accomplished by using *-inch square bars, which have an area of .1406. The spacing may therefore be 1406= 6.0 inches. As referred to later, there should also be a few bars laid perpendicular to the main reinforcing bars, or parallel with the I-concrete beams, so as to prevent shrinkage. The required amount of this steel is not readily calculable. Since the I-concrete beams are 6 feet apart; if we place two lines of *-inch square bars spaced 2 feet apart, parallel with the I-concrete beams, there will then be reinforcing steel in a direction parallel with the I-concrete beams at distances apart not greater than 2 feet, since the I-concrete beams themselves will prevent shrinkage immediately around them. The working unit-compressions for even the best grade of concrete are seldom allowed exceeding 600 pounds per square inch. An inspection of Fig. 93 will show that the curve from the point o to the point indicating a pressure of 600 pounds, although really a parabola, is so nearly a straight line that there is but little error in considering it to be straight. On this account, many formulae for the strength of reinforced concrete have been developed on the basis of a uniform modulus of elasticity for the concrete. This is virtually the same as assuming that q equals zero in Equation 16. The other equations which are derived from equations involving q, must also be correspondingly modified.

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