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The moment of resistance of a concrete 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): = cbkd (d —x). 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. 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: 1 500500X12 = 2 16,000 (500) < 12 + 16,000) = .0043 From Table XV we find by interpolation that, for r = 12, and p = .0043, k = .273. Then X = -Icc1 = .091d; and (d—x) = .909d. Substituting these values in either formula of Equation 29, we have: M 62 bd2. 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 concrete beam to some definite maximum loading. The real meaning of this is best illustrated by a numerical example using another percentage. Assume that the percentage of steel is exactly doubled, or that p = 2 >< .0043 = .0086. From Table XV, for r = 12, and p = .0086, we find k=.362; x =.121d; and (d—x) = .879d. Substituting these values in both forms of Equation 29, we have: M0 = 80 bd2; and, M= 121 W2. The interpretation of these two equations, and also of the equation found above (M = 62 bd2), is as follows: Assume a concrete 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 bd2, 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 fiber. 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 80 bd2, 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 bcl2, then the steel would be 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 concrete. 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 80 bd2. 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 concrete, the possible necessity for keeping the dimensions of the concrete 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 be 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 be 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 E0, which also means a small value for c, the permissible compression stress.

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