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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.

**Are You in Newton ****New Hampshire****? Do You
Need Concrete Cutting?**

**We Are Your Local
Concrete Cutter**

**Call 603-622-4441**

**We Service Newton NH
and all surrounding Cities & Towns**