Substituting q = 3 in Equation 16, we have: The various values for the ratio of the moduli of elasticity (r) are discussed in the succeeding section. The values of k for various values of r and p, and for the uniform value of q = 3, have been com puted in the following tabular form. Five values have been chosen for r, in conjunction with nine values of p, varying by 0.2 per cent and covering the entire practicable range of p, on the basis of which values k has been worked out in the tabular form. Usually the value of k can be determined directly from the table. By interpolating between two values in the table, any required value within the limits of ordinary practice can be determined with all necessary accuracy.
265. Ratio of Moduli. Theoretically there is an indefinite number of values of r, the ratio of the moduli of elasticity of the steel and the concrete. The modulus for steel is fairly constant at about 29,000,000 or 30,000,000. The value of the initial modulus for con crete varies according to the quality of the concrete, from 1,500,000 to 3,000,000 for stone concrete. An average value for cinder concrete is about 750,000. Some experimental values for stone concrete have fallen somewhat lower than 1,500,000, while others have reached 4,000,000 and even more. We may probably use the following values with the constant value of 29,000,000 for the steel.
The value given above for 1:6:12 concrete is mentioned only because the value r = 20 is sometimes used with the weaker grades of concrete, and the value of approximately 1,450,000 for the elasticity of such concrete has been found by experimenters. The use of such a lean concrete is hardly to be recommended, because of its unrelia bility. Considering the variability in cinder concrete, the even value of r = 40 is justifiable, rather than the precise value 38.67.
266. Percentage of Steel. The previous calculations have been made as if the percentage of the steel might be varied almost in definitely. While there is considerable freedom of choice, there are limitations beyond which it is useless to pass; and there is always a most economical percentage, depending on the conditions. We have already determined that : _ 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 moduli. 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 com pressive stress of 2,700 pounds (c'), which we may consider in con junction 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.
Example: What percentage of steel would be required for ordinary stone concrete, with r = 15, c =2,000, and s = 55,000? ANS. 0.95 per cent.
267. Resisting Moment. 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 centroid of compression of the concrete, times the total compression of the con crete, or, otherwise, times the tension in the steel. The compression
in the concrete and the tension in the steel are equal, and it is there fore 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 dimen sions of the concrete and ol the percentage of the reinforcing steel, it may be more convenient to write the equation : m=pbds (d — x) (20) From Equation 12 we have the total compression in the concrete. Multiplying this by the distance from the steel to the centroid 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 = s for ultimate stresses in the concrete, the equation becomes: Rio 7 — 12 c'bkd (d —x) (22) 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.
26S. 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 moduli of elasticity (r) of 15. The ultimate strength of the concrete (c') is assumed as 2.000.
Answer. From Table XIII, p = .012, and r = 15, k = .490; x = .357 kd = .175 d; d — x = .825 d. From Equation 22 we have: la= X 2,000 X .490 X 7 X 10 X 8.25=330,137 inch-pounds. 12 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 = .S4 square inch. Therefore the unit-stress in the steel would equal 40,017 ÷ .S4 = 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.
From Equation 20, assuming s = 55,000, we have: