From

Equation 19 we have the formula that the resisting moment at any point in the beam equals the area of the steel, times the unit tensile stress in the steel, times the distance from the steel to the centroid of compression of the steel, which is the distance d - x. We may compute the moment in the beam at two points at a unit-distance apart. The area of the steel is the same in each equation, and d - x is substantially the same in each case; and therefore the difference of moment, divided by (d - x), will evidently equal the difference in the unit-stress in the steel, times the area of the steel. To express this in an equation, we may say, denoting the difference in the moment by d M, and the difference in the unit-stress in the steel by d s: dM = A x ds.

(d-x)

But A X d s is evidently equal to the actual difference in tension in the steel, measured in pounds. It is the amount of tension which must be transferred to the concrete in that unit-length of the beam. But the computation of the difference of moments at two sections that are only a unit-distance apart, is a comparatively tedious operation, which, fortunately, is unnecessary. Theoretical mechanics teaches us that the difference in the moment at two consecutive sections of the beam is measured by the total vertical shear in the beam at that point. The shear is very easily and readily computable; and therefore the required amount of tension to be transferred from the steel to the concrete can readily be computed. A numerical illustration may be given as follows: Suppose that we have a beam which, with its load, weighs 20,000 pounds, on a span of 20 feet. Using ultimate values, for which we multiply the loading by 4, we have an ultimate loading of 80,000 pounds. Therefore,

Mo = Wo l = 80,000 x 240 = 2,400, 000

Using the constants previously chosen for 1:3:5 concrete, and therefore utilizing Equation 23, we have this moment equal to 397 b d2. Therefore b d2 = 6,045.

If we assume 6 = 15 inches; d = 20.1 inches; then d - x = .86d = 17.3 inches. The area of steel equals:

A = .0084 b d = 2.53 square inches. We know from the laws of mechanics, that the moment diagram for a beam which is uniformly loaded is a parabola, and that the ordinate to this curve at a point one inch from the abutment will, in the above case, equal (119/120 )2 of the ordinate at the abutment. This ordinate is measured by the maximum moment at the center, multiplied by the factor (119/120)2 = 14,161 = .9834; therefore the actual moment

14,400 at a point one inch from the abutment = (1.00 - .9834) = .0166 of the moment at the center. But .0166 X 2,400,000 = 39,840.

But our ultimate loading being 80,000 pounds, we know that the shear at a point in the middle of this one-inch length equals the shear at the abutment, minus the load on this first 1/2 inch, which is 1/240 of 40,000 (or 167) pounds. The shear at this point is therefore 40,000 - 167 (or 39,833) pounds. This agrees with the above value 39,840, as closely as the decimals used in our calculations will permit.

The value of d - x is somewhat larger when the moment is very small than when it is at its ultimate value. But the difference is comparatively small, is on the safe side, and it need not make any material difference in our calculations. Therefore, dividing 39,840 by 17.3, we have 2,303 pounds as the difference in tension in the steel in the last inch at the abutment. Of course this does not literally mean the last inch in the length of the beam, since, if the net span were 20 feet, the actual length of the beam would be considerably greater. The area of the steel as computed above is 2.53 square inches. Assuming that this is furnished by five 3/4-inch square bars, the surfaces of these five bars per inch of length equals 15 square inches. Dividing 2,303 by 15, we have 153 pounds per square inch as the required adhesion between the steel and the concrete. While this is not greater than the adhesion usually found between concrete and steel, it is somewhat risky to depend on this; and therefore the bars are usually bent so that they run diagonally upward, and thus furnish a very great increase in the strength of the beam, which prevents the beam from failing at the ends. Tests have shown that beams which are reinforced by bars only running through the lower part of the beam without being turned up, or without using any stirrups, will usually fail at the ends, long before the transverse moment, which they possess at their center, has been fully developed.