When a definite load, such as a weight carried by a column, is to be supported on a subsoil whose bearing power has been estimated at some definite figure, the required area of the footing becomes a perfectly definite quantity, regardless of the method of construction of the footing. But with the area of the footing once determined, it is possible to effect considerable economy in the construction of the footing, by the use of reinforced concrete. An ordinary footing of masonry is usually made in a pyramidal form, although the sides will be stepped off instead of being made sloping. It may be approximately stated that the depth of the footing below the base of the column, when ordinary masonry is used, must be practically equal to the width of the footing. The offsets in the masonry cannot ordinarily be made any greater than the heights of the various steps. Such a plan requires an excessive amount of masonry.

Fig. 109. Detail of Floor Pane

Fig. 109. Detail of Floor-Pane.

A footing of reinforced concrete consists essentially of a slab, which is placed no deeper in the ground than is necessary to obtain a proper pressure from the subsoil. In the simplest case, the column is placed in the middle of the footing, and thus acts as a concentrated load in the middle of the plate (Fig. 110). The mechanics of such a problem are somewhat similar to those of a slab supported on four sides and carrying a concentrated load in the center, with the very important exception, that the resistance, instead of being applied merely at the edges of the slab, is uniformly distributed over the entire surface. Since the column has a considerable area, and the slab merely overlaps the column on all sides, the common method is to consider the overlapping on each side to be an inverted cantilever carrying a uniformly distributed load, which is in this case an upward pressure. The maximum moment evidently occurs immediately below each vertical face of the column. At the extreme outer edge of the slab the moment is evidently zero, and the thickness of the slab may therefore be reduced considerably at the outer edge. The depth of the slab, and the amount of reinforcement, which is of course placed near the bottom, can be determined according to the usual rules for obtaining a moment. This can best be illustrated numerically.

Example. Assume that a load of 252,000 pounds is to be carried by a column, on a soil which consists of hard, firm gravel. Such soil will ordinarily safely carry a load of 7,000 pounds per square foot. On this basis, the area of the footing must be 36 square feet, and therefore a footing 6 feet square will answer the purpose. A concrete column 24 inches square will safely carry such a loading. Placing such a column in the middle of a footing will leave an offset 2 feet broad outside each face of the column. We may consider a section of the footing made by passing a vertical plane through one face of the column. This leaves a block of the footing 6 feet long and 2 feet wide, on which there is an upward pressure of 12 X 7,000 = 84,000 pounds. The center of pressure is 12 inches from the section, and the moment is therefore 12 X 84,000 = 1,008,000 inch-pounds. Multiplying this by 4, we have 4,032,000 inch-pounds as the ultimate moment. Applying Equation 21, we place this equal to 397 bd2, in which b = 72 inches. Solving this for d, we have d - 11.9 inches. A total thickness of 15 inches would therefore answer the purpose. The amount of steel required per inch of width = .0084 d = .0084 X 11.9 = .100 square inch of steel per inch of width. Therefore 3/4-inch bars spaced 5.6 inches apart will serve the purpose. A similar reinforcing of bars should be placed perpendicularly to these bars.

Fig. 110. Simple Footing of Reinforced Concrete.

Fig. 110. Simple Footing of Reinforced Concrete.

The above very simple solution would be theoretically accurate in the case of an offset 2 feet wide for the footing of a wall of indefinite length, assuming that the upward pressure was 7,000 pounds per square foot. The development of such a moment uniformly along the section of our square footing, implies a resistance to bending near the outer edges of the slab which will not actually be obtained. The moment will certainly be greater under the edges of the column. On the other hand. we have used bars in both directions. The bars passing under the column in each direction are just such as are required to withstand the moment produced by the pressure on that part of the footing directly in front of each face of the column. It may be considered that the other bars have their function in tying the two systems into one plate whose several parts mutually support one another. If further justification of such a method is needed, it may be said that experience has shown that it practically fulfils its purpose.

A more effective method of reinforcing a simple footing is shown in Fig. 111. Two sets of the reinforcing bars are at a-a and b-b, and are placed only under the column. To develop the strength of the corners of the footings, bars are placed diagonally across the footing, as at c-c and d-d. In designing this footing, the projections of the footing beyond the column are treated as free cantilever beams, or by the method discussed above. The maximum shear occurs near the center; and therefore, if it is necessary to take care of this shear by means of reinforcement, it should be provided by using stirrups.

Example. Assume that a load of 300,000 pounds is to be carried by a column 28 inches square, on a soil that will safely carry a load of 6,000 pounds per square foot. What should be the dimensions of the footing and the size and spacing of the reinforcing bars? The bars are to be placed diagonally as well as directly across the footing, as illustrated in Fig. 111. Also investigate the shear.

Solid ion. The load of 300,000 pounds will evidently require an area of 50 square feet. The sides of the square footing will evidently be 7.07 feet, or, say, 85 inches; and the offset on each side of the 28-inch column is 28.5 inches. The area of each cantilever wing which is straight out from the column is 28.5 X 28 = 798 square inches = 5.54 square feet. The load is therefore 5.54 X 6,000 = 33,250 pounds. Its lever-arm is one-half of 28.5 inches, or 14.25 inches. The moment is therefore 473,812 inch-pounds. Adopting the straight-line formula Mc = 80bd2, on the basis that p = .0086,we may write the equation:

473,812 = 80 X 28 X d2, from which, d2 = 211; d = 14.5; A = pbd = .0086 X 28 X 14.5 = 3.59 square inches.

This area of metal may be furnished by eight 3/4-inch round bars, and therefore there should be eight 3/4-inch round bars spaced about 3.5 inches apart under the column in both directions.

The mechanics of the reinforcement of the corner sections, which are each 28.5 inches square, is exceedingly complicated in its precise theory. The following approximation to it is probably sufficiently precise. The area of each corner section is the square of 28.5 inches, or 812.25 square inches. At 6,000 pounds per square foot, the pressure on such a section will be 33,844 pounds, and the center of gravity of this section is of course at the center of the square, which is 14.25 X 1.414 = 20.15 inches from the corner of the column. A bar immediately under this diagonal line would have a lever-arm of 20.15 inches. A bar parallel to it would have the same lever-arm from the middle of the bar to the point were it passes under the column. Therefore, if we consider that this entire pressure of 33,844 pounds has an average lever-arm of 20.15 inches, we have a moment of 681,957 inch-pounds. Using, as before, the moment equation Mc = 80bd2, we may transpose this equation to read b = Mc 80d2 Then,

Fig. 111. Reinforcement for Footing.

Fig. 111. Reinforcement for Footing.

A = pbd = p M /80d1 d = p M/80d

= . 0086 x 681.957

80x 14.5

= 5.06 square inches.

This area of steel will be furnished by five 1 1/8-inch round bars. The diagonal reinforcement will therefore consist of five 1 1/8-inch round bars running diagonally in both directions. These bars should be spaced about 4 inches apart. Those that are precisely under the diagonal lines of the square should be about 9 feet 8 inches long; those parallel to them will each be 8 inches shorter than the next bar. Shear. The total load of this column is 300,000 pounds. The shear in the footing is of course a maximum immediately under the edges of the column. The perimeter of the column is four times 28 inches, or 112 inches. The thickness of the footing is something greater than the value found above for d (14.5 inches), and we shall therefore make it, say, 18 inches. This will mean that the surface area which would need to be punched out if the column were to shear its way through the footing would be 18 X 112 inches, or 2,016 square inches. Since the area of the column is approximately one-ninth of the area of the footing, the shearing force is about eight-ninths of the total load on a column, or it is eight-ninths of 300,000 pounds, which is 266,667 pounds. Dividing this by 2,016, we have about 130 pounds per square inch as the shearing force on the concrete of the footing, ignoring the assistance from the 26 bars in the footing. There is therefore no occasion to provide for shear in such a footing. The intensity of the shear decreases from the maximum value just given, to zero at the edges of the footing.