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 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:

M = As(d-x) ..(19)

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:

M=p b d s(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:

M0= 9/8 (1-1/3q) qc'bkd (d-x) .(21)

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 = 2/3 for ultimate stresses in the concrete, the equation becomes:

M0 = 7/12 c'bkd (d-x)..(22)

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 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 Jed = .175 d; d - x = .825 d. From Equation 22 we have:

M0= 7/12 X 2,000 X .490 X 7 X 10 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. From Equation 20, assuming s = 55,000, we have:

M0=.012 X 7 X 10 X 55,000 X 8.25 = 372,900 inch-pounds.

If the beam were actually stressed with this moment, the total compression in the concrete would equal 372,900 8.25 = 45,200 pounds. From Equation 13 we have:

45,200 = 7/12 c'bkd = 7/12 c' X 7 X .490 X 10.

Solving for c', c' = 45,200 20.008 = 2,205 pounds, which is considerably more than that assumed - 2,000.

The practical interpretation of the above is that if the beam is tested by Equation 22, indicating an ultimate moment of 330,137 inch-pounds, and the actual proposed loading, multiplied by its factor of safety, does not have a moment which exceeds this value, the compression in the concrete will not be more than 2,000 pounds per square inch, while the tension in the steel will be not greater than 47,640 pounds per square inch (ultimate value), which is safe but uneconomical. On the other hand, if Equation 20 were employed, indicating an ultimate moment of 372,900 pounds, and the ultimate loading of the beam seemed to require this moment, the steel would be all right, but the concrete would have an ultimate compression of 2,205 pounds, which would be dangerous for that grade of concrete. Therefore, as a general rule, whenever the percentage of steel has been assumed, both equations (20 and 22) should be tested. The lowest ultimate moment should be the limit which should not be exceeded by the ultimate moment of the actual loading, for the use of the higher value will mean an excessive stress in either the concrete or the steel.

Example 2. What will be the ultimate resisting moment of a 5-inch slab made of a high quality of concrete (1:2:4), using the most economical percentage of steel?

Answer. For this quality of concrete, r = 10; the ultimate compressive strength of the concrete is 2,700; and the ultimate tension in the steel is assumed at 55,000. Substituting these values in Equation 18, we find that the economical percentage of steel is 1.21. Interpolating this value of p in Table XIII, considering that r = 10, we have k = .424. Substituting this value of k in Equation 14, we find that x = .151 d. In the case of the 5-inch slab, we shall assume that the center of gravity of the steel is placed 1 inch from the bottom of the slab. Therefore d = 4 inches. For a slab of indefinite width, we shall assume that b = 12 inches. Therefore our computed value for the ultimate resisting moment gives the moment of a strip of the slab one foot wide, and the computed amount of the steel is the amount of steel per foot of width of the slab.

Substituting these various values in Equation 20, we find as the value of the ultimate resisting moment:

Mo = .0121 X 12 X 4 X 55,000 X .849 X 4 = 108,482 inch-pounds

The area of steel required for each foot of width is:

A = .0121 X 12 X 4 = .5808 square inch.

This equals .0484 square inch per inch of width. Since a 1/2-inch square bar has an area of .25 square inch, we may provide the rein-

.25 forcement by using 1/2-inch square bars spaced .0484 = 5.17 inches,

or, say, 5 1/4 inches.

Example 3. A very instructive comparison may be made by considering a 5-inch slab with d = 4 inches, but made of 1:3:5 concrete. In this case we call r - 12; c = 2,000; and s (as before) = 55,000. By the same method as before, we obtain p = .0084; k = .395; and therefore x = .141 d. Substituting these values in Equation 20, we have:

Mo = .0084 X 12 X 4 X 55,000 X .859 X 4= 76,197 inch-pounds.

The area of steel per foot of width is:

A = .0084 X 12 X 4 = .4032 square inch.

This would require 1/2-inch square bars spaced 7.33 inches. Although the amount of steel required in this slab is considerably less than was required in the previous case, the ultimate moment of the slab is also very much less. In fact the reduction of strength is very nearly in proportion to the reduction in the amount of steel. Therefore, it must be observed that, although the percentage of steel used with high-grade concrete is considerably higher, the thickness of the concrete will be considerably less; and in spite of the fact that the percentage of steel may be higher, its absolute amount for a slab of equal strength may be approximately the same.

Example 4. Another instructive principle may be learned by determining the required thickness of a slab made of 1:3:5 concrete, which shall have the same ultimate strength as the high-grade concrete mentioned in example 2. In other words, its ultimate moment per foot of width must equal 10S,482 inch-pounds. The values of r, c, and s are the same as in example 3, and therefore the value of p must be the same as in example 3; therefore p = .00S4. Since r and p are the same as in example 3, k again equals .395, and therefore x = .141 d. We therefore have from Equation 20:

M0 = 108,482 = .0084 X 12 X d X 55,000 X .859 X d.

Solving this equation for d, we find d2 = 22.78; and d = 4.77. The area of the steel A = p b d = .0084 X 12 X 4.77 = .481. This is considerably less than the area of steel per foot of width as computed in example 2 for a slab of equal strength. On the other hand, the slab of 1:3:5 concrete will require about 15 per cent more concrete. It will also weigh about 10 pounds per square foot more than the thinner slab, which will reduce by that amount the permissible live load. The determination of the relative economy of the two kinds of concrete will therefore depend somewhat on the relative price of the concrete and the steel. The difference in the total cost of the two methods is usually not large; and abnormal variation in the price of cement or steel may be sufficient to turn the scale one way or the other.