and solving for d, which may be facilitated by the use of a table of squares, we have:

If the 1.911 sq. ins. is divided among 4 rods, the area of each will

If square rods are employed, the required dimensions will be

=

0.478 (nearly)

4-11/16 inch rods, since 11/16 X 11/16 d = 21 inches is the effective depth or depth to the plane of the steel. At least 2 inches of concrete should be placed below the steel for protection and bond. Hence the total depth of the beam must be 21+ 2 23 inches.

=

If a heavy rock concrete weighing 144 pounds per cu. ft., including the steel, is employed, the weight of the beam 23 ins. deep, 13 ins. wide, and 14 feet long, will be

This is 814 pounds less than the assumed weight, and if the dimensions are again computed, using the actual weight of 4,186 pounds in place of the assumed weight of 5,000 pounds, it will be found to make a difference of half an inch in the required depth of the beam.

An inspection of the above computation for weight reveals a quick method for obtaining the weight of heavy rock concrete; viz., multiply together the total depth in inches by the breadth in inches by the length of the beam in feet.

The above method of designing a beam will probably impress the novice as faulty in that too much is assumed in advance. A very little practice will, however, enable him to estimate the probable weight very much more closely than was done in the above example, where the beam was purposely overestimated by 20 per cent in order to show that such an overestimate has very little effect on the design, and even if the first estimate should be extremely wide of the mark, two trials at the most should be all that would be required to determine the proper dimensions.

In assuming the percentage of lower flange steel in advance, the designer has two things to consider; first the most economical

percentage; and second, whether he wishes to make the compression or tension half of his beam the stronger. Probably .007 is the most economical percentage, as less than this amount of steel unduly increases the volume of concrete, while more than .007 affects unfavorably the cost of the steel. Below .or the steel will probably be weaker than the concrete at a breaking load, while above or the steel is likely to prove the stronger. Near this point either the steel or the concrete may be the first to fail if the beam is tested to destruction, depending chiefly on the materials and workmanship employed in mixing and placing the concrete. The designer should be sufficiently familiar with the quality of the work so that he can fix the percentage of steel at such a rate that the steel will begin to stretch before the concrete commences to crumple, thus producing deflection and giving warning in advance, in case the beam should be loaded beyond its capacity. In addition to this percentage, upper flange steel over the supports and stirrups should in general be provided.

In assuming the breadth and computing the effective depth, after the design has reached the stage where the product b d2 = a known number, several trials may be necessary to give the best proportions for the beam. For economy a beam is made as narrow as possible, but there are practical limits to decreasing the breadth which must not be encroached upon. Thus a beam should not be narrower than 1/24 of the span. It must be wide enough to provide at least 1 1/2 diameters and preferably 2 between the reinforcing bars and between the bars and sides of the beam. Moreover, the breadth should not be less than half of the depth, excepting for very large beams. Probably the best width is between 1/2 and 3/4 of the effective depth, d.

How to Design Reinforced Concrete Slabs.-For the purpose of design, a reinforced concrete slab placed as a continuous sheet over several girders and carrying a uniformly distributed load, may be treated as though the slab was divided into narrow strips, each having a width equal to the spacing of the reinforcing bars, and a length equal to the distance between the supporting girders.

The slab can then be designed by the following formulas:

Formulas (5) and (6) are identical in form with those employed for beams, and differ only in the coefficient of l, 1/10 being used instead of 1/8. The nomenclature is also the same, and if a percentage of steel different from .007, is desired, the corresponding denominator for formula (5) can be obtained from Table XVI in the same way as for beams. Such a slab requires top reinforcement extending over the girders for at least one-fourth of the span, on both sides of the girder; or if expanded metal or other fabric is employed the fabric must be placed so that in the centre of the span it will sag to near the bottom of the slab, while over the supports it will be near the top. The sectional area of lower flange steel may be obtained from formula (6), when seven-tenths per cent is used, or the coefficient .007 may be varied to suit the requirements.

Example.-Design a reinforced concrete slab supported by beams spaced 8 feet apart, which may be used to sustain a uniform load of 125 pounds per square foot, exclusive of its own weight.

Solution. Assume a steel percentage of .007, a spacing of the reinforcing bars of 6 inches; and the weight of a strip 6 inches wide, and 8 feet long at 240 pounds.

By spacing the bars 6 inches apart, the breadth, b, becomes 6 inches, and the external load, W, at 125 pounds per sq. ft., will be:

W = 8 ft. X 6/12 ft. X 125 lbs. per sq. ft.

500 pounds

If I inch of concrete is placed below the steel, the required thickness

From (6) the area of steel will be

A

=

.007 X 4 X 6

.168 sq. ins.

This may be obtained from 1-1/2 inch round rod.

When fabric is employed instead of steel rods, a strip 1 foot wide is taken as the basis of the design, or b = 12 inches, while formula (6) will give the sectional area of fabric required for the 1-foot strip, for seven-tenths per cent of steel.

In general, a reinforced concrete slab should not be less than 3 inches thick and should have at least 3/4 inch of concrete below the steel.

In an oblong slab the steel is placed crosswise from girder to girder. In a square slab supported on four girders, equidistant from each other, the rods are placed both ways. When reinforced in this manner, the same amount of steel is used as for the oblong slab, but there is a saving in concrete, as the concrete for a square slab need only be designed for half the load.

In an oblong slab, a few rods should also be placed longitudinally to prevent temperature cracks and to serve as binders for the main tension bars which run crosswise between the supporting girders.

Tables for Use in Designing Beams and Slabs.-Such tables may be divided into two classes: (a) those which give the required dimensions without computation, and (b) those which are used to facilitate computation by saving arithmetical labor. Tables XVII, XVIII, and XIX are of the latter and Table XX of the former class.

Table XVII is a table of squares for facilitating the computation of beam depths. Table XVIII gives the weight per lineal foot of reinforced concrete beams at 144 pounds per cubic foot, and is used for estimating the weight of beams. Table XVIII also shows comparative costs of beams at $10.00 per cu. yd. This is for the purpose of comparing the cost of beams of different proportions of depth to breadth and of different percentages of steel, in order to employ those which are most economical. Table XIX gives the sectional areas of round and square bars, and their weights and cost at the rate of 2 cents per pound. This is also convenient for making a comparison of costs. The costs given in Tables XVIII and XIX do not represent the actual costs, which may be 50 per cent more or less for any given structure. They are relative costs for gauging the

relative economy of different beams having equal strength or capacity.

Table XX, which is reproduced with slight modifications, by courtesy of the Atlas Portland Cement Co., from their book on the utilization of "Concrete in Factory Construction," gives the proper dimensions for beams and slabs that will carry uniformly distributed floor or roof loads of 125, 50, and 30 pounds respectively, per square foot. These beams, if checked over, by the straight line formulas (10) and (11), of Chapter XVIII, will be found to average about seven-tenths per cent of steel, to have a fibre stress in the concrete of about 500 pounds per square inch, and in the steel of between 12,000 and 15,000 pounds per square inch.

TABLE XVII.-BEAM DEPTHS AND THEIR SQUARES.

Example, Involving Use of Tables.-Compute the cost of beams spaced 8 feet apart, and having a span of 12 feet which will support a 6-inch slab of concrete in addition to a floor load of 140 pounds per square foot.

Solution.-Surface area, 12 X 8

13,440 lbs.

=

X 144 lbs. per cu. ft. 6,912 lbs.

Estimated weight of beam, Table XVIII, 12 x 24 ins. (288 X 12) 3,456 lbs. 13,440 + 6,912 + 3,456 23,808 lbs.

=