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Concrete Cutting Sawing Salem NH New Hampshire

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Although it would be difficult to develop any rule for the proper spacing between bars without making assumptions which are perhaps doubtful, the following empirical rule is frequently adopted by designers: The spacing between bars, center to center, should be two and a-quarter times the diameter of the bars. Fire insurance and municipal specifications usually require that there shall be two inches clear outside of the steel. This means that the concrete beam shall be four inches wider than the net width from out to out of the extreme bars. The data given in Table XVIII will therefore be found very convenient, since, when it is desired to use a certain number of bars of given size, a glance at the table will show immediately whether it is possible to space them in one row; and, if this is not possible, the necessary arrangement can be very readily designed. For example, assume that six 1-inch bars are to be used in a concrete beam. The table shows immediately that the required width of the concrete beam (following the rule) will be 14.72 inches; but if, for any reason, a concrete beam 11 inches wide is considered preferable, the table shows that four i-inch bars may be placed side by side, leaving two bars to be placed in an upper row.

Following the same rule regarding the spacing of the bars in vertical rows, the distance from center to center of the two rows should be 2.25 >< .875 = 1.97 inches, showing that the rows should be, say, two inches apart center to center. It should also be noted that the plane of the center of gravity of this steel is at two-fifths of the distance between the bars. Assume that a 5-inch concrete slab is supporting a load on concrete beams spaced 8 feet apart, the concrete beams haying a span of 20 feet. Assume that the moment of the concrete beam has been computed as 900,000 inch-pounds. What will be the dimensions of the concrete beam if the concrete is not to have a compression greater than 500 pounds per square inch and the tension of the steel is not to be greater than 16,000 pounds per square inch? There is an indefinite number of solutions to this problem. There are several terms in Equation 35 which are mutually dependent; it is therefore impracticable to obtain directly the depth of the concrete beam on the basis of assuming the other quantities; therefore it is only possible to assume figures which experience shows will give approximately accurate results, and then test these figures to see whether all the conditions are satisfied. Within limitations, we may assume the amount of steel to be used, and determine the depth of concrete beam which will satisfy the other conditions together with that of the assumed area of steel. For example, we shall assume that six 1-inch square bars having an area of 4.59 square inches will be a suitable reinforcement for this concrete beam. We shall also assume as a trial figure that x = 1.5.

Substituting these values in the second formula of Equation 35, we may write the second formula: 900,000        4.59 X 16,000 (d-1.5). Solving for d, we find that d = 13.75. If we test this value by means of Equation 36, we shall find that, substituting the values of I, c, r, and s in Equation 36, the resulting value of d equals 18.33. This shows that if we make the depth of the concrete beam only 13.75, the neutral axis will probably be within the concrete slab, and the problem comes under Case 3, to which we must apply Equation 29. Dividing the area of the steel, 4.50, by (b' X d), we have the value of p equals .00348. Interpolating with this value of p in Table XV, we find that when r = 12, k = 2.50; lcd = 3.44; x = 1.15; and d - x = 12.6. Substituting these values in Equation 29, we find that the moment 900,000 = 2,082c, or that c = 432 pounds per square inch. This shows that the unit-compression of the concrete is safely within the required figure.

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