As an illustration, using the values in Example 2, Article 291, of Jh = 1,350,000; s =-- 16,000; (d = (26.5 2) = 24.5, the resulting value of A = 3 44 square inches, which is larger than the more precise value previously computed.
Equation 37 is particularly applicable when the neutral axis is in the rib. Under this condition, the average pressure on the concrete of the slab is always greater than c, or at least it is never less than As before explained, the average pressure just equals lie when the neutral axis is at the bottom of the slab. We may therefore say that the total pressure on the slab is always greater than ?; c b t. We therefore write the approximate equation: As before, the values obtained from this equation are safe, but are unnecessarily so. Applying them to Example 2, Article 291, by substituting = 1,350,000, b' = 60, t = 4, and (d = 24.5, we compute c = 459. But we know that this approximate value of c is greater than the true value; and if this value is safe, then the true value is certainly safe. The more accurate value of c, computed in Article 291, is 352. If the value of c in Equation 38 is assumed, and the value of d is omputed, the result is a depth of beam un necessarily great.
If the beam is so shallow that we may know, even without the test of Equation 36, that the neutral axis is certainly within the slab, then we may know that the center of pressure is certainly less than A t from the top of the slab, and that the lever-arm is certainly less than (d it); and we may therefore modify Equation 37 to read: Applying this to Example 1 of Article 291, and substituting = 900,000, s = 16,000, (d. a t) = (13.75 1.67) = 12.0S, we find that A = 4.65, instead of the 4.59 previously computed. This again illustrates that the formula gives an excessively safe value, although in this case the difference is small.
Equations 37 and 38 should be considered as a pair which are applied according as the steel or the concrete is the determining feature. When the percentage of steel is assumed (as is usual), both equations should be used to test whether the unit-stresses in both the steel and the concrete are safe. It is impracticable to form a simple approximate equation corresponding to Equation 39, which will express the moment as a function of the compression in the con crete. Fortunately it is unnecessary, since, when the neutral axis
is within the slab, there is always an abundance of compressive strength.
293. Shearing Stresses between Beam and Slab. Every solu tion for T-beam construction should be tested at least to the extent of knowing that there is no danger of failure on account of the shear between the beam and the slab, either on the horizontal plane at the lower edge of the slab, or in the two vertical planes along the two sides of the beam. Let us consider a T-beam such as is illustrated in Fig. 106. In the loWer part of the figure is represented one-half of the length of the flange, which is considered to have been separated from the rib. Following the usual method of considering this as a free body in space, acted on by external forces and by such internal forces as are necessary to produce equilibrium, we find that it is acted on at the left end by the abutment reaction, which is a vertical force, and also by a vertical load on top. We may consider P' to represent the summation of all compressive forces acting on the flanges at the center of the beam. In order to produce equilibrium, there must be a shearing force acting on the under side of the flange. We represent this force by Since these two forces are the only horizontal forces, or forces with horizontal components, which are acting on this free body in space, P' must equal Let us consider z to represent the shearing force per unit of area. We know from the laws of me chanics, that, with a uniformly distributed load on the beam, the shearing force is maximum at the ends of the beam, and diminishes uniformly towards the center, where it is zero. Therefore the average value of the unit-shear for the half-length of the beam, must equal z. As before, we represent the width.of the rib by b. For conven ience in future computations, we shall consider L to represent the length of the beam, measured in feet. All other dimensions are measured in inches. Therefore the total shearing force along the lower side of the flange, will be: There is also a possibility that a beam may fail in case the flange (or the slab) is too thin; but the slab is always reinforced by bars which are transverse to the beam, and the slab will be placed on both sides of the beam, giving two shearing surfaces.