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We have already determined that: Substituting the above value of Ic in this equation, we have, after considerable reduction: The above equation shows that we cannot select the percentage of steel at random, since it evidently depends on the selected stresses for the steel and concrete, and also on the ratio of their module. For example, consider a high-grade concrete (1:2:4) whose modulus of elasticity is considered to be 2,900,000, and which has a limiting compressive stress of 2,700 pounds (c'), which we may consider in conjunction with the limiting stress of 55,000 pounds in the steel. The values of c, s, and r are therefore 2,700, 55,000, and 10 respectively. Substituting these values in Equation 18, we compute p = .012. What percentage of steel would be required for ordinary stone concrete, with r = 15, c =2,000, and s = 55,000 ANS 0.95 percent. The moment which resists the action of the external forces is evidently measured by the product of the distance from the center of gravity of the steel to the cancroids of compression of the concrete, times the total compression of the concrete, or, otherwise, times the tension in the steel. The compression in the concrete and the tension in the steel are equal, and it is therefore only a matter of convenience to express this product in terms of the tension in the steel. Therefore, adopting the notation already mentioned, we may write the formula: But since the computations are frequently made in terms of the dimensions of the concrete and of the percentage of the reinforcing steel, it may be more convenient to write the equation: From Equation 12 we have the total compression in the concrete. Multiplying this by the distance from the steel to the cancroids of compression (d - x), we have another equation for the moment: This equation is perfectly general, except that it depends on the assumption as to the form of the stress-strain diagram as described in Article 260. On the assumption that q = 'for ultimate stresses in the concrete, the equation becomes: When the percentage of steel used agrees with that computed from Equation 18, then Equations 20 and 22 will give identically the same results; but when the percentage of steel is selected arbitrarily, as is frequently done, then the proposed section should be tested by both equations. When the percentage of steel is larger than that required by Equation 18, the concrete will be compressed more than is intended before the steel attains its normal tension. On the other hand, a lower percentage of steel will require a higher unit-tension in the steel before the concrete attains its normal compression. When the discrepancy between the percentage of steel assumed and the true economical value is very great, the stress in the steel (or the concrete) may become dangerously high when the stress in the other element (on which the computation may have been made) is only normal.

268. Example 1. What is the ultimate resisting moment of a concrete beam made of 1:3:5 concrete, which is 7 inches wide, 10 inches deep to the reinforcement, and which uses 1.2 per cent of reinforcement? The concrete is supposed to have a ratio for the module of elasticity (r) of 15. The ultimate strength of the concrete is assumed as 2.000.

Answer. From Table XIII, p = .012, and r = 15, Ic = .490; x = .357 lcd = .175 d; d - x = .825 d. From Equation 22 we have: M0=>< 2,000 x .490 X 7 X10 x 8.25=330,137 inch-pounds. The total compression in the concrete is the continued product of all the factors except the last, and equals 40,017. But this equals the tension in the steel, whose area = pbd = .012 x 7 X 10 .84 square inch. Therefore the unit-stress in the steel would equal 40)017 ± .84 = 47,640 pounds per square inch. This is considerably less than the usual ultimate of 55,000, and shows that the percentage of steel is considerably in excess of the normal value.

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