To prevent the footing from shearing vertically at the face of the stem, there is a shearing resistance of 12 X 12 X 25 = 3,600 lb. per linear foot of wall, while the shearing force is 3,212 lb.; and hence there is no danger of failure. Since there is no danger of failure by shear at the face of the wall, there is none at the back.
Temperature Reinforcement. To prevent unsightly tem perature cracks, the wall should be provided either with contraction joints (§ 385-87) to localize the cracks, or with longitudinal rein forcement sufficient to resist temperature stresses and thereby equalize the strain between different sections along the length of the wall and cause it to stretch as a homogeneous material (§ 503-6). The percentage of steel required to prevent temperature cracks will depend upon probable range of temperature, the thickness of the wall, and the position of the exposed surface. In § 504, it was shown that a change of temperature of 100° F. in the concrete would require 0.5 per cent of steel having an elastic limit of 60,000 lb. per sq. in. It is usually assumed that high-carbon steel equal to 0.3 to 0.4 per cent of the cross section of the wall is sufficient to prevent objection able contraction cracks in the North Central States. The tempera ture reinforcement should be placed near the exposed face; and an equal amount is required both hori zontally and vertically.
Design of a Counterforted Reinforced-Concrete Retaining Wall. A wall of this type consists of a thin vertical curtain wall supported at intervals by vertical ribs or counterforts, both the curtain wall and the counterforts resting upon and being connected to a base plate or footing. The curtain wall is usually only 6 or 8 inches thick at the top and 10 or 12 at the bottom; and the counterforts are usually of about the same di mensions as the curtain wall, and are spaced 5 to 10 feet center to center. For high walls the counter forted type is more economical of material than the cantilever type; but the cost of constructing the forms is more, the net result being, however, in favor of the counter forted type for walls more than about 20 ft. high.
It will be assumed (1) that a wall 28 feet high is to be de signed, (2) that the face of the wall is to be brought as close as pos sible to the property line, and (3) that the stability is to be equal to that of a standard solid wall. It will be assumed also that both the
standard solid wall and the proposed wall rest upon piles, as is nearly necessary if the structure stands upon compressible soil and the foot ing can not project in front. Fig. 115 shows the trial dimensions of the proposed counterforted wall which is to have the same stability as the standard plain concrete wall of the New York Central Rail road shown in Fig. 116, page 527.
The wall shown in Fig. 116 contains 205.7 cu. ft. of concrete per linear ft., and its weight at 150 lb. per cu. ft. is 30,850 lb. The weight of the earth vertically above the back of the wall is 7,890 lb. The sum of the moments of these two weights about the left-hand end of the middle third of the base is 80,969 ft.-lb. which will be assumed to be the moment of the earth pressure. If the earth pressure be con sidered that due to a liquid weighing w' lb. per cu. ft., then the above moment, 80,969 = w' h' X * h = * w' h'. Solving gives w' = 22.2 lb. per cu. ft., which means that the solid wall can sustain the pressure of a liquid weighing 22.2 lb. per cu. ft. with an approximate factor of safety against overturning of 3. The problem then is to design a reinforced concrete wall that will support the pressure of a liquid weighing 22 lb. per cu. ft.
Stability of Proposed Design. The total weight of earth and concrete per bay = 306,590 lb., and its center of gravity is 6.67 ft. from C. The total horizontal pressure above B against the ver tical plane through DB = w'hZb =i X 22 X X 7.5 = 55,770 lb. The tangent of the angle which the resultant makes with the vertical = 55,700 ± 306,590 = 0.182. The distance to the left of G where the resultant pierces the base of the footing = 10.83 X 0.182 = 1.97 ft.; or the distance from C = 6.67 — 1.97 = 4.70 ft.; and the distance from the center of the footing = 7.00 — 4.70 = 2.30 ft. This shows that the center of pressure is practically at the limit of the middle third, and hence the approximate factor of safety (eq. 13, page 468) against overturning is 3.