TT' / 4,200 X 6 X 12 = 37,S00 inch-pounds.
Placing this numerical value of M0 = 397 b as in Equation 23, we have 37,S00 = 397 b In this case, b = 12 inches. Substituting this value of b, we solve for and obtain = 7.93, and d = 2.S2 inches. Allowing an extra inch below the steel, this will allow us to use a 4-inch slab. Theoretically we could make it a little less. Practically this figure should be chosen. The required steel, from Equation 23, equals .00S4 bd. Taking b = 1, we have the required steel per inch of width of the slab = .00S4 X 2.S2 = .0237 square inch. If we use square bars which have a cross-sectional area of .23 square inch, we may space the bars -- = 10 inches. This .0237 reinforcement could also be accomplished by using '-inch square bars, which have an area of .1406. The spacing may therefore be .1406 .0237 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 beams, so as to prevent shrinkage. The required amount of this steel is not readily calculable. Since the I-beams are 6 feet apart, if we place two lines of -;--inch square bars spaced 2 feet apart, parallel with the I-beams, there will then be reinforcing steel in a direction parallel with the I-beams at distances apart not greater than 2 feet, since the I-beams themselves will prevent shrinkage immediately around them.
270. Straight-Line Formulae. The working unit-compressions for even the best grade of concrete are seldom allowed to exceed 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 formulre 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 equa tions involving q, must also be correspondingly modified.
Adopting the same notation as in Article 256, we may say that the triangle ninN in Fig. 07 represents the compressive forces; that the area of the triangle measures the summation of those forces; and, assuming that in this case c = inn, the summation is: cLkd _ . (25) The center of gravity of the triangle, which is the centroid of com pression of the concrete, is at of the height of the triangle (kd) from the compression face of the concrete. The same value is obtained by making q = 0 in the equation above Equation 14, which gives us: 3 kd (26) The similarity of this equation to Equation 18 is readily apparent, the difference being due only to the elimination of the effect of q.
The moment of resistance of a beam equals the total tension in the steel, or the total compression in the concrete (which are equal), times (d — x). Therefore we have the choice of two values (as before): If the economical percentage p has already been determined from Equation 28, then either equation may be used, as most convenient, since they will give identical results. If the percentage has been arbitrarily chosen, then the least value must be determined, as was described in Article 267.
271. Determination of Values for Frequent Use. For 1: 3: 5 concrete, using as before r = 12, and with a working value for c = 500, and s = 16,000, we find from Equation 28 that the economical percentage of steel equals: From 'Fable XV we find by interpolation that, for r — 12, and p =
.0043, k = .273. Then (see Equation 26): The percentage of steel computed from Equation 28 has been called the most economical percentage, because it is the percentage which will develop the maximum allowed stress in the concrete and the steel at the same time, or by the loading of the beam to some definite maximum loading. The real meaning of this is best illus trated by a numerical example using another percentage. Assume that the percentage of steel is exactly doubled, or that p = 2 X .0043 = .00S6. From Table XV, for r = 12, and p = .00S6, we find x .121d; and (d— x) = .879d. Substituting these values in both forms of Equation 29, we have: The interpretation of these two equations, and also of the equation found above (,1/ = 62 /AP), is as follows: Assume a beam of definite dimensions b and d, and made of concrete whose modulus of elasticity is that of the modulus of elasticity of the reinforcing steel; assume that it is reinforced with steel having a cross-sectional area = .0043 bd. Then, when it is loaded with a load which will develop a moment of 62 the tension in the steel will equal 16,000 pounds per square inch, and the compression in the concrete will equal 500 pounds per square inch at the outer fibre. Assume that the area of the steel is exactly doubled. One effect of this is to lower the neutral axis (k is increased from .273 to .362), and more of the concrete is available for compression. The load may be increased about 29 per cent, or until the moment equals SO bcP, before the compression in the concrete reaches 500 pounds per square inch. Under these conditions the steel has a tension of about 10,600 pounds per square inch, and its full strength is not utilized. If the load were increased until the moment was 121 then the steel would he stressed to 16,000 pounds per square inch, but the concrete would be compressed to over 750 pounds, which would of course be unsafe with such a grade of con crete. If the compression in the concrete is to be limited to 500 pounds per square inch, then the load must be limited to that which will give a moment of SO Even for this the steel is doubled in order to increase the load 29 per cent. Whether this is justifiable, depends on several circumstances—the relative cost of steel and con crete, the possible necessity for keeping the dimensions of the beam within certain limits, etc. Usually a much larger ratio of steel than 0.43 per cent is used; 1.0 per cent is far more common; but when such is used, it means that the strength of the steel cannot he fully utilized unless the concrete can stand high compression. A larger value of r will indicate higher values of k, which will indicate higher moments; but r cannot he selected at pleasure. It depends on the character of the concrete used; and, with E constant, a large value of r means a small value for E,, which also means a small value for c, the permissible compression stress. Whenever the per centage 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 he tested, and that one giving the lower moment should be used.