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Therefore,
if the shearing strength is sufficient, according to the rule, for a plain concrete
beam, it is certainly sufficient for the T-concrete beam. In the first example
of Article 291, the total load on the concrete beam is 30,000 pounds. Therefore
the maximum shear V at the end of the concrete beam is 15,000 pounds. In this
particular case, d — x = 12.25. For this concrete beam, d= 13.75 inches, and b
= 11 inches. Substituting these values in Equation 31, we have: V+15,000 and V=IIX
12.25 111 pounds per square inch. Although this is probably a very safe stress
for direct shearing, it is more than double the allowable direct tension due to
the diagonal stresses; and therefore ample reinforcement must be provided. If
only two of the i-inch bars are turned at an angle of
450 at the end, these two bars will have an area of 1.54 square inches, and
will have a working tensile strength (at the unit-stress of 16,000 pounds) of
24,640 pounds. This is more than the total vertical shear at the ends of the concrete
beam; and we may therefore consider that the concrete beam is protected against
this form of failure. Assume a concrete floor construction as outlined in
skeleton form in Fig. 107.

The concrete columns are spaced 16 feet by 20 feet.
Girders which support the alternate rows of concrete beams connect the concrete
columns in the 16-foot direction. The live load on the concrete floor is 150
pounds per square foot. The concrete is to be a 1:2:4 mixtures, with r = 10 and
c'= 600 require the proper dimensions for the girders, concrete beams, and
concrete slab. The load on the girders may be computed in either one of two
ways, both of which give the same results. We must consider that each concrete
beam supports an area of 8 feet by 20 feet. We may therefore consider that
girder d supports the load of b (on a concrete floor area 8 ft. by 20 ft.) as a
concentrated load in the center. Or, we may consider that, ignoring the concrete
beams, the concrete beam a girder supports a uniformly distributed load on an
area 16 ft. by 20 ft. The moment in either case we estimate that a 5-inch concrete
slab (or d = 4) will carry the load. This will weigh 60 pounds per square foot,
and make a total live and dead load of 210 pounds per square foot. A strip one
foot wide and 8 feet long will carry a total load of 1,680 pounds, and its
moment will be - x 1,680 X 96 = 20,160 inch-pounds. Using the first half of
Equation 29, we can substitute the known values, and say that: 20,160 = X 600 X
12 X .358d x .881 d=1,135d? d2==17.76 and d = 4.21. In this case the span of
the concrete slab is considered as the distance from center to center of the concrete
beams.

This is evidently more exact than to use the net span (which equals
eight feet, less the still unknown width of concrete beam), since the true span
is the distance between the centers of pressure on the two concrete beams. It
is probable that the true span (really indeterminable) will be somewhat less
than 8 feet, which would probably justify using the round value of d = 4
inches, and the concrete slab thickness as 5 inches, as first assumed. The area
of the steel per inch of width of the concrete slab = pbd
= .01 X 1 X 4.21 = .0421 square inch. Using k-inch round bars whose area equals
.1963 square inch; the required spacing of the bars will be .1963 — .0421 =
4.66 inches. Practically this would be called 4- inches. The load on a concrete
beam is that on an area of 8 feet by 20 feet, and equals 8 x 20 X 210 = 33,600
pounds for live and dead load. As a rough trial value, we shall assume that the
concrete beam will be 12 inches wide and 15 inches deep below the concrete slab,
or a volume of 1 X 1.25 >< 20 cubic feet = 25 cubic feet, which will
weigh 3,750 pounds. Adding this, we have 37,350 pounds as the total live and
dead load carried by each concrete beam.

**Are You in South Hampton ****New
Hampshire****? Do You Need Concrete Cutting?**

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Concrete Cutter**

**Call 603-622-4441**

**We Service South
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