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 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 maxi mum shear near the abutments. This is illustrated, although with out definite calculations, in Fig. 101.
2S0. Detailed Design of a Plain Beam. This will be illustrated by a numerical example. A beam having a span of 1S feet supports one side of a 6-inch slab S feet wide which carries a live load of 200 pounds per square foot. In addition, a special piece of machinery, weighing 2,400 pounds, is located on the slab so near the middle of the beam that we shall consider it to be a concentrated load at the center of the beam. The floor area carried by the beam is 18 feet by 4 feet = 72 square feet. Adding 3 inches to the 6 inches thickness of the slab as an allowance for the weight of the beam, we have 9 X 12 = 10S pounds per square foot for the dead weight of the floor. With a factor of 2 for dead load, this equals 216. Using a factor of 4 on the live load (200), we have 500 pounds per square foot. Then the ultimate load on the beam, due to these sources, is (216 + SOO) 72 = 73,152 pounds. So far as its effect on moment is concerned, the concentrated load of 2,400 pounds at the center would have the same effect as 4,800 pounds uniformly distributed. As it is a piece of vibrating machinery, we shall use a factor of 82X (0), and thus have an ultimate effect of 6 X 4,800 = 28,800 pounds. Adding this to 73,152, we have 101,952 pounds as the equivalent, ultimate, uniformly distributed load. Then, In order to reduce as much as possible the size and weight of this beam, we shall use 1: 2: 4 concrete, and therefore apply Equation 24: A still better combination would be a deeper and narrower beam with b = 12 inches, and d = 20.15 inches. With this combination, the required area of the steel will equal: A = .0121 X bd = .0121 X 12 X 20.15 = 2.03 square inches.
This can be supplied by eight bars inch square.
The total ultimate load as determined above, is 101,952 pounds. One-half of this gives the maximum shear at the ends, or 50,976 pounds. Applying Equation 31, we have, since d — x = .85 d = 17 inches: As already discussed in previous cases, the ends of the beam must be reinforced against diagonal tension, since the above value of v is too great, even as an ultimate value, for such stress. Therefore the
ends of the beam must be reinforced by turning the bars up, or by the use of stirrups. The beam must therefore be reinforced about as shown in Fig. 102. Although the concentrated center load in this case is comparatively too small to require any change in the design, it should not be forgotten that a concentrated load may cause the shear to change so rapidly that it might require special provision for it by means of stirrups in the center of the beam, where there is ordinarily no reinforcement which will assist shearing stresses.
2S1. Effect of Quality of Steel. There is one very radical dif ference between the behavior of a concrete-steel structure and that of a structure composed entirely of steel, such as a truss bridge. A truss bridge may be overloaded with a load which momentarily passes the elastic limit, and yet the bridge will not necessarily fail, nor cause the truss to be so injured that it is useless and must be immediately replaced. The truss might sag a little, but no immediate failure is imminent. On this account, the factor of safety on truss bridges is usually computed on the basis of the ultimate strength.
A concrete-steel structure acts very differently. As has already been explained, the intimate union of the concrete and the steel at all points along the length of the bar (and not merely at the ends), is an absolute essential for stability. If the elastic limit of the steel has been exceeded owing to an overload, then the union between the concrete and the steel has unquestionably been destroyed, provided that union depends on mere adhesion. Even if that union is assisted by a mechanical bond, the distortion of the steel has broken that bond to some extent, although it will still require a very considerable force to pull the bar through the concrete. It is therefore necessary that the elastic limit of the steel should be considered the virtual ultimate so far as the strength of the steel is concerned. It is accordingly con sidered advisable, as already explained, to multiply all working loads by the desired factor of safety (usually taken as 4), and then to pro portion the steel and concrete so that such an ultimate load will produce crushing in the upper fibre of the concrete, and at the same time will stress the steel to its elastic limit. On this basis, economy in the use of steel requires that the elastic limit should he made as high as possible.