This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.
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 diagonal stresses in the concrete between the steel and the neutral axis. The student must accept without proof some of the following statements 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 ÷ bd. It may be proved that v, the intensity of the vertical shear per square inch, is: v = V /b(d-x)..(31)
In the above case, the ultimate total shear V in the last inch at the end of the beam, is 39,840 pounds. Then, v = 39,840 153.5 pounds per square inch, 15x17.3
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-sectional 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 generally 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.
Fig. 99. Intensity of Shear at Various Points in Height of Beam.