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Concrete Cutting Sawing Sandown NH New Hampshire

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Substituting the known values in the second part of Equation 29, we find that the stress in the steel s equals about 15,500 pounds per square inch. Assume that a concrete floor is loaded so that the total weight of live and dead load is 200 pounds per square foot; assume that the T-concrete concrete beams are to be 5 feet apart, and that the concrete slab is to be 4 inches thick; assume that the span of the concrete beams is 30 feet. We therefore have an area of 150 square feet to be supported by each concrete beam, which will give a total load of 30,000 pounds on each concrete beam. The moment at the center of such a concrete beam will therefore be equal to the total load, multiplied by one-eighth of the span (expressed, in inches), and the moment is therefore 1,350,000 inch-pounds. As a trial value, we shall assume that the concrete beam is to be reinforced with six i-inch bars, which have an area of 3.37 square inches. Substituting this value of the area in the second part of Equation 35, and assuming that s = 16,000 pounds per square inch, we find as an approximate value for d - x, that it will equal 25 inches.

This is very much greater than the value of d that would be found from substituting the proper values in Equation 36, so that we know at once that the problem must be solved by the methods of Case 1. For a 4-inch concrete slab, the value of x must be somewhere between 1.33 and 2.0. As a trial value, we may call it 1.5, and this means that d will equal 26.5. Assuming that this concrete slab is to be made of concrete using a value for r = 12, we know all the values in Equation 34, and may solve for lcd, which we find to equal 5.54 inches. As a check on the approximations made above, we may- substitute this value of lcd, and also the value of t in Equation 33, and obtain a more precise value of x, which we find to equal 1.62. Substituting the value of the moment and the other known quantities in the upper formula of Equation 35, we may solve for the value of c, and obtain the value that c = 352 pounds per square inch. This value for c is so very moderate that it would probably be economy to assume a lower value for the area of the steel, and increase the unit-compression in the concrete; but this solution will not be here worked out.

A great deal of T-concrete beam computation is done on the basis that the center of pressure of the concrete is at the middle of the concrete slab, and therefore that the lever-arm of the steel ci - - t. From these assumptions we may write the approximate formula: M>As (d—t) If the values of M, and s are known or assumed, we may assume a reasonable value for either A or (ci - I) and calculate the corresponding value of the other. On the assumption that the concrete slab takes all the compression, the distance between the steel and the center of compression of the concrete varies between (ci - t) and (d < c) which is the approximate value when the concrete beam becomes so small that it merges into the concrete slab. The smaller value (ci - t) is the absolute limit which is never reached. Therefore the lever-arm is always larger than (ci - t). Therefore, if we use Equation 37 to compute the area of steel A for a definite moment LVI, and unit steel tension s, the resulting value of A for an assumed depth ci, or the resulting value of ci for an assumed area A, will be larger than necessary. In either case the result is safe, but uneconomically so. As an illustration, using the values in Example 2, Article 291, of M, = 1,350,000; s = 16,000; (d - t) = (26.5 - 2) = 24.5, the resulting value of A = 3 44 square inches, which is larger than the more precise value previously computed. Equation 37 is particularly applicable when the neutral axis is in the rib.

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