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Concrete Cutting Sawing Nottingham NH New Hampshire

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The area of the steel as computed above is 2.53 square inches. Assuming that this is furnished by five 3-inch square reinforcing bars, the concrete 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 concrete beam, which prevents the concrete beam from failing at the ends. Tests have shown that concrete beams which are reinforced by bars only running through the lower part of the concrete 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. Concrete beams which are tested to destruction frequently fail at the ends of the concrete 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 concrete 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-- 605d equal V -- b d, It may be proved that v, the intensity of the vertical shear per square inch, is: In the above case, the ultimate total shear V in the last inch at the end of the beam, is 39. 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. 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 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 8 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 = 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.

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

We Are Your Local Concrete Cutter

Call 603-622-4441

We Service Nottingham NH and all surrounding Cities & Towns