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As fast as the bars can be spared from the bottom of the concrete beam, they may be turned up diagonally so that there are at every section of the concrete 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 concrete beam, it would be far better to use nine k-inch bars than four 1-inch bars. In either case, the steel has the 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 six (6), 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 had. Then, M. = - W. 1           x 101,952 x 216. 2,752,704. In order to reduce as much as possible the size and weight of this concrete beam, we shall use 1: 2: 4 concrete, and therefore apply Equation 24: 2,752,704 & 565bd'; bd' = 4,872. If b = 16 inches, d2 = 304.5, and d = 17.5 inches. A still better combination would be a deeper and narrower concrete 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 2.93 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: V 50,796 = b (d—x) = 12 >< 17 = 249 pounds per square inch. As already discussed in previous cases, the ends of the concrete 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 concrete beam must be reinforced by turning the bars up, or by the use of stirrups. The concrete 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 concrete beam, where there is ordinarily no reinforcement which will assist shearing stresses. There is one very radical difference between the behavior of a concrete-steel concrete structure and that of a concrete 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, and not 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 concrete 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. I

Are You in Peterborough New Hampshire? Do You Need Concrete Cutting?

We Are Your Local Concrete Cutter

Call 603-622-4441

We Service Peterborough NH and all surrounding Cities & Towns