But our ultimate loading being S0,000 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 inch, which is 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 i-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 pre vents 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.
27S. Distribution of Vertical Shears. Beams which are tested to destruction frequently fail at the ends of the beams, long before the transverse strength at the center has been fully developed. Even if the bond between the steel and the concrete is amply strong for the requirements, the beam may fail on account of the shearing or diag onal stresses in the concrete between the steel and the neutral axis. The student must accept without proof some of the following state ments regarding the distribution of the shear.
The intensity of the shear of various points in the height of the beam, may be represented by the diagram in Fig. 99. If we ignore the tension in the concrete due to transverse bending, the shear will be uniform between the steel and the neutral axis. Above the neutral axis, the shear will diminish toward the top of the beam, the curve being parabolic.
If the distribution of the shear were uniform throughout the section, we might say that the shear per square inch would equal V b d, It may be proved that v, the intensity of the vertical,shear per square inch, is: tr(31) v b (d — x) In the above ease, the ultimate total shear V in the last inch at the end of the beam, is 39,840 pounds. Then, 39,840 v = = 153.5 pounds per square inch, ' 15 X 17.3 The agreement of this numerical value of the unit-intensity of the vertical shear with the required bond between the concrete and the steel, is due to the accidental agreement of the width of the beam (15 inches) with the superficial area of the bars per inch of length of the beam (15 square inches). If other bars of the same cross-sec tional area, but with greater or less superficial surface, had been selected for the reinforcement, even this accidental agreement would not have been found.
The actual strength of concrete in shear is usually far greater than this. The failure of beams which fail at the ends when loaded with loads far within their capacity for transverse strength, is gener ally due to the secondary stresses. The computation of these stresses is a complicated problem in Mechanics; but it may be proved that if we ignore the tension in the concrete due to bending stresses, the diagonal tension per unit of area equals the vertical shear per unit of area (v). But concrete which may stand a shearing stress of 1,000 pounds per square inch will probably fail under a direct tension of 200 pounds per square inch. The diagonal stress has the nature of a direct tension. In the above case the beam probably would not fail by this method of failure, since concrete can usually stand a tension up to 200 pounds per square inch; but such beams, when they are not diagonally reinforced, frequently fail in that way before their ultimate loads are reached.