The economy of a retaining wall of reinforced concrete lies in the fact that by adopting a skeleton form of construction and utilizing the tensional and transverse strength which may be obtained from reinforced concrete, a wall may be built, of which the volume of concrete is, in some cases, not more than one-third the volume of a retaining wall of plain concrete which would answer the same purpose. Although the cost of reinforced concrete per cubic foot will be somewhat greater than that of plain concrete, it sometimes happens that such walls can be constructed for one-half the cost of plain concrete walls. The general outline of a reinforced-concrete retaining wall is similar to the letter L, the base of which is a base-plate made as wide as (and generally a little wider than) the width' usually considered necessary for a plain concrete wall. As a general rule, the width of the base should be about one-half the height. The face of the wall is made of a comparatively thin plate whose thickness is governed by certain principles, as explained later. At intervals of 10 feet, more or less, the base-plate and the face are connected by buttresses. These buttresses are very strongly fastened by tie-bars to both the base-plate and the face-plate. The stress in the buttresses is almost exclusively tension. The pressure of the earth tends to force the face-plate outward; and therefore the faceplate must be designed on the basis of a vertical slab subjected to transverse stresses which are maximum at the bottom and which reduce to zero at the top.

If the wall is "surcharged" (which means that the earth at the top of the wall is not level, but runs back at a slope), then the faceplate will have transverse stresses even at the top. The base-plate is held down by the pressure of the superimposed earth. The buttresses must transmit the bursting pressure on the face of the wall backward and downward to the base-plate. The base-plate must therefore be designed by the same method as a horizontal slab carrying a load equal and opposite to the upward pull in each buttress. If the base-plate extends in front of the face of the wall, thus forming INTERIOR VIEW IN ONE OF THE MANUFACTURING BUILDINGS OF THE GEORGE N. PIERCE COMPANY, BUFFALO, N. Y.

A Modern Factory Building of Concrete Construction, Lighted from Sawtooth Windows an extended toe, as is frequently done with considerable economy and advantage, even that toe must be designed to withstand transverse bending at the wall line, and also shearing at that point. The application of these principles can best be understood by an illustration.

## 301. Numerical Example

Assume that it is required to design a retaining wall to withstand an ordinary earthwork pressure of 20 feet, the earth being level on top. We are at once confronted with the determination of the actual lateral pressure of the earthwork. Unfortunately, this is an exceedingly uncertain quantity, depending upon the nature of the soil, upon its angle of repose, and particularly upon its condition whether wet or dry. The angle of repose is the largest angle with the horizontal at which' the material will stand without sliding down. A moment's consideration will show that this angle depends very largely on the condition of the material, whether wet or dry, etc. On this account any great refinement in these calculations is utterly useless.

Assuming that the back face of the wall is vertical, or practically so; that the upper surface of the earth is horizontal; and that the angle of repose of the material is 30°, the total pressure of the wall equals 1/6 w h2, in which h is the total height of the wall, and w is the weight per unit-volume of the earth. If the angle of repose is steeper than this, the pressure will be less. If the angle of repose is less than this, the fraction 1/6 will be larger, but the unit-weight of the material will probably be smaller. Assuming the weight at the somewhat excessive figure of 96 pounds per cubic foot, we can then say, as an ordinary rule, that the total pressure of the earth on a vertical strip of the wall one foot wide will equal 16 h2, in which h is the height of the wall in feet. The average pressure, therefore, equals 16 h; and the maximum pressure at a depth of h feet equals 32 h. Applying this figure to our numerical example, we have a total pressure on a vertical strip one foot wide, of 16 X 202 = 6,400 pounds. The pressure at a depth of 20 feet = 32 X 20 = 640 pounds.

It is usual to compute the thickness and reinforcement of a strip one foot wide running horizontally between two buttresses. Practically the strip at the bottom is very strongly reinforced by the baseplate, which runs at right angles to it; but if we design a strip at the bottom of the wall without allowing for its support from the baseplate, and then design all the strips toward the top of the wall in the same proportion, the upper strips will have their proper design, while the lower strip merely has an excess of strength. We shall assume, in this case, that the buttresses are spaced 15 feet center to center. Then the load on a horizontal strip of face-plate 12 inches high, 15 feet long, and 19 feet 6 inches from the top, will be 15 X 19.5 X 32, or 9,360 pounds. Multiplying this by 4, we have an ultimate load of 37,440 pounds. The span in inches equals 180. Then,

Mo = 37,440 x 180/ 8 = 842,400 inch-pounds

Placing this equal to 397 bd2, in which 6 = 12 inches, we find that d2 = 176.8, and d = 13.3 inches. At one-half the height of the wall, the moment will equal one-half of the above, and the required thickness d would be 9.4 inches. The actual thickness at the bottom, including that required outside of the reinforcement, would therefore make the thickness of the wall about 16 inches at the bottom. At one-half the height, the thickness must be about 12 inches. Using a uniform taper, this would mean a thickness of 8 inches at the top.

The reinforcement at the bottom would equal .0084 X 13.3 = .112 square inch of metal per inch of height. Such reinforcement could be obtained by using 3/4-inch bars spaced 5 inches apart. The reinforcement at the center of the height would be .0084 X 9.4 = .079 square inch per inch of width. This could be obtained by using 5/8-inch bars about 5 inches apart, or by using 3/4-inch bars about 7 inches apart. The selection and spacing of bars can thus be made for the entire height. While there is no method of making a definite calculation for the steel required in a vertical direction, it may be advisable to use 1/2-inch bars spaced about 18 inches apart.