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Concrete Cutting Sawing Newton NH New Hampshire

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Under L = 8, in Table XVI, we find that 1,356 comes between 1,241 and 1,501, showing that a concrete slab with an effective thickness d of about 51 inches will have this ultimate carrying capacity. The total thickness of the concrete slab should therefore be about 6 inches. The table also shows that 2- 1- bars spaced about 51 inches apart will serve for the reinforcement. We might also provide the reinforcement by *-inch square bars spaced a little over 3 inches apart; but it would probably be better policy to use the half-inch bars, especially since the 1-inch bars will cost somewhat more per pound. It is too much to expect of workmen that bars will be accurately spaced when their distance apart is expressed in fractions of an inch. But it is a comparatively simple matter to require the workmen to space the bars evenly, provided it is accurately computed how many bars should be laid in a given width of concrete slab. For example, in the above case, a panel of the concrete flooring which is, say, 20 feet wide, should have a definite number of bars; 20 feet = 240 inches, and 240 -- 5.75 = 41.7. We shall call this 42, and instruct the workmen to distribute 42 bars equally in the panel 20 feet wide. The workmen can do this without even using a foot-rule, and can adjust them to an even spacing with sufficient accuracy for the purpose. In Table XVII has been computed for convenience the ultimate total load on rectangular concrete beams made of average concrete (1: 3:5) and with a width of 1 inch. For other widths, multiply by the width of the concrete beam. Since M0 = W01; and since by Equation 23, for this grade of concrete, M0 = 397 b d2; and since for a computation of concrete beams 1 inch wide, b = 1, we may write 1 W01 = 397 d2. For 1 we shall substitute 12 L. Making this substitution and solving for W0, we have W. = 265 d2 ± L. Since b = 1, A, the area of steel per inch of width of the concrete beam = .0084 d. What is the ultimate total load on a simple concrete beam having a depth of 16 inches to the reinforcement, 12 inches wide, and having a span of 20 feet? Looking in Table XVII, under L = 20, and opposite d = 16, we find that a concrete beam 1 inch wide will sustain a total load of 3,392 pounds. For it width of 12 inches, the total ultimate load will be 12 >< 3,392 40,704 pounds. At 144 pounds per cubic foot, the concrete beam will weigh 3,840 pounds. Using a factor of 2 on this, we shall have 7,680 pounds, which, subtracted from 40,704, gives 33,024. Dividing this by 4, we have 8,256 lbs. as the allowable live load on such a concrete beam. The previous discussion has considered merely the tension and compression in the upper and lower sides of the concrete beam. A plain, simple concrete beam resting freely on two end supports, has neither tension nor compression in the fibers at the ends of the concrete beam. The horizontal tension and compression, found at or near the center of the concrete beam, entirely disappear by the time the end of the concrete beam is reached. This is done by transferring the tensile stress in the steel at the bottom of the concrete beam, to the compression fibers in the top of the concrete beam, by means of the intermediate concrete. This is, in fact, the main use of the concrete in the lower part of the concrete beam. It is therefore necessary that the bond between the concrete and the steel shall be sufficiently great to withstand the tendency to slip. The required strength of this bond is evidently equal to the difference in the tension in the steel per unit of length. For example, suppose that we are considering a bar 1 inch square in the middle of the length of a concrete beam. Suppose that the bar is under an actual tension of 15,000 pounds per square inch. Since the bar is 1 inch square, the actual total tension is 15,000 pounds.

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