The failure of a beam by actual shear is almost unknown. The failures usually ascribed to shear are generally caused by diagonal tension. A solution of the very simple Equation 31 will indicate the intensity of the vertical shear.

The relation of crushing strength to shearing strength is expressed by the equation:

Unit shearing; strength z = c'/2 tanθ in which z is the unit shearing strength, and 0 is the angle of rupture under direct compression. This angle is usually considered to be 60°; for such a value the shearing strength would equal c' 3.464. When 6 = 45°, the shearing strength would equal one-half of the crushing strength, and this agrees very closely with the results of tests made by Professor Spofford. But the shearing strength is considered to be a far less reliable quantity than the crushing strength; and therefore dependence is not placed on shear, even for ultimate loading, to a greater value than about one-half of the above value; or,

Unit shearing strength z = c' 6.928.

Usually the unit-intensity of the vertical shear (even for ultimate loads) is less than this. But this ignores the assistance furnished by the bars. Actual failure would require that the bars must crush the concrete under them. When, as is usual, there are bars passing obliquely through the section, a considerable portion of the shear is carried by direct tension in the bars.

It seems impracticable to develop a rational formula for the amount of assistance furnished by these diagonal bars, unless we make assumptions which are doubtful and which therefore vitiate the reliability of the whole calculation. Therefore the rules which have been suggested for a prevention of this form of failure are wholly empirical. Mr. E. L. Ransome uses a rule for spacing vertical stirrups, made of wires or 1/4-inch rods, as follows:

KENILWORTH AVENUE FACADE OF UNITY CHURCH, OAK PARK, ILL.

KENILWORTH AVENUE FACADE OF UNITY CHURCH, OAK PARK, ILL.

Frank Lloyd Wright, Architect, Oak Park, 111.

Built in 1907, Entirely of Concrete. Cost, $40,000. Only Flat Roofs are Used in the Structure - an Unusual Treatment for a Church.

For Plan, See Opposite Page.

PLAN OF THE TEMPLE FLOOR. UNITY CHURCH, OAK PARK, ILL.

PLAN OF THE TEMPLE FLOOR. UNITY CHURCH, OAK PARK, ILL.

Frank Lloyd Wright, Architect, Oak Park, 111. Exterior View Shown on Opposite Page.

The first stirrup is placed at a distance from the end of the beam equal to one-fourth the depth of the beam; the second is at a distance of one-half the depth beyond the first stirrup; the third, three-fourths of the depth beyond the second; and the fourth, a distance equal to the depth of the beam beyond the third (see.Fig. 100). This empirical rule agrees with the theory, in the respect that the stirrups are closer at the ends of the beam, where the shear is greatest. The four stirrups extend for a distance from the end equal to 2 1/2 times the depth of the beam. Usually this is a sufficient distance; but some"systems" use stirrups throughout the length of the beam. On very short beams, the shear changes so rapidly that at 2 1/2 times the depth from the end of the beam the shear is not generally so great as to produce dangerous stresses. With a very long beam, the change in the shear is correspondingly more gradual; and it is possible that stirrups or some other device must be used for a greater actual distance from the end, although for a less proportional distance.

Fig. 100. Spacing of Stirrups.

Fig. 100. Spacing of Stirrups.

When the diagonal reinforcement is accomplished by bending up the bars at an angle of about 45°, the bending should be done so that there is at all sections a sufficient area of steel in the lower part of the beam to withstand the transverse moment at that section. As fast as the bars can be spared from the bottom of the beam, they may be turned up diagonally so that there are at every section of the beam one or more bars which would be cut diagonally by such a section. On this account it is far better to use a larger number of bars, than a smaller number of the same area. For example, if it were required that there shall be 2.25 square inches of steel for the section at the middle of the beam, it would be far better to use nine -2-inch bars than four 3/4-inch bars. In either case, the steel has the same area and the same weight. The nine 1/2-inch bars give a much better distribution of the metal. The superficial area of the nine 1/2-inch bars is 18 square inches per linear inch of the beam, while the area of the four 3/4-inch bars is only 12 square inches per inch of length. But an even greater advantage is furnished by the fact that we have nine bars instead of four, which may be bent upward (and bent more easily than the 3/4-inch bars) as fast as they can be spared from the bottom of the beam. In this way the shear near the end of the beam may be much more effectually and easily provided for.

Since the shear is greatest at the ends of the beam, more bars should be reserved for turning up near the ends. For example, in the above case of the nine bars, one or two bars might be turned up at about the quarter-points of the beam. One or two more might be turned up at a distance equal to, or a little less than, the depth of the beam from the quarter-points toward the abutments. Others would be turned up at intermediate points; at the abutments there should be at least two, or perhaps three, diagonal bars, to take up the maximum shear near the abutments. This is illustrated, although without definite calculations, in Fig. 101.

Fig. 101. Bars Turned Up to Take Up Shear near Ends of Beam.

Fig. 101. Bars Turned Up to Take Up Shear near Ends of Beam.