The above methods of calculation may be somewhat simplified by the determination, once for all, of constants which are in frequent use. For example, a very large amount of work is being done, using 1:•'!: 5 concrete. Sometimes engineers will use the formulŠ developed on the basis of 1:3:5 concrete, even when it is known that a richer mixture will be used. Although such a practice is not economical, the error is on the side of safety; and it makes some allowance for the fact that a mixture which is nominally richer may not have any greater strength than the values used for the 1:3:5 mixture, on account of defective workmanship or inferior cement or sand.' Some of the constants for use with 1:3:5 mixture and 1:2:4 mixture will now be worked out.

HOUSE AT 123 EAST 73d STREET, NEW YORK CITY

HOUSE AT 123 EAST 73d STREET, NEW YORK CITY.

R. Burnside Potter, of Robertson & Potter, Architect, New York.

Front of A. B. C. Brick. Door and Fence, Wood, Painted; Walls and lintels. Cremo Marble; Cornice and Dormers, Copper, Painted; Roofs of Slate. Cost, about $70,000. An Excellent Example of a Simple, Dignified Treatment for a Narrow City Lot. The Treatment of the Roof, to Give a Low Effect, is Particularly Good. For Plans, See Opposite Page.

GROUND FLOOR.

GROUND FLOOR.

FIRST FLOOR.

FIRST FLOOR.

PLANS OF HOUSE AT 123 EAST 73d STREET, NEW YORK CITY.

R. Burnside Potter, of Robertson & Potter, Architect, New York.

Finish Floors from Ground to First Floor, Teak; above that, North Carolina Pine. Drawing Room Paneled Completely in Wood, Painted; and Walls throughout Rest of House Hung in Silk and Other Materials. Front View of House Exterior Shown on Opposite Page.

For the 1:3:5 mixture, r = 12; c = 2,000; and we shall assume s = 55,000. On the basis of such values, the economical percentage of steel is .84 per cent. Under these conditions, k will always be .395; and x will equal .141 d. Therefore the term (d - x) will always equal .859 d, or, say, .86 d, which is close enough for a working value. Since the above values for c and s represent the ultimate values, the resulting moment is the ultimate moment, which we shall call M0. Therefore, for 1:3:5 concrete, we have the constant values:

M0

= .0084 X bd X 55,000 X .86d

= 397 bd2

.(23)

A

= .0084 bd

(d-x)

= .86d

Similarly we can compute a corresponding value for 1:2:4 concrete, using the values previously allowed for this grade:

M0

= .565 b d2

...(24)

A

= .0121 bd

(d-x)

= .86d

Numerical Example. A flooring with a live-load capacity of 150 pounds per square foot, is to be constructed on I-beams spaced 6 feet center to center, using 1:3:5 concrete. What thickness of slab will be required, and how much steel must be used?

Answer. Using the approximate estimate, based on experience, that such a slab will weigh about 50 pounds per square foot, Ave can compute the ultimate load by multiplying the live load, 150, by four, and the dead load, 50, by two, and obtain a total ultimate load of 700 pounds per square foot. A strip 1 foot wide and 6 feet long

(between the beams) will therefore carry a total load of 700 X 6 = 4,200 pounds. Considering this as a simple beam, we have: Mo = wol /8 = 4,200 x 6 x 12/8 = 37,800 ,inch-pounds

Placing this numerical value of M0 = 397 b d2, as in Equation 23, we have 37,800 = 397 b d2. In this case, b = 12 inches. Substituting this value of b, we solve for d2, and obtain d2 = 7.93, and d = 2.82 inches. Allowing an extra inch below the steel, this will allow us to use a 4-inch slab. Theoretically we could make it a little less. Practically this figure should be chosen. The required steel, from Equation 23, equals .0084 bd. Taking b = 1, we have the required steel per inch of width of the slab = .0084 X 2.82 = .0237 square inch. If we use 1/2-inch square bars which have a cross-sectional area .25 of .25 square inch, we may space the bars .0237 = 10 inches. This reinforcement could also be accomplished by using 3/8-inch square bars, which have an area of .1406. The spacing may therefore be.1406 = .0237 6.0 Inches

As referred to later, there should also be a few bars laid perpendicular to the main reinforcing bars, or parallel with the I-beams, so as to prevent shrinkage. The required amount of this steel is not readily calculable. Since the I-beams are 6 feet apart, if we place two lines of 3/8-inch square bars spaced 2 feet apart, parallel with the I-beams, there will then be reinforcing steel in a direction parallel with the I-beams at distances apart not greater than 2 feet, since the I-beams themselves will prevent shrinkage immediately around them.