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The
load is uniformly distributed; and the moment: Mx
37,350 X 240 1,120,500 inch-pounds. We shall assume that the concrete beam is
to have a depth ci to the reinforcement, of 22
inches, and shall utilize Equation 39 to obtain an approximate value for the
area. Substituting the known quantities in Equation 39, we have: 1,120,500 = A
>< 16,000 x (22-1.67) - A = 3.44 square inches. For T-concrete concrete
beams with very wide concrete slabs and great depth of concrete beam, the
percentage of steel is always very small. In this case, p = 3.44 ± (96 X 22) =
.00163. Such a value is beyond the range of those given in Table XV, and
therefore we must compute the value of k from Equation 27; and we find that k =
.165; kd = 3.63, which shows that the neutral axis is
within the concrete slab; x = - kd = 1.21, and
therefore (ci - x) = 20.79. Substituting these values
in the upper part of Equation 29 in order to find the value of c, we find that
c = 309 pounds per square inch. Substituting the known values in the second
half of Equation 29, in order to obtain a more precise value of s, we find that
s = 15,737 pounds per square inch. The required area (3.44 square inches) of
the bars will be afforded by six f-inch round bars (6 X .60 = 3.60) with
considerable to spare.

From Table XVIII we find that six 1-inch bars (either
square or round), if placed in one row, would require a concrete beam 14.72
inches wide. This is undesirably wide, and so we shall use four bars in the
lower row and two above, and make the concrete beam 11 inches wide. This will
add nearly an inch to the depth, and the total depth will be 22 + 3, or 25
inches. The concrete below the concrete slab is therefore 11 inches wide by 20
inches deep, instead of 12 inches wide by 15 inches deep, as assumed when
computing the dead load. The section of 220 square inches will therefore weigh
more than the suggested section of 180 square inches; but the difference in
dead load weight H is so small that it is unnecessary to alter the
calculations, especially since the unit-stresses in the concrete and steel are
both lower than the working limits. It should also be noted that the span of
these concrete beams was considered as 20 feet, which is the distance from
center to center of the concrete columns (or of the girders). This is certainly
more nearly correct than to use the net span between the concrete columns (or
girders), which is yet unknown, since neither the concrete columns nor the
girders are yet designed. There is probably some margin of safety in using the
span as 20 feet. The load on one concrete beam is computed above as 37,350
pounds. The load on the girder is therefore the equivalent of this load
concentrated at the center, or of double the load (74,700 pounds) uniformly
distributed. Assuming for a trial value that the girder will be 12 inches by 22
inches below the concrete slab, its weight for sixteen feet will be 4,392, or
say 4,400 pounds.

This gives a total of 79,100 pounds as the equivalent total
live and dead load uniformly distributed over the girder. Its moment in the
center therefore equals X 79,100 X 192 = 1,898,400 inch-pounds. The width of
the concrete slab in this case is almost indefinite, being twenty feet, or
forty-eight times the thickness of the concrete slab. We shall therefore assume
that the compression is confined to a width of fifteen times the concrete slab
thickness, or that b' = 75 inches. Assume for a trial value that d = 25 inches;
then from Equation 39, if s = 16,000, we find that A = 5.08 square inches. Then
p = .0027; and, from Equation 27, k = .207, and lcd =
5.175. This shows that the neutral axis is below the concrete slab, and that it
belongs to Case 1, Article 286. Checking the computation of lcd
from Equation 34, we compute lcd = 5.18, which is
probably the more correct value because computed more directly.

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

**We Are Your Local
Concrete Cutter**

**Call 603-622-4441**

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